1. Fisika-Studi dan Pengajaran I. Judul II. Ahmad Saripudin III. S. Riyadi

PUSAT PERBUKUAN Departemen Pendidikan Nasional

Hak Cipta Pada Departemen Pendidikan Nasional dilindungi oleh Undang-Undang

Mudah dan Aktif Belajar Fisika untuk Kelas XII Sekolah Menengah Atas/Madrasah Aliyah Program Ilmu Pengetahuan Alam

530.07 DUD m

Penulis Penyunting Pewajah Isi Ilustrator Pewajah Sampul

: : : : :

Dudi Indrajit Ahmad Saripudin Neni Yuliati S. Riyadi A. Purnama

Ukuran Buku

: 21 x 29,7cm

DUDI Indrajit Mudah dan Aktif Belajar Fisika 3 :untuk Kelas XII Sekolah Menengah Atas / Madrasah Aliyah Program Ilmu Pengetahuan Alam / penulis, Dudi Indrajit, ; penyunting, Ahmad Saripudin, ; illustrator, S. Riyadi. — Jakarta : Pusat Perbukuan, Departemen Pendidikan Nasional, 2009. vi, 306 hlm, : ilus. ; 30 cm Bibliografi : hlm. 306 Indeks ISBN 978-979-068-816-2 (No. Jil Lengkap) ISBN 978-979-068-931-2 1. Fisika-Studi dan Pengajaran I. Judul II. Ahmad Saripudin III. S. Riyadi

Hak Cipta Buku ini dibeli oleh Departemen Pendidikan Nasional dari Penerbit Setia Purna Inves, PT Diterbitkan oleh Pusat Perbukuan Departemen Pendidikan Nasional Tahun 2009 Diperbanyak oleh ....

Kata Pengantar Puji syukur kami panjatkan ke hadirat Allah SWT, berkat rahmat dan karunia-Nya, Pemerintah, dalam hal ini, Departemen Pendidikan Nasional, pada tahun 2009, telah membeli hak cipta buku teks pelajaran ini dari penulis/penerbit untuk disebarluaskan kepada masyarakat melalui situs internet (website) Jaringan Pendidikan Nasional. Buku teks pelajaran ini telah dinilai oleh Badan Standar Nasional Pendidikan dan telah ditetapkan sebagai buku teks pelajaran yang memenuhi syarat kelayakan untuk digunakan dalam proses pembelajaran melalui Peraturan Menteri Pendidikan Nasional Nomor 22 Tahun 2007 tanggal 25 Juni 2007. Kami menyampaikan penghargaan yang setinggi-tingginya kepada para penulis/penerbit yang telah berkenan mengalihkan hak cipta karyanya kepada Departemen Pendidikan Nasional untuk digunakan secara luas oleh para siswa dan guru di seluruh Indonesia. Buku-buku teks pelajaran yang telah dialihkan hak ciptanya kepada Departemen Pendidikan Nasional ini, dapat diunduh (down load), digandakan, dicetak, dialihmediakan, atau difotokopi oleh masyarakat. Namun, untuk penggandaan yang bersifat komersial harga penjualannya harus memenuhi ketentuan yang ditetapkan oleh Pemerintah. Diharapkan bahwa buku teks pelajaran ini akan lebih mudah diakses sehingga siswa dan guru di seluruh Indonesia maupun sekolah Indonesia yang berada di luar negeri dapat memanfaatkan sumber belajar ini. Kami berharap, semua pihak dapat mendukung kebijakan ini. Kepada para siswa kami ucapkan selamat belajar dan manfaatkanlah buku ini sebaik-baiknya. Kami menyadari bahwa buku ini masih perlu ditingkatkan mutunya. Oleh karena itu, saran dan kritik sangat kami harapkan. Jakarta, Juni 2009 Kepala Pusat Perbukuan

iii

Kata Pengantar Fisika adalah salah satu rumpun ilmu sains yang mempelajari alam semesta. Ruang lingkup ilmu Fisika sangat luas, mulai dari atom yang berdimensi nanometer hingga jagat raya yang berdimensi tahunan cahaya. Dalam kehidupan sehari-hari, banyak ditemukan aplikasi ilmu Fisika, baik berupa fenomena-fenomena di alam atau rekayasa teknologi. Oleh karena itu, Fisika memiliki tingkat urgensitas yang tinggi karena merupakan dasar untuk penguasaan teknologi di masa depan. Sesuai dengan misi penerbit untuk memberikan kontribusi yang nyata bagi kemajuan ilmu pengetahuan maka penulis dan penerbit merealisasikan tanggung jawab tersebut dengan menyediakan buku bahan ajar Fisika yang berkualitas, sesuai dengan tuntutan kurikulum yang berlaku saat ini. Buku ini disusun dengan mengutamakan pendekatan secara inkuiri (eksperimen) dan disajikan secara sistematis, komunikatif, dan integratif, serta adanya keruntutan rangkaian (bab dengan subbab, antarsubbab dalam bab, antaralenia dalam subbab). Sebelum mempelajari materi, sebaiknya Anda terlebih dahulu membaca bagian Advanced Organizer yang terdapat pada halaman awal setiap bab agar Anda dapat mengetahui isi bab secara umum. Pada awal setiap bab, disajikan pula Tes Kompetensi Awal sebagai evaluasi materi prasyarat untuk mempelajari bab yang bersangkutan. Di akhir setiap bab, terdapat Rangkuman, Peta Konsep, dan Refleksi yang bertujuan lebih meningkatkan pemahaman Anda tentang materi yang telah dipelajari dengan memunculkan umpan balik untuk evaluasi diri. Buku ini dilengkapi juga dengan beberapa materi, tugas, dan soal pengayaan, di antaranya Informasi untuk Anda (Information for You), Tantangan untuk Anda, Mari Mencari Tahu, Tugas Anda, Pembahasan Soal, dan Tokoh yang dapat memperluas pengetahuan materi Fisika yang sedang dipelajari. Untuk menguji pemahaman Anda terhadap materi yang telah dipelajari diberikan Tes Kompentensi Subbab pada setiap akhir subbab, Tes Kompetensi Bab pada setiap akhir bab, dan Tes Kompetensi Fisika Semester pada setiap akhir semester. Selain itu, pada akhir buku juga diberikan Tes Kompetensi Akhir untuk menguji pemahaman materi Fisika selama satu tahun ajaran. Semua tes kompetensi tersebut merupakan sarana mengevaluasi pemahaman dan melatih kemampuan menerapkan konsep/prinsip Fisika yang berkaitan dengan materi yang telah dipelajari. Adapun Kunci Jawaban (nomor ganjil) kami sajikan untuk memudahkan Anda dalam mengevaluasi hasil jawaban. Untuk menumbuhkan daya kreativitas, kemampuan psikomotorik, dan cara berpikir ilmiah, kami sajikan Aktivitas Fisika dan Proyek Semester yang menuntut peran aktif Anda dalam melakukan kegiatan tersebut. Demikianlah persembahan kami untuk dunia pendidikan.

Bandung, Mei 2007

Penerbit

Panduan untuk Pembaca Materi-materi pembelajaran pada buku ini berdasarkan kurikulum yang berlaku saat ini dan disajikan secara sistematis, komunikatif, dan integratif. Di setiap awal bab, dilengkapi gambar pembuka pelajaran, bertujuan memberikan gambaran materi pembelajaran yang akan dibahas, dan mengajarkan siswa konsep berpikir kontekstual sekaligus merangsang cara berpikir kontekstual. Selain itu, buku ini juga ditata dengan format yang menarik dan didukung dengan foto dan ilustrasi yang representatif. Penggunaan bahasa yang sederhana, sesuai dengan tingkatan kognitif siswa sehingga membuat pembaca lebih mudah memahaminya. Buku Fisika untuk SMA Kelas XII ini terdiri atas sepuluh bab, yaitu Gejala Gelombang, Gelombang Cahaya, Gelombang Bunyi, Medan Listrik, Potensial Listrik, dan Kapasitor, Medan Magnetik, Induksi Elektromagnetik, Radiasi Benda Hitam, Fisika Atom, Teori Relativitas Khusus, serta Fisika Inti dan Radioaktivitas. Untuk lebih jelasnya, perhatikan petunjuk untuk pembaca berikut. (1) Judul Bab, disesuaikan dengan tema materi dalam bab. (2) Hasil yang harus Anda capai, tujuan umum yang harus Anda capai pada bab yang Anda pelajari. 4 (3) Setelah mempelajari bab ini, Anda harus mampu, kemampuan yang harus Anda kuasai setelah mempelajari bab. 18 (4) Gambar Pembuka Bab, disajikan untuk mengetahui contoh manfaat dari materi yang akan dipelajari. 1 2 19 (5) Advanced Organizer, uraian singkat tentang isi bab untuk menumbuhkan motivasi belajar dan mengarahkan Anda agar lebih 3 fokus terhadap isi bab. 5 (6) Tes Kompetensi Awal, merupakan soal prasyarat yang harus 20 Anda pahami sebelum memasuki materi pembelajaran. (7) Materi Pembelajaran, disajikan secara sistematis, komunikatif, integratif, dan sesuai dengan perkembangan ilmu dan teknologi terkini (up to date). (8) Gambar dan Ilustrasi, sesuai dengan materi dalam bab yang 6 disajikan secara proporsional dan harmonis. 21 (9) Contoh Soal, berisi contoh dan penyelesaian soal. (10) Tugas Anda, berisi tugas atau latihan soal yang berkaitan 22 dengan materi tersebut. 7 (11) Pembahasan Soal, berisi contoh soal yang berasal dari Ebtanas, UAN, UMPTN, atau SPMB. 8 23 (12) Mari Mencari Tahu, tugas mencari informasi yang bertujuan menumbuhkan rasa ingin tahu dan mendorong siswa untuk mencari informasi lebih jauh. (13) Aktivitas Fisika, kegiatan yang dilakukan secara berkelompok untuk mengembangkan kecakapan hidup Anda. (14) Ingatlah, catatan atau hal-hal penting yang perlu Anda ketahui. (15) Informasi untuk Anda (Information for You), berisi pengayaan mengenai informasi dan aplikasi materi, disajikan dalam 2 bahasa 9 10 (bilingual). (16) Tantangan untuk Anda, berisi soal-soal yang disajikan 12 24 dengan tingkat kesulitan lebih tinggi. 11 (17) Kata Kunci (18) Tokoh, berisi tokoh Fisika penggagas ide baru dan pekerja 13 keras sehingga akan menumbuhkan semangat inovatif/kreatif, etos kerja, dan mengembangkan kecakapan hidup Anda. (19) Tes Kompetensi Subbab, berisi soal-soal untuk mengevaluasi penguasaan materi subbab. (20) Rangkuman (21) Peta Konsep 14 (22) Refleksi, sebagai umpan balik bagi siswa setelah mempelajari materi di akhir pembelajaran tiap bab. 25 (23) Tes Kompetensi Bab, berisi soal-soal untuk mengevaluasi 15 penguasaan materi bab. (24) Proyek Semester, kegiatan percobaan untuk meningkatkan pemahaman konsep Fisika dan memotivasi Anda untuk menggali 16 informasi, memanfaatkan informasi, dan menyelesaikan masalah. 26 (25) Tes Kompetensi Fisika Semester, berisi soal-soal untuk mengevaluasi penguasaan materi selama satu semester. 17 (26) Tes Kompetensi Akhir, berisi soal-soal untuk mengevaluasi penguasaan materi selama satu tahun ajaran.

v

Daftar Isi Kata Sambutan • iii Kata Pengantar • iv Panduan untuk Pembaca • v

Bab 1 Gejala Gelombang • 1 A. B.

Pemahaman Gelombang • 2 Gelombang Berjalan dan Gelombang Stasioner • 7 C. Sifat-Sifat Gelombang • 19 Rangkuman • 26 Peta Konsep • 27 Refleksi • 27 Tes Kompetensi Bab 1 • 28

Bab 2 Gelombang Cahaya • 31 A. Interferensi Cahaya • 32 B. Difraksi Cahaya • 37 C. Polarisasi Cahaya • 42 Rangkuman • 46 Peta Konsep • 46 Refleksi • 46 Tes Kompetensi Bab 2 • 47

Bab 4 Medan Listrik, Potensial Listrik, dan Kapasitor • 75 A. B. C.

Muatan Listrik • 76 Medan Listrik • 80 Energi Potensial Listrik dan Potensial Listrik • 87 D. Kapasitor • 92 Rangkuman • 100 Peta Konsep • 101 Refleksi • 101 Tes Kompetensi Bab 4 • 102

Bab 3 Gelombang Bunyi • 49 A. B. C.

Sifat Dasar Gelombang Bunyi • 50 Cepat Rambat Bunyi • 51 Unsur Bunyi dan Pemanfaatan Gelombang Bunyi • 53 D. Sifat-Sifat Gelombang Bunyi • 55 E. Dawai dan Pipa Organa sebagai Sumber Bunyi • 58 F. Energi Gelombang Bunyi • 66 Rangkuman • 70 Peta Konsep • 71 Refleksi • 71 Tes Kompetensi Bab 3 • 72

Bab 5 Medan Magnetik • 105 A. Medan Magnetik • 106 B. Gaya Magnetik • 114 C. Aplikasi Gaya Magnetik • 120 D. Sifat Magnet Bahan • 122 Rangkuman • 124 Peta Konsep • 125 Refleksi • 125 Tes Kompetensi Bab 5 • 126

vi

Bab 6 Induksi Elektromagnetik • 131 A.

Gejala Induksi Elektromagnetik • 132 B. Aplikasi Induksi Elektromagnetik Faraday • 142 C. Arus dan Tegangan Listrik BolakBalik • 147 Rangkuman • 165 Peta Konsep • 166 Refleksi • 166 Tes Kompetensi Bab 6 • 167 Proyek Semester 1 • 169 Tes Kompetensi Fisika Semester 1 • 171

Bab 9 Teori Relativitas Khusus • 229

Bab 7 Radiasi Benda Hitam • 175

A. Semua Gerak Bersifat Relatif • 230 B. Teori Relativitas Einstein • 235 Rangkuman • 252 Peta Konsep • 253 Refleksi • 253 Tes Kompetensi Bab 9 • 254

A.

Pengertian Radiasi Benda Hitam • 176 B. Dualisme Gelombang Partikel • 183 Rangkuman • 192 Peta Konsep • 193 Refleksi • 193 Tes Kompetensi Bab 7 • 194

Bab 10 Fisika Inti dan Radioaktivitas • 257

Bab 8 Fisika Atom • 197

A. B. C. D.

Inti Atom • 258 Radioaktivitas • 264 Reaksi Inti • 272 Reaktor Nuklir, Bom Nuklir, dan Radioisotop • 277 Rangkuman • 283 Peta Konsep • 284 Refleksi • 284 Tes Kompetensi Bab 10 • 285 Proyek Semester 2 • 287 Tes Kompetensi Fisika Semester 2 • 288

A. Evolusi Model Atom • 198 B. Atom Berelektron Banyak • 210 Rangkuman • 224 Peta Konsep • 225 Refleksi • 225 Tes Kompetensi Bab 8 • 226

Tes Kompetensi Akhir • 291 Kunci Jawaban • 295 Apendiks • 300 Senarai • 303 Indeks • 304 Daftar Pustaka • 306

vii

Bab

1 Sumber: Science Encyclopedia, 2001

Riak pada air kolam terjadi karena perambatan getaran akibat jatuhnya batu pada air kolam tersebut.

Gejala Gelombang Hasil yang harus Anda capai: menerapkan konsep dan prinsip gejala gelombang dalam menyelesaikan masalah. Setelah mempelajari bab ini, Anda harus mampu: mendeskripsikan gejala dan ciri-ciri gelombang secara umum.

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A. Pemahaman Gelombang B. Gelombang Berjalan dan Gelombang Stasioner C. Sifat-Sifat Gelombang

Gejala Gelombang

1

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Tokoh Frank Oppenheimer (1912–1985)

Frank Oppenheimer lahir di New York pada tahun 1912. Di Exploratorium dia mendemonstrasikan pengamatannya bahwa gerakan pendulum bila diproyeksikan pada sabuk kertas bergerak membentuk lintasan grafik sinusoidal. Sumber: Conceptual Physics, 1998

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Aktivitas Fisika 1.1 Jenis Gelombang Berdasarkan Arah Getarannya Tujuan Percobaan Mengamati gelombang transversal dan gelombang longitudinal. Alat-Alat Percobaan Sebuah slinki (pegas plastik) yang panjangnya 3 meter. Langkah-Langkah Percobaan 1. Pegang kuat-kuat salah satu ujung slinki oleh teman Anda atau diikatkan pada tiang. 2. Rentangkan sesuai panjangnya dan getarkan ujung yang satu dengan satu kali hentakan naik turun dari posisi setimbang dan kembali ke posisi setimbang (posisi saat tangan Anda diam). 3. Ulangi langkah 2, dan getarkan ujung slinki tersebut terus menerus naik turun, kemudian amati perambatan gelombang sepanjang slinki.

2

Mudah dan Aktif Belajar Fisika untuk Kelas XII

Tugas Anda 1.1 4. 5.

6.

Letakkan slinki di atas lantai licin, kemudian dengan bantuan teman Anda pegang salah satu ujungnya. Hentakkan salah satu ujung pegas dengan satu kali dorongan dan satu kali tarikan ke posisi semula. Amati rapatan dan regangan yang merambat sepanjang slinki.

Ulangi langkah 5, namun dorongan dan tarikannya dilakukan secara terusmenerus. Amati perambatan dan regangan sepanjang slinki.

Selain dalam medium slinki, dapatkah Anda mencari medium lain untuk menggambarkan gelombang longitudinal?

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arah getar

0 Gambar 1.1 Berdasarkan arah getarnya, gelombang mekanik dikelompokkan menjadi: (a) Gelombang transversal; (b) Gelombang logitudinal.

Gambar 1.2 Gelombang pada tali.

Gejala Gelombang

3

puncak

1. Panjang, Frekuensi, Periode, dan Cepat Rambat Gelombang

lembah panjang gelombang

/ rapatan

renggangan

panjang gelombang

0 Gambar 1.3 (a) Panjang gelombang pada gelombang transversal. (b) Panjang gelombang pada gelombang longitudinal.

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:3;0/6 03?B?BA/<
!/?/9
A3?@30BA
27/<8/<5
53:=;0/<5
 
@3>3?A7 27AB<8B99/<
>/2/
'+
  '/2/
53:=;0/<5
:=<57AB27/<8/<5
53:=;0/<5
 
272347<7@79/< @30/5/7
8/?/9
E/<5
@/;/
23<5/<
8/?/9
/B@/A
?3<55/<5/<
E/<5 03?239/A/<
/A/B
8/?/9
/B@/A
?/>/A/<
E/<5
03?239/A/<
@3>3?A7 >/2/
'+
  !/?/9
E/<5
27A3;>B6
53:=;0/<5
@3A7/>
@/AB
@/AB/<
D/9AB
27/A
?/;0/A
53:=;0/<5
!
@32/<59/<
0/<E/9<E/
53:=;0/<5
E/<5 A3?03
@/AB
@/AB/<
D/9AB
273?:B9/<
BB6
8/?/9 @/AB
>/<8/<5
53:=;0/<5

2. Persamaan Umum Gelombang 3@/?/<03@/?/<
/>/
E/<5
27;7:797
@30B/6
53:=;0/<5
<2/
>/@A7 ;/@76
7<5/A
0/6D/
@3A7/>
53:=;0/<5
;3;7:797
4?39B3<@7

>3?7=23
  2/<
13>/A
?/;0/A
53:=;0/<5
!
B0B<5/<
/
H

"3A3?/<5/< ! 
13>/A
?/;0/A
53:=;0/<5
; @ 
>/<8/<5
53:=;0/<5
; 


>3?7=23
53:=;0/<5
@

y



B

F

bukit gelombang

E

C

x

F' G

A B'

H

D lembah gelombang

Gambar 1.4

Panjang gelombang ( ) pada grafik simpangan (y) terhadap arah rambat (x).

 
4?39B3<@7
53:=;0/<5
F '3?6/A79/<
'+

/;0/?
A3?@30BA
;3?B>/9/<
>?=E39@7
2/?7 @7;>/<5/<
#
2/<
>3?>7<2/6/<
"
2/?7
@30B/6
53:=;0/<5
3/7
03/<8/<5
53:=;0/<5
/2/:/6
8/?/9
2/?7 
93

/A/B
8/?/9
2/?7

93

;>:7AB2=

/2/:/6
<7:/7
@7;>/<5/< A3?03@/?
E/<5
2/>/A
271/>/7
>/?A793:
;327B;
E/7AB
8/?/9
2/?7

93

/A/B 2/?7

93

'B<1/9
53:=;0/<5
/2/:/6
A7A79
A3?A7<557
>/2/
53:=;0/<5 E/7AB
A7A79

2/<
A7A79

/@/?
53:=;0/<5
/2/:/6
139B<5/<
/A/B
:3;0/6 A3?3<2/6
E/7AB
A7A79

2/<
A7A79

2/>B<
@7;>B:
/2/:/6
A7A79A7A79
>/2/ 53:=;0/<5
93A79/
;3<1/>/7
93@3A7;0/<5/<
/A/B
@7;>/<5/<<E/
<=:

Contoh 1.1 3:=;0/<5
>/2/
A/:7
;3?/;0/A
23<5/<
9313>/A/<

; @
!79/
8/?/9
/B<1/9 53:=;0/<5
E/<5
03?B?BA/<
/2/:/6

;3A3?
A3/<8/<5
53:=;0/<5 2 = 36 m 0 4?39B3<@7
53:=;0/<5
A3?@30BA 0 793A/6B7 !


; @ !/?/9
/B<1/9


;

4

Mudah dan Aktif Belajar Fisika untuk Kelas XII

/

0





 !/27
>/<8/<5
53:=;0/<5
A/:7

; !


 



!  



F 

Pembahasan Soal



!/27
4?39B3<@7
53:=;0/<5
A/:7
/2/:/6

F

Contoh 1.2 7
>3?;B9//<
@B<5/7
A3?2/>/A
2B/
0B/6
5/0B@
E/<5
A3?>7@/6
@38/B6

1;
"32B/<E/ B<1/9
53:=;0/<5 @32/<59/<
5/0B@
E/<5
:/7<
03?/2/
27
2/@/?
53:=;0/<5
7
//A
@/AB
0B97A
!79/
4?39B3<@7
53:=;0/<5
/7?
A3?@30BA
/2/:/6

F
A3/A ?/;0/A
53:=;0/<5
>/2/
/7? 1,5 0 793A/6B7




F 3<5/<
;3;3?6/A79/<
5/;0/?
27>3?=:36   
1;




 
1; ! 




1;
I

F 

1; @ 

; @ !/27
13>/A
?/;0/A
53:=;0/<5
>/2/
/7?
/2/:/6

; @

Gelombang air laut menyebabkan permukaan air naik turun dengan periode 2 s. Jika jarak antara dua puncak gelombang 5 m, gelombang akan mencapai jarak 10 m dalam waktu .... a. 1 s b. 2 s c. 3 s d. 4 s e. 5 s UMPTN 2001 Pembahasan: T =2s

5 m

jarak antara dua puncak = satu gelombang

5m = 2,5 m/s  T 2s s = vt 10 m = (2,5 m/s)t

v=

t=

10 m =4s 2,5 m/s Jawaban: d

3. Energi Gelombang 7
/D/:
0/0
7<7
<2/
A3:/6
;3<53A/6B7
0/6D/
53:=;0/<5
276/@7:9/< =:36
53A/?/<
E/<5
27?/;0/A9/<
,/2/
>35/@
<3?57
A3?@30BA
27AB:7@9/<
@30/5/7
03?79BA     23<5/<



 @30/5/7
9=<@A/:7AB2= *3:/<8BA<E/
>3?6/A79/<
5/;0/?
03?79BA m   x

Gambar 1.5 Gerakan tangan dirambatkan oleh tali dalam bentuk gelombang.

'+

;3?B>/9/<
@30B/6
;=23:
53:=;0/<5
A/:7
3:=;0/<5 A/:7
A3?@30BA
A3?8/27
9/?3:7AB2=

2/<
4?39B3<@7
@B2BA

3?/9/< A/<5/<
7<7
;3<E30/09/<
@3A7/>
3:3;3<
A/:7
;3<5/:/;7
53?/9
6/?;=<79  /2/:/6
;/@@/
2/?7
@35;3<
9317:
A/:7

;3?B>/9/<
?/>/A
;/@@/
:7<3/? A/:7
2/< " ;3?B>/9/<
>/<8/<5
@35;3<
A/:7
3<5/<
23;797/<
3<3?57 53:=;0/<5
2/>/A
27AB:7@9/<
@30/5/7
03?79BA         

Tugas Anda 1.2 Embusan angin kencang membuat Sears Building di Chicago goyang dan selanjutnya bergetar dengan frekuensi sebesar 0,1 Hz. Berapa periode getarannya?

Gejala Gelombang

5

     "       "

H 

*3:/<8BA<E/
>3?6/A79/<
'+

03?79BA P1

/

t1 x

Gambar 1.6 (a) Pada waktu t1, gelombang tali mencapai titik P1 dan memiliki energi karena gerak harmonik elemen-elemennya. (b) Pada saat waktu bertambah sebesar t , gelombang tali menjalar sejauh v t .

Tantangan untuk Anda Grafik simpangan terhadap waktu sebuah gelombang yang memiliki kecepatan 0,05 m/s seperti pada gambar berikut.

t1 +  t

x

v t

'+

;3;>3?:76/A9/<
@30B/6
53:=;0/<5
E/<5
;3<8/:/?
93 9//A
;3<1/>/7
A7A79
'
*3A7/>
@35;3< A/:7
>/2/
0/57/<
97?7
'
;3;7:797
3<3?57  E/<5
2703?79/<
=:36
 +,'( 1 
*3A3:/6   
53:=;0/<5
A3:/6
;3<8/:/?
@38/B6
"

!  
@3>3?A7 A3?:76/A
>/2/
'+


<3?57
E/<5
27A?/<@;7@79/<
;3:/:B7
' @3:/;/

/2/:/6      "

     ! 

y(cm) 10

t(s)

H 

*3:/<8BA<E/
8B;:/6
3<3?57
E/<5
27A?/<@;7@79/<
@3A7/>
@/AB/<
D/9AB 27@30BA
:/8B
3<3?57
/A/B
@/;/
23<5/<
2/E/
E/<5
27A?/<@;7@79/<
*31/?/ ;/A3;/A7@
27AB:7@
@30/5/7
03?79BA



150 cm Berdasarkan gambar tersebut, tentukan: a. amplitudo (A); b. periode (T); c. frekuensi gelombang (f).

P1

0

     ! 

H

/?7
 +,'(
1
<2/
2/>/A
;3:76/A
0/6D/
2/E/
E/<5
27A?/<@ ;7@79/<
03?0/<27<5
:B?B@
23<5/<
9B/2?/A
/;>:7AB2=
9B/2?/A
4?39B3<@7 2/<
:/8B
53:=;0/<5

Contoh 1.3 *30B/6
53:=;0/<5
>/<8/<5<E/

1;
2/<
/;>:7AB2=<E/

1;
03?53?/9
23<5/< 9313>/A/<

; @
27
@3>/<8/<5
A/:7
E/<5
>/<8/<5<E/

;
2/<
;/@@/<E/

5 / 3?/>/9/6
3<3?57
A=A/:
53:=;0/<5
>/2/
A/:7 0 3?/>/9/6
2/E/
E/<5
27?/;0/A9/<
A/:7
A3?@30BA 0 793A/6B7 

1; 

;  

1;  

5?/; ! 

; @

6

Mudah dan Aktif Belajar Fisika untuk Kelas XII

)/>/A
;/@@/
:7<3/?
A/:7
/2/:/6   95    

I  H 95 ; ; ?39B3<@7
@B2BA
53:=;0/<5
/2/:/6   ; @ 


 

  ?/2 @

  ; / <3?57
A=A/:
53:=;0/<5
>/2/
A/:7
/2/:/6

Kata Kunci

     "  /E/
E/<5
27A?/<@;7@79/<
;3:/:B7
@30B/6
A7A79
>/2/
A/:7
/2/:/6






 


I
 H
95 ;

?/2 @
  
;

;


! 0



• • • • • • • • • • • • •

gelombang mekanik gelombang elektromagnetik gelombang transversal gelombang longitudinal panjang gelombang periode gelombang frekuensi gelombang simpul laju energi gelombang daya gelombang cepat rambat gelombang rapat massa linear energi total gelombang

     ! 




 


I
 H
95 ;
?/2 @   
; 
; @

 


Tes Kompetensi Subbab

A

 +$%(&"
&'
.%.
&-#"(  

 

!3:/@9/<
83<7@83<7@
53:=;0/<5
03?2/@/?9/<
@B;03? 2/<
/?/6
53A/?<E/ 3:=;0/<5
/7?
:/BA
03?53?/9
23<5/<
9313>/A/<

; @ !/?/9
/B<1/9
53:=;0/<5
E/<5
03?239/A/< /2/:/6

;
+3/<8/<5
53:=;0/<5 0 >3?7=23 1 4?39B3<@7
53:=;0/<5 3?79/<
2B/
/:/@/<
;3<5/>/
53:=;0/<5
/7?
:/BA /;>:7AB2=<E/
@3;/97<
9317:
879/
@3;/97<
8/B6
2/?7 @B;03?
53:=;0/<5 B/
0B/6
5/0B@
03?8/?/9

;
"32B/
5/0B@
A3?@30BA 03?/2/
27
/A/@
>3?;B9//<
/7?
"32B/
5/0B@
03?53?/9 3<8/:/?/<
53:=;0/<5 /7?
*/AB
5/0B@
03?/2/
27
/A/@
0B97A
53:=;0/<5
2/< @/AB
:/57
03?/2/
27
:3;0/6
53:=;0/<5
7
/


932B/
5/0B@
A3?@30BA
A3?2/>/A
@30B/6
0B97A 53:=;0/<5
!79/
13>/A
?/;0/A
53:=;0/<5

; @ A3

F
2/<
/;>:7AB2=

1; / 7AB<5:/6
>3<5B?/<5/<
3<3?57
879/
4?39B3<@7<E/ A3A/>
2/<
/;>:7AB2=<E/
03?B0/6
;3<8/27

1; 0 7AB<5:/6
3<3?57
53:=;0/<5
A3?@30BA
879/ 1



4?39B3<@7
2/<
/;>:7AB2=<E/
;3<8/27

9/:7
2/?7 2 @3;B:/ *B/AB
53:=;0/<5
A?/<@C3?@/:
E/<5
;3;7:797
4?39B3<@7
9F
;3<553A/?9/<
>/?A793:
03?;/@@/

;5 23<5/<
/;>:7AB2=

;;
!79/

  
67AB<5:/6 03@/?
3<3?57
53A/?
E/<5
27;7:797
=:36
@B;03?
53A/? 53:=;0/<5
A3?@30BA

B. Gelombang Berjalan dan Gelombang Stasioner $/@76
7<5/A9/6
<2/
1/?/
;3;0B/A
53:=;0/<5
A?/<@C3?@/: ;3<55B/2/
A/:7
A3?@30BA
A3?8/27
53:=;0/<5 E/<5
27A/<2/7
23<5/<
A3?03/<8/<5
A/:7
!79/
A/<5/<
<2/
;/;>B
;3<553?/99/<
B8B<5
A/:7
AB?B< /<8/<5
A/:7
/5/7;/3?6/A79/<
53:=;0/<5
A3?@30BA
;3;7:797
/;>:7AB2=
A3A/> 3:=;0/<5
@3>3?A7
7<7
27@30BA
53:=;0/<5
03?8/:/<

Gejala Gelombang

7

1. Gelombang Berjalan y



O

v

P

x

*3BA/@
A/:7
>/<8/<5
2753A/?9/<
@367<55/
>/2/
A/:7
;3?/;0/A
53:=;0/<5 A?/<@C3?@/:
93
/?/6
@B;0B"
>=@7A74
@3>3?A7
A3?:76/A
>/2/
'+
 3:=;0/<5
03?8/:/<
E/<5
276/@7:9/<
27;B:/7
2/?7
A7A79
&
E/7AB
A7A79
/@/: ?/;0/A/<
53:=;0/<5
*//A
A7A79
&
A3:/6
03?53A/?

@39=<
@7;>/<5/<
53A/? A7A79
&
A3?6/2/>
D/9AB
53:=;0/<5
;3;33?@/;//<
@7;>/<5/<
53A/? 6/?;=<7@
*31/?/
;/A3;/A7@
>3?@/;//<<E/
27AB:7@9/<
@30/5/7
03?79BA

#   @7< 

Gambar 1.7 Gelombang merambat ke arah sumbu-x positif.

H

"3A3?/<5/< # 
@7;>/<5/<
53:=;0/<5
;  
/;>:7AB2=
53:=;0/<5
;


9313>/A/<
@B2BA
?/2 @  
D/9AB
@

&:36
9/?3/A
8B5/
27AB:7@

#   @7<


 /A/B #   @7< 

H

3:=;0/<5
;3?/;0/A
;B:/7
2/?7
A7A79
&
93
@B;0B"
>=@7A74
;33?:B9/<
53:=;0/<5
B
! " 03?53A/?
@3:/;/

@39=<
;/9/
A7A79

/9/<
03?53A/?
@3:/;/

H

@39=< ! '3?@/;//<
@7;>/<5/<
53:=;0/<5
27
A7A79
'
/2/:/6

 " #   @7<      !

H

 +,'(
1
03?:/9B
B=@7A74
2/>B<
>3?@/;//<
53:=;0/<5
E/<5
;3?/;0/A
2/?7
@B;0B" >=@7A74
;3B@/A
9==?27
 " #   @7<      !

Ingatlah y =  A sin
t  kx  +A = awal getaran gelombang ke atas –A = awal getaran gelombang ke bawah +k = gelombang merambat ke kiri –k = gelombang merambat ke kanan

8

H

'/2/
 +,'(
1
2/<
 +,'(
1
/;>:7AB2=

03?6/?5/ >=@7A74
9/?3
/D/:
53A/?/<
27;B:/7
2/?7
A7A79
@3A7;0/<5
& 03?53?/9
93
/A/@
!79/
53?/9/<
27;B:/7
2/?7
A7A79
@3A7;0/<5
&
03?53?/9 93
0/D/6
;/9/
/;>:7AB2=

03?6/?5/
<35/A74
@367<55/
>3?@/;//< @7;>/<5/<
53:=;0/<5
27
'
@31/?/
B;B;
;3<8/27

 " #    @7<      !

H

/:/;
03/A
27AB:7@
@30/5/7
03?79BA #     @7<

  
" 

23<5/<

/2/:/6
<7:/7
2/?7


Mudah dan Aktif Belajar Fisika untuk Kelas XII



/A/B
 !

H 

"3A3?/<5/< # 
@7;>/<5/<
53:=;0/<5
;


9313>/A/<
@B2BA
53A/?/<
?/2 @ " 
>=@7@7
A7A79
'
27B9B?
2/?7
A7A79
/@/:
&
;  
07:/<5/<
53:=;0/<5  
/;>:7AB2=
;  
D/9AB
A7A79
/@/:
03?53A/?
@  
4?39B3<@7
53A/?/< 53:=;0/<5
F

Tugas Anda 1.3 Gelombang air laut merupakan gelombang berjalan. Energi gelombang air laut ini dapat digunakan untuk menggerakkan generator listrik. Bukan hal yang mustahil teknologi ini suatu saat diterapkan di Indonesia. Menurut Anda, kira-kira di daerah laut manakah teknologi ini dapat diterapkan? Coba Anda diskusikan alasannya bersama teman-teman.

a. Sudut Fase, Fase, dan Beda Fase <2/
A3:/6
;3<53A/6B7
0/6D/
53:=;0/<5
03?8/:/<
;3;7:797 >3?@/;//<
@7;>/<5/<
@3>3?A7
>/2/
 +,'(
1
,3?@/;//<<E/
@30/5/7
03?79BA

  " #   @7<
  "    @7<
     /?7
>3?@/;//<
A3?@30BA
03@/?
@B2BA
4/@3<E/
/2/:/6

  "   
    

H

  "       

H 

2/<
03@/?
4/@3<E/
/2/:/6

@367<55/
6B0B<5/<
//A
27AB:7@ @30/5/7
03?79BA

  

 

'3?6/A79/<
'+

+7A79

E/<5
03?8/?/9
"
2/?7
A7A79
/@/: 53A/?/<
&
A3:/6
03?53A/?
@3:/;/

@39=<
;3;7:797
4/@3
 
'/2/
/D/: 53A/?/<
E/<5
@/;/
A7A79

03?8/?/9
"
2/?7
A7A79
/@/:
53A/?/<
&
2/< ;3;7:797
4/@3
 
32/
4/@3
A7A79

2/<

/2/:/6

 "   "                     

"  " " 

y

H 

A

B x

O x1 x2

Gambar 1.8 Titik A yang berjarak x1 dan titik B yang berjarak x2 dari O. Kedua titik memiliki beda fase  .

H

Mari Mencari Tahu $/@76
7<5/A9/6
<2/
A3/:7<5
2/6@E/A
E/<5
;3<7;>/
@/B2/?/ @/B2/?/
97A/
27
136
2/<
@397A/?<E/
>/2/

3@3;03?



 &)'(!
,.('# ;3:B:B6:/3?A7
0/5/7;//
>3<E30/0>3<E30/0<E/
9/>/< 2/<
27
;/3?
Gejala Gelombang

9

Informasi untuk Anda Satuan frekuensi adalah hertz (Hz), diambil dari nama Heinrich Hertz, orang yang pertama kali mendemonstrasikan gelombang radio pada 1886. Satu getaran per sekon disebut 1 hertz; dua getaran per sekon dinamakan 2 hertz, dan selanjutnya.

Information for You The unit of frequency is called the hertz (Hz), after Heinrich Hertz, who demonstrated radio in 1886. One vibration per second is 1 hertz; two vibrations per second is 2 hertz, and so on.

Contoh 1.4 '3?@/;//<
@30B/6
53:=;0/<5
03?8/:/<
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Tantangan untuk Anda Sebuah gelombang merambat dari sumber S ke kanan dengan laju 8 m/s, frekuensi 16 Hz, dan amplitudo 4 cm. Gelombang tersebut melalui titik P yang 1 berjarak 9 m dari S. Jika S telah 2 1 bergetar 1 S dan arah gerak 4 pertamanya ke atas, hitunglah simpangan titik P pada saat itu.

10

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gelombang pantul

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gelombang pantul

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gelombang datang

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13

Tugas Anda 1.4 Rumus syarat terbentuknya simpul dan perut dapat juga ditulis seperti berikut. 1 x = (2n – 1) 4 untuk simpul dan 1 x = n 2 untuk perut. Coba Anda diskusikan dengan teman, bagaimanakah penurunannya?




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Tantangan untuk Anda Gelombang stasioner dapat terjadi karena superposisi gelombang datang dan gelombang pantul oleh ujung bebas. Titik simpul yang kesepuluh berjarak 1,52 m dari ujung bebasnya. Jika frekuensi gelombang itu 50 Hz, berapakah cepat rambat gelombangnya?

14

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Mudah dan Aktif Belajar Fisika untuk Kelas XII

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Kata Kunci • • • • • • • • • •

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15

3. Kecepatan Gelombang Stasioner !79/
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Aktivitas Fisika 1.2 Percobaan Melde Tujuan Percobaan Mengamati gelombang stasioner. Alat-Alat Percobaan 1. Kawat tipis (2 buah) yang berbeda massanya 2. Beban 3. Pembangkit getaran (vibrator) 4. Katrol Langkah-Langkah Percobaan 1. Susunlah alat-alat seperti pada gambar berikut. vibrator gelombang

Pembahasan Soal Tali yang panjangnya 5 m bertegangan 2 N digetarkan sehingga terbentuk gelombang stasioner. Jika massa tali 625 × 10–3 kg, maka cepat rambat gelombang tali adalah .... a. 2 m/s b. 5 m/s c. 6 m/s d. 10 m/s e. 40 m/s UMPTN 1996 Pembahasan: Diketahui:
= 5 m; F = 2 N; m = 625 × 10–3 kg. Menurut Melde kecepatan rambat gelombang tali adalah F v = 

F
m 2N 5m = 6,25 10 -3 kg = 40 m/s =

P

2. 3.

4. 5.

P

S

beban

Hidupkan pembangkit getaran dengan menghubungkannya ke sumber tegangan sehingga pada tali terbentuk gelombang seperti pada gambar. Jika belum terbentuk gelombang stasioner, ubahlah berat beban yang tergantung pada ujung katrol sehingga pada suatu saat terbentuk gelombang stasioner. Amatilah dengan saksama gelombang yang terjadi dan tulislah kesimpulan Anda dari kegiatan tersebut. Lakukan langkah 1 sampai 4 untuk kawat kedua yang berbeda massanya.

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16

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Mudah dan Aktif Belajar Fisika untuk Kelas XII

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Tugas Anda 1.5 Gelombang stasioner terbentuk jika superposisi terjadi dari dua gelombang yang frekuensi dan amplitudo sama tapi fasenya berlawanan. Jika frekuensi dan amplitudo dibuat tidak sama, gelombang apa yang terjadi?

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8/?/9

1;
2/<



03?/2/
>/2/
8/?/9

1;
2/?7
B8B<5
>3;/
Contoh 1.10 '/2/
>3?1=0//<
$3:23
A3?033?BA
53:=;0/<5
2/?7
@3/<8/<5<E/ 
;
!79/
>/2/
B8B<5
9/D/A
275//<8/<5
53:=;0/<5
>/2/
@3/A
?/;0/A
53:=;0/<5 2 4?39B3<@7
@B;03?
53A/? 0 793A/6B7  

; 030/< 

5


I
 H
95 @3
Tantangan untuk Anda Kawat untuk saluran transmisi listrik yang massanya 40 kg diikat antara dua menara tegangan tinggi yang jaraknya 200 m. Salah satu ujung kawat dipukul oleh teknisi yang berada di salah satu menara, sehingga timbul gelombang yang merambat ke menara yang lain. Jika gelombang pantul terdeteksi setelah 10 s, berapa tegangan kawat?

Gejala Gelombang

17

/ 3
0 '/<8/<5
53:=;0/<5
27>3?=:36
2/?7
>3?@/;//<     
;




;

!/27
>/<8/<5
53:=;0/<5
>/2/
@3
  @3
 
030/<
I




I
 H
I




%

  




  
; @    !/27
13>/A
?/;0/A
53:=;0/<5
/2/:/6

; @ 2 ?39B3<@7
@B;03?
53A/? !   



  F  !/27
4?39B3<@7
@B;03?
53A/?
/2/:/6

F

! 


Tes Kompetensi Subbab

B

 +$%(&"
&'
.%.
&-#"( 





18

*3BA/@
A/:7
E/<5
>/<8/<5<E/



1;
@/:/6
@/AB B8B<5<E/
2753A/?9/<
6/?;=<79
:7AB2=

1;
@32/<59/<
B8B<5  :/7<<E/
A3?79/A
3A/?/<
;3?/;0/A
23<5/<
:/8B

1; @ +3:7AB2=
53:=;0/<5
27
A7A79
E/<5
03?8/?/9

1; 2/?7
A7A79
/@/:
53A/?/< 0 @7;>/<5/<
53:=;0/<5
27
A7A79
A3?@30BA
@3A3:/6
A/:7 03?53A/?
@3:/;/

@  1 :3A/9
@7;>B:
93
2/<
>3?BA
93
2/?7
A7A79
/@/: 53A/?/< *3BA/@
A/:7
>/<8/<5<E/

;
*/AB
B8B<5
A/:7
2779/A
93 27<27<5
2/<
B8B<5
A/:7
:/7<<E/
2753A/?9/<
23<5/< /;>:7AB2=

1;
@3?A/
>3?7=23
 
@
@367<55/ 53:=;0/<5
>/2/
A/:7
2/>/A
;3?/;0/A
23<5/<
93:/8B/< 
1; @
*30B/6
A7A79
'
>/2/
8/?/9
"


;
2/?7
B8B<5 >3;/3?@/;//<
@7;>/<5/<
53:=;0/<5
2/A/<5
2/< 53:=;0/<5
>/3?@/;//<
@7;>/<5/<
@B>3?>=@7@7
53:=;0/<5
27 A7A79
' */:/6
@/AB
B8B<5
2/?7
@3BA/@
A/:7
E/<5
;3;7:797
>/<8/<5 
1;
2753A/?9/<
6/?;=<79

Mudah dan Aktif Belajar Fisika untuk Kelas XII

/









3?/>/
>/<8/<5
53:=;0/<5
E/<5
03?53?/9
>/2/ A/:7
879/
>3?BA
93
03?8/?/9

1;
2/?7
A7A79
/@/: 53A/?/< 0 7
;/B:
93
27B9B?
2/?7
A7A79
/@/: 53A/?/< *30B/6
53:=;0/<5
E/<5
4?39B3<@7<E/



F
;3;7:797 9313>/A/<

; @ / 3?/>/9/6
8/?/9
//9/6
>3?032//<
@B2BA
4/@3
@B/AB
A7A79
E/<5 03?53A/?
2/:/;
@3:/<5
D/9AB
 H
@39=< *30B/6
53:=;0/<5
;3?/;0/A
2/?7
A7A79
&
93
A7A79
' 23<5/<
13>/A
?/;0/A

; @
4?39B3<@7

F
/;>:7AB2= 
1;
2/<
8/?/9
&'


;
+3/<5/<
A7A79
' @//A
&
A3:/6
03?53A/?
@3:/;/

@ +3/A
?/;0/A
53:=;0/<5 
; @ *3BA/@
A/:7
E/<5
>/<8/<5<E/

;
27?3
*3A3:/6
A3?8/27
53:=;0/<5
@A/@7=<3? A3?<E/A/
>3?BA
93
03?8/?/9

;
2/?7
A7A79
/@/: 53A/?/< / 3?/>/
>/<8/<5
53:=;0/<5
E/<5
A3?8/27 0 7AB<5
:3A/9
@7;>B:
93
27B9B?
2/?7
A7A79
/@/: 53A/?/<

C. Sifat-Sifat Gelombang 1. Pemantulan 38/:/
>3;//A
<2/
:76/A
2/:/; 93672B>/<
@36/?76/?7
$7@/:<E/
>3;//A
;3:76/A
0/E/<5/<
>/2/
13?;7<
38/:/
>3;//A
>B:/
27/;/A7
>/2/
53:=;0/<5
A/:7
'/2/
B8B<5
A/:7
E/<5
97A/
79/A9/< /9/<
A3?:76/A
>3;/3?A7
>/2/
'+
  @32/<59/<
B3;//A
;3<5/;/A7
>3;//2/
A/<597
/7?
A/<597
?7/9 B9B;
>3;//3;//A
27:76/A
>/2/
'+

*31/?/
;/A3;/A7@
>3?<E/A//< A3?@30BA
2/>/A
27?B;B@9/<
2/:/;
033?@/;//<
03?79BA

  

/

0

Gambar 1.13 Pemantulan gelombang pada tali: (a) ujung terikat; dan (b) ujung bebas. gelombang datang

H 

"3A3?/<5/<  1

@B2BA
2/A/<5  2

@B2BA
>/
i

2

a. Pemantulan Gelombang Lingkaran oleh Bidang Datar /5/7;/3;//=:/
E/<5
@/;/
23<5/<
53:=;0/<5
2/A/<5 '/2/
'+
 
+7A79
&
;3?B>/9/<
@B;03?
53:=;0/<5
2/A/<5 23<5/<
;3;3?6/A79/<
6B9B;
>3;//3?=:36
0/E/<5/<
&
27
A7A79
 +7A79

/2/:/6
@B;03?
53:=;0/<5
>//B@/A
27


r normal muka gelombang pantul

Gambar 1.14 Hukum pemantulan

O

Aktivitas Fisika 1.3 Pemantulan Gelombang Tujuan Percobaan Mengamati pemantulan gelombang pada tangki riak. Alat-Alat Percobaan 1. Wadah bak air 2. Papan 3. Kaca

sinar gelombang datang sinar gelombang pantul 1

A

Sumber: Physics for ‘O’ Level, 1999

Gambar 1.15 4. 5.

Lampu atau OHP Air

Langkah-Langkah Percobaan 1. Buatlah semacam bak air (tangki riak) dengan kaca sebagai alasnya. 2. Letakkan bak tersebut di bawah sinar lampu OHP, kemudian letakkan sebuah kertas putih di bawah bak. 3. Buatlah riak kecil dengan cara mencelupkan salah satu jari Anda di tengahtengah bak. Perhatikan yang terjadi. 4. Masukkan papan tersebut di ujung bak, lalu gerakkan secara perlahan dalam arah horizontal sehingga terbentuk riak. 5. Jelaskan hasil yang diperoleh dan sebutkan perbedaan kedua riak tersebut. Kemudian, jelaskan pula bagaimana dengan pemantulan gelombang pada dinding.

Pemantulan gelombang lingkaran oleh bidang datar.

Sumber: Physics for ‘O’ Level, 1999

Gambar 1.16 Tangki riak yang ditempatkan di atas OHP.

Gejala Gelombang

19

2. Pembiasan Gelombang ,3;07/@/<
53:=;0/<5
:3A/99/<:/6
93>7<5
53:/@ A30/:
27
@30/57/<
2/@/?
A/<597
?7/9
53:=;0/<5
/5/?
932B/
072/<5
;3;7:797 932/:/;/<
/7?
E/<5
03?032/
'3?032//<
932/:/;/<
/7?
A3?@30BA
A3?:76/A >/2/
'+

03?79BA gelombang setelah dibiaskan

2

batas

tempat dangkal

Gambar 1.17

keping gelas tebal

1

gelombang datang

meja kaca tangki riak

Pembiasan pada tangki riak yang kedalamannya berbeda.

/

Sumber: Dokumentasi Penerbit

0 sinar datang

1 n1

/?7
'+

A3?:76/A
0/6D/
>/<8/<5
53:=;0/<5
27
A3;>/A
E/<5 :3076
2/:/;
:3076
03@/?
2/?7>/2/
>/<8/<5
53:=;0/<5
27
A3;>/A
E/<5 2/<59/:
   
@367<55/
13>/A
?/;0/A
53:=;0/<5<E/
>B<
03?032/
3<5/< 23;797/<
>3?B0/6/<
>/<8/<5
53:=;0/<5
;3<E30/09/<
A3?8/27<E/ >3;03:=9/<
53:=;0/<5
'3;03:=9/<
53:=;0/<5
273;07/@/< ;07::/6
@30/A/<5
0=:>=7<
:/:B
;/@B99/<
93
2/:/;
53:/@
9/1/
E/<5 03?7@7
/7?
03<7<5
=:>=7<
A3?@30BA
/9/<
A3?:76/A
03<59=9
'3?7@A7D/
7<7 27@30BA
>3?7@A7D/
  
 
=3?7@A7D/
>3;07/@/<
E/<5 :/7<
/2/:/6
43<=;33:/<57 '3;07/@/<
A3?8/27
/970/A
/2/<E/
>3?032//<
7<239@
07/@
'3?032//< 7<239@
07/@
;3<E30/09/<
>3?B0/6/<
13>/A
?/;0/A
53:=;0/<5
'/2/ >3?7@A7D/
0=:>=7<
27
2/:/;
53:/@
1/6/E/
E/<5
27>///A
?/;0/A
53:=;0/<5
03?B0/6B0/6 B9B;
E/<5
;3<55/;0/?9/<
43<=;33;07/@/<
'3?6/A79/<
'+

B9B;
*<3::7B@ ;3<E/A/9/<
0/6D/  @7/2/
@/AB
072/<5
2/A/?  @7
5/?7@
<=?;/:
2/<
@7
5/?7@
<=?;/:
2/<
;3;33?@/;//<
03?79BA

n2


@7<



@7<


2 sinar bias n2 > n1

Gambar 1.18 (a) Pembiasan terjadi pada bolpoin yang dicelupkan ke dalam air; (b) Skema perjalanan sinar pada proses pembiasan.

20

H 

"3A3?/<5/<  
13>/A
?/;0/A
1/6/E/
27
?B/<5
6/;>/
; @ ! 
13>/A
?/;0/A
1/6/E/
93A79/
;3:3D/A7
;327B;  
7<239@
07/@
;BA:/9
0/6/< '3?6/A79/<
 +,'(
1 
*7;0=:
 
2/;

;3?B>/9/< 9=<@A/B<
2
/2/:/6
@B2BA
E/<5
2703
Mudah dan Aktif Belajar Fisika untuk Kelas XII

8./5
,3+<
+,<96>=
<>+=>
,+2+8
7/;>:+5+8
:/;,+8.381+8
5/-/:+=+8 -+2+A+
.+6+7
;>+81
?+5>7
 
.+8
5/-/:+=+8
-+2+A+
.3
.+6+7
7/.3>7 ,+2+8
=/;=
/81+8
./7353+8
38./5<
,3+<
.38A+=+5+8
./81+8 :/;<+7++8





 D 

Tugas Anda 1.6 Menurut Anda, mengapa gelombang air laut yang begitu besar ketika sampai di pantai menjadi kecil? Apakah terjadi hal yang sama untuk laut yang berbatasan dengan daratan yang bertebing karang curam?

35+
)*%&
/
.3<><>8
5/7,+63
>8=>5
7/7,+8.3815+8
<>.>= ,3+<
  
.+81+8
<>.>=
.+=+81
  
.3:/;96/2
:/;<+7++8
,/;35>=



<38 



<38   

 D 

/;.+<+;5+8
)*%&
/
.+:+=
.35/=+2>3
,+2@+
,/<+;8A+ <>.>=
,3+<
,/;1+8=>81
:+.+
<>.>=
.+=+81
  
38./5<
,3+<
7/.3>7 :/;=+7+
 
.+8
38./5<
,3+<
7/.3>7
5/.>+


 35+

<+7+
./81+8
 

+5+8
<+7+
./81+8
 
%+.+
5/+.++8
383 ;/0;+5<3
=3.+5
7/8A/,+,5+8
:/7,/695+8
<38+;
(38+;
.+=+81
+5+8
.3=/;>< 5+8
=+8:+
:/7,/6955+8
+;+2
2 1+;3<
:/;:+84+81+8
<38+;
.+=+81
23
1+;3<
89;7+6
.+8
1+;3<
:/;:+8 4+81+8
<38+;
.+=+81

sinar datang

sinar datang

garis normal

garis normal sinar datang

1

1

1

n1

n1

n1 n2

2

sinar bias

n2

n2

2

2

sinar bias

Gambar 1.19

sinar bias n2 = n 1

n2 > n 1

+

,

Pembiasan sinar bergantung pada indeks bias medium. n2 < n 1

-

Contoh 1.8 (/,>+2
1/697,+81
.+=+81
:+.+
,3.+81
,+=+<
+8=+;+
.>+
7/.3>7
./81+8
<>.>=
.+=+81 C
35+
:/;,+8.381+8
+8=+;+
38./5<
,3+<
7/.3>7
:/;=+7+
.+8
7/.3>7
5/.>+
+.+6+2


=/8=>5+8
<>.>=
,3+<8A+
.+8
6>53<5+8
:/7,3+<+8
1/697,+81
=/;=

-

35/=+2>3 <>.>=
.+=+81


C

38./5<
,3+<
 



Gejala Gelombang

21

, ">53<+8
:/7,3+<+8
1/697,+81

+ (>.>=
,3+<  <38   
<38 

 

 <38 

37°


 <38 

 




medium 1 medium 2

   
<38
   

53°

3. Polarisasi %96+;3<+<3
7/;>:+5+8
:/;3<=3@+
A+81
7/7:/;632+=5+8
-3;3
52+< 1/697,+81
=;+86+8
:/7,3+<+8
,3+<
5/7,+;
+63<+<35+8
./81+8
7/7,+A+815+8 1/697,+81
=;+8=+<
=+63
%/;2+=35+8
%)
 celah

Gambar 1.20 Gelombang tali yang terpolarisasi.

gelombang terpolarisasi

+ gelombang tidak terpolarisasi

, Gambar 1.21 Polarisasi gelombang tali.

22

gelombang terpolarisasi

(/>=+<
=+63
.31/=+;5+8
./81+8
7/6/@+=3
+2
-/6+2

.3= /697,+81
A+81
5/6>+;
.+;3
=+63
.3=
1/697,+81
=/;:96+;3<+<3
638/+; *8=>5
6/,32
7/7+2+73
:96+;3<+<3
+7+=36+2
%)
 %)

7/7:/;632+=5+8
1/697,+81
=;+8+
-/6+2
A+81
:9<3<38A+
<+7+
.+8
1/697,+81
=+63
=/;8A+=+ 6969<
:+.+
5/.>+
-/6+2
=/;=
%)

7/7:/;632+=5+8 1/697,+81
=;+8+
-/6+2
A+81
<+6381 7/8A36+81
=/1+5
6>;><
)/;8A+=+
1/697,+81
6969<
:+.+
-/6+2
A+81 :/;=+7+
A+3=>
-/6+2
A+81

-/6+2
A+81
=/1+5
6>;><
./81+8
+;+2
1/=+; /;.+<+;5+8
%)

.+8
%)

.+:+= .3<37:>65+8
,+2@+
2+8A+
1/697,+81
=;+823
-/6+2 /697,+81
A+81
.+:+=
6969<
.+;3
-/6+2
2+8A+
7/736353
<+=>
+;+2
1/=+; /697,+81
A+81
./7353+8
.3=
1/697,+81
:96+;3<+<3
(38+;
+6+73 5+8
=/;7+<>5
<38+;
=/;:96+;3<+<3
(37,96
>8=>5 <38+;
A+81
=3.+5
=/;:96+;3<+<3
+.+6+2
<38+;
A+81
=/;:96+;3<+<3
+.+6+2


Mudah dan Aktif Belajar Fisika untuk Kelas XII


8=>5 

cahaya alami yang datang getaran vertikal diserap sebagian

getaran horizontal diserap sempurna oleh polaroid

Gambar 1.22 cahaya diteruskan terpolarisasi linear

Cahaya tak terpolarisasi dilewatkan pada sebuah kristal.

4. Interferensi Gelombang (/6+38
:/7+8=>6+8
.+8
:/7,3+<+8
<30+=
6+38
.+;3
1/697,+81
+.+6+2 38=/;0/;/8<3
(30+=
38=/;0/;/8<3
:+.+
1/697,+81
+.+6+2
:96+
:/81>+=+8 38=/;0/;/8<3
598<=;>5=30
.+8
:96+
:/81236+81+8
38=/;0/;/8<3
./<=;>5=30 7>5+
1/697,+81
.+;3
.>+
1/697,+81
A+81
<+6381
,/;=/7>
*8=>5
7/632+= :96+
383
6/,32
4/6+<
:/;2+=35+8
%)

,/;35>= interferensi konstruktif

Gambar 1.23 interferensi destruktif

Interferensi gelombang air

Sumber: www.physics.umd.edu

8.+
:>8
.+:+=
7/6+5>5+8
:/;-9,++8
>8=>5
7/632+=
1/4+6+ 38=/;0/;/8<3
1/697,+81
7,366+2
+2
=+8153
;3+5
5/7>.3+8
,>+=6+2 -/6+2
1+8.+
:+.+
+2
:+:+8
#+<>55+8
:+:+8
=/;=
5/
.+6+7 =+8153

.3
+=+<
=+8153
;3+5
+3;
!/7> .3+8
,>+=6+2
<>7,/;
1/697,+81
./81+8
-+;+
7/8-/6>:5+8
<+6+2
<+=> 4+;3
8.+
5/
.+6+7
+3;
.3
<+6+2
<+=>
>4>81
=+8153
"32+=6+2
:96+
38=/; 0/;/8<3
:+.+
5/;=+<
.3
,+@+2
=+8153
=/;=
.3
,+@+2
>4>81
=+8153
6+38 %/7+<+81+8
:+:+8
,/;-/6+2
=+.3
+=+8
.>+ <>7,/;
1/697,+81
%+.+
.38.381
=+8153
+5+8
=+7:+5
:/81>+=+8
.+8 :/81236+81+8
1/697,+81
+53,+=
,/;=/7>8A+
5/.>+
1/697,+81
=/;= %/81>+=+8
=/;4+.3
5+;/8+
+.+8A+
5/<+7++8
0++
1/697,+81 !/<+7++8
0++
1/697,+81
A+81
7/736353
0;/5>/8<3
.+8 <37:+81+8
A+81
<+7+
%/81236+81+8
=/;4+.3
5+;/8+
+.+8A+
5/=3.+5<+7++8 0++
1/697,+81 7/736353
0;/5>/8<3
A+81
<+7+
8+7>8
<37:+81+88A+
<+6381
,/;6+@+8+8

+

+

=

=

+

,

Gambar 1.24 Interferensi dua gelombang (a) interferensi konstruktif (menguatkan) dan (b) interferensi destruktif (menghilangkan).

5. Difraksi Gelombang 30;+5<3
.+6+7
,+2+<+
;+8 %/;3<=3@+
.30;+5<3
+=+>
:/6/8=>;+8
1/697,+81
.+:+=
.3+7+=3
./81+8 7/811>8+5+8
=+8153
;3+5
.+8
:+:+8
,/;-/6+2
$+7>8
1/697,+81
A+81 =37,>6
.3
+3;
=3.+5
.32+<365+8
96/2
1/=+;+8
=+81+8
7/6+385+8
.+;3
1/;+5+8 7+4>
7>8.>;
+2
:+:+8
5+
1/697,+81
,>5+8
,/;,/8=>5 63815+;+8
35+
8.+
7/6+5>5+8
2+6
=/;=
8.+
+5+8
7/8.+:+=5+8

Gejala Gelombang

23

,+2@+
+;
.+;3
:+:+8
,/;-/6+2
,/8=>5
7>5+ 1/697,+81
=3.+5
5
1/697,+81
A+81
:/;=+7+
%96+
A+81
+5+8 8.+
.+:+=3
.3=>84>55+8
96/2
%)
  celah

Gambar 1.25 Difraksi pada gelombang air gelombang air

Sumber: Physics for Scientist & Engineers with Modern Physics, 2000

gelombang air setelah melewati celah

6. Efek Doppler

s

5 4 3

2

1

+

vp

vs s 5 4321

P

, Gambar 1.26 Sumber gelombang yang bergerak pada permukaan air menyebabkan panjang gelombang lebih pendek sesuai arah gerak.

35+
8.+
7/7,>+=
1/=+;+8
:+.+
:/;7>5++8
+3;
7/811>8+5+8 +2
+6+=
:/811/=+;
 
:+.+
:/;7>5++8
+3;
=/;=
+5+8 =/;,/8=>5
1/697,+811/697,+81
A+81
7/;+7,+=
5/
84>55+8
%)
 
:+
A+81
=/;4+.3
435+
?3,;+=9;
.3=+;35
5/ 5+8+8
35+
?3,;+=9;
.3=+;35
5/
5+8+8
+5+8
=/;,/8=>5
63815+;+8
1/697,+81 +3;
<+=
.+;3
5+ 1/697,+81
A+81
.32+<365+8
,/;1/;+5
5/
5+8+8
8-+5
5/
:>8-+5
1/697,+81
A+81
,/;./5+=+8
.3
7,/; 1/697,+81
A+81
,/;1/;+5
7/8A/,+,5+8
:/;>,+2+8
:+84+81
1/697,+81 %+84+81
1/697,+81
A+81
,/;>,+2
+5+8
7/8A/,+,5+8
:/;>,+2+8
0;/5>/8<3 %/;3<=3@+
:/;>,+2+8
0;/5>/8<3
1/697,+81
+53,+=
:/81+;>2
1/;+5
;/6+=30 +8=+;+
<>7,/;
1/697,+81
.+8
:/81+7+=
.38+7+5+8
/0/5
9::6/; 8.+
.+:+=
7/81+7+=3
/0/5
9::6/;
5/=35+
,/;.3;3
.3
:38113;
4+6+8 .+8
+.+
79,36
7/8>4>
+=+>
7/8./5+=3
8.+
,>8A3
79,36
=/;=
=/; ./81+;
6+
435+
79,36
=/;=
7/84+>23 8.+
,>8A3
79,36
=/;=
=/;./81+;
= 5+63
:/;=+7+
.3+7+=3
96/2
<=;3+
 )!*+!&
'(($)   D  
,+2+8
,>8A3
79,36
7/84+.3
=38113
+=+>
;/8.+2
=/;= .3,+2+8
0;/5>/8<3
!/=35+
79,36
=/;=
7/8./5+=3 8.+
0;/5>/8<38A+
=
,/; 1/;+5
7/84+>23
8.+
0;/5>/8<38A+
,+2+8
0;/5>/8<3 =/;=
.+:+=
.34/6+<5+8
= !/=35+
<>7,/;
,>8A3
.+8
:/8./81+;
.3+7
1/697,+81
,>8A3
A+81 .3:+8-+;5+8
96/2
<>7,/;
,>8A3
7/736353
:+84+81
1/697,+81
A+81
<+7+ ./81+8
1/697,+81
,>8A3
A+81
.3=/;37+
:/8./81+;
,/;6+5>
:/;<+7++8          $+7>8
:+.+
<++=
<>7,/;
,>8A3
,/;1/;+5
.+8
:/8./81+;
.3+7
7+5+ :+84+81
1/697,+81
A+81
.3=/;37+
:/8./81+;
+.+6+2

 

24

Mudah dan Aktif Belajar Fisika untuk Kelas XII

   

/8<3
A+81
.3=/;37+
:/8./81+;
+.+6+2
=

           

 D +

*8=>5
5+<><
7,/;
,>8A3
.3+7
.+8
:/8./81+; ,/;1/;+5
:+84+81
1/697,+81
A+81
.3=/;37+
:/8./81+;
+.+6+2

 

 

Tokoh

/8<3
A+81
.3=/;37+
:/8./81+;
+.+6+2
=

 

          

 D ,

(/-+;+
>7>7
:/;<+7++8
/0/5
9::6/;
>8=>5
<>7,/;
,>8A3
.+8 :/8./81+;
,/;1/;+5
,/;6+5>
:/;<+7++8

 

     



Christian Johann Doppler (1803–1853)

 D 

!/=/;+81+8  
0;/5>/8<3
A+81
.3=/;37+
:/8./81+;
B  
0;/5>/8<3
.+;3
<>.>=
,>8A3
B  
5/-/:+=+8
1/697,+81
,>8A3
7<  
5/-/:+=+8
1/;+5
:/8./81+;
7<  
5/-/:+=+8
1/;+5
<>7,/;
,>8A3
7< %/;4+843+8
=+8.+
A+81
.31>8+5+8
.+6+7
:/;<+7++8
+.+6+2
=

 35+
:/8./81+;
7/8./5+=3
<>7,/;
:/8./81+;
:9<3=30  35+
:/8./81+;
7/84+>23
<>7,/;
:/8./81+;
8/1+=30  35+
<>7,/;
7/8./5+=3
:/8./81+;
<>7,/;
8/1+=30

 35+
<>7,/;
7/84+>23
:/8./81+;
<>7,/;
:9<3=30

Christian Johann Doppler adalah fisikawan yang lahir di Salzburg, Austria. Dia belajar di Vienna dan menjadi seorang profesor Fisika (1851). Dia terkenal dengan pengamatannya tentang variasi frekuensi gelombang suara dan gelombang cahaya karena pengaruh kecepatan gerak relatif antara sumber dan pengamat. Sumber: www. allbiographic.com

Contoh 1.11 (/,>+2
79,36
,/;1/;+5
./81+8
5/-/:+=+8

574+7
<+7,36
7/7,>8A35+8
56+5<98 ./81+8
0;/5>/8<3

B
+;3
+;+2
,/;6+@+8+8
5+8
0;/5>/8<3
,>8A3
56+5<98
A+81
.3./81+;
96/2 :/81/8.+;+
=
435+
-/:+=
;+7,+=
,>8A3
.3
>.+;+
+.+6+2

7< - 35/=+2>3 

574+7


7< 79,36  :/8./1+; 

574+7


7<  >.+;+ 

7< 56+5<98
79,36 

B  :/8./81+;




>.+;+  :/8./81+; >.+;+  79,36

56+5<98
79,36


7<


7< 

7<


7<  
B


B

+.3
:/81/8.+;+
8A3
56+5<98
./81+8
0;/5>/8<3

B

Gejala Gelombang

25

Contoh 1.12

Kata Kunci • • • • • • • • • • • • • •

!/;/=+
+:3
,/;1/;+5
7/83811+65+8
<=+<3>8
./81+8
5/-/:+=+8

574+7
(/9;+81 7+<383<
7/7,>8A35+8
:/6>3=8A+
./81+8
0;/5>/8<3

B
35+
-/:+=
;+7,+=
,>8A3 .3
>.+;+

7<
,/;+:+
0;/5>/8<3
A+81
.3./81+;
96/2
:/=>1+<
.3
<=+<3>8
=/;= - 35/=+2>3
5/;/=+
+:3



574+7


7<
>.+;+


7<
:/8./81+;


7<

refleksi tangki riak sudut datang sudut pantul garis normal refraksi fatamorgana indeks bias sudut bias polarisasi interferensi interferensi konstruktif interferensi destruktif efek Doppler

Tes Kompetensi Subbab

 :/8./81+; 


>.+;+  :/8./81+; >.+;+  5/;/=+
+:3

:/6>3=


7<


7<  
B
7<


7<



B +.3
0;/5>/8<3
A+81
.3./81+;
96/2
:/=>1+<
.3
<=+<3>8
=/;=
+.+6+2

B

C

)"#&$
$%
,#,
$+! &







 

(/,>+2
1/697,+81
.+=+81
.+;3
>.+;+
./81+8
<>.>= .+=+81
 C
7+<>5
5/
.+6+7
+3;
.+8
.3,3+<5+8
./81+8 <>.>=
,3+<
C
35+
:+84+81
1/697,+81
=/;=
.3
>.+;+ 
-7
=/8=>5+8
:+84+81
1/697,+81
.3
.+6+7
+3; (/,>+2
1/697,+81
.+=+81
:+.+
,3.+81
,+=+<
+8=+;+ .>+
7/.3>7
./81+8
<>.>=
.+=+81
C
35+
38./5<
,3+<

;/6+=30
7/.3>7
.>+
=/;2+.+:
7/.3>7
<+=>
+.+6+2
  =/8=>5+8
<>.>=
,3+<8A+
.+8
6>53<5+8
:/7,3+<+8 1/697,+818A+ /697,+81
.+=+81
.+;3
:/;7>5++8
+3;
A+81
.+6+7
5/ :/;7>5++8
+3;
A+81
.+815+6
./81+8
<>.>=
.+=+81
 C 35+
-/:+=
;+7,+=
1/697,+81
:+.+
:/;7>5++8
+3;
A+81 .+6+7
.+8
:/;7>5++8
+3;
A+81
.+815+6

7<
.+8

7< =/8=>5+8
<>.>=
,3+<8A+ /6+<5+8
+:+
A+81
.37+5<>.
:96+;3<+<3 (/,>+2
+7,>6+8<
,/;1/;+5
./81+8
5/-/:+=+8

7< <+7,36
7/7,>8A35+8
<3;38/8A+
:+.+
0;/5>/8<3

B (/9;+81
:/81/7>.3
=;>5
,/;1/;+5
,/;6+@+8+8
+;+2 ./81+8
5/-/:+=+8
 
7<
7/8./81+;
,>8A3
<3;38/





+7,>6+8<
)/8=>5+8
0;/5>/8<3
A+81
.3./81+; :/81/7>.3
<++= + 5/.>+
79,36
<+6381
7/8./5+=3 , 5/.>+
79,36
<+6381
7/84+>23 !/=35+
,/;.3;3
.3
=;9=9+;
8A3 ,/;0;/5>/8<3

B
.+;3
<3;38/
79,36
:963<3
A+81 7/8./5+=38A+
(/=/6+2
79,36
:963<3
7/6/@+=38A+ 0;/5>/8<3
,>8A3
A+81
=/;./81+;
7/84+.3

B
/;+:+ 5/6+4>+8
79,36
:963<3
=/;= !/-/:+=+8
,>8A3
.3
>.+;+



7< (/,>+2
<>7,/;
,>8A3
7/736353
0;/5>/8<3

B ,/;1/;+5
./81+8
5/-/:+=+8

7<
7/8./5+=3
5+8
,/<+;
0;/5>/8<3
A+81 .3./81+;
96/2
:/81+7+=
435+ + =3.+5
+.+
+8138 , =/;.+:+=
+8138
A+81
,/;1/;+5
./81+8
5/-/:+=+8 
7<
7,/;
,>8A3
.+8 - =/;.+:+=
+8138
A+81
,/;1/;+5
./81+8
5/-/:+=+8 
7<
,/;6+@+8+8
+;+2
./81+8
+;+2
1/;+5
<>7,/; ,>8A3
/:+=
;+7,+=
,>8A3
.3
>.+;+

7<

Rangkuman

  



26

/;.+<+;5+8
<>7,/;8A+
1/697,+81
.35/697:955+8 7/84+.3
.>+
7+-+7
A+3=>
1/697,+81
7/5+835
.+8 1/697,+81
/6/5=;97+18/=35 /;.+<+;5+8
+;+2
1/=+;8A+
1/697,+81
.3,/.+5+8 7/84+.3
1/697,+81
=;+8.38+6 /<+;
-/:+=
;+7,+=
1/697,+81
,/;,+8.381
6>;>< ./81+8
:+84+81
1/697,+81
.+8
,/;,+8.381
=/;,+635 ./81+8
@+5=>
>8=>5
7/8/7:>2
<+=>
1/697,+81    /<+;8A+
.+A+
A+81
.3=;+8<73<35+8
7/6+6>3
<>+=> 1/697,+81
,/;,+8.381
6>;><
./81+8
5>+.;+= +7:63=>.9
5>+.;+=
0;/5>/8<3
.+8
6+4>
1/697,+81

Mudah dan Aktif Belajar Fisika untuk Kelas XII





        (37:+81+8
1/697,+81
,/;4+6+8
.3
<>+=>
=3=35 .38A+=+5+8
./81+8
:/;<+7++8 



    <38 
       <38 
       



%/;<+7++8
5/-/:+=+8
+2
=3=35
.+6+7
<>+=> 1/697,+81
=/;2+.+:
@+5=>
+.+6+2
=>;>8+8
:/;=+7+ %
A+81
.38A+=+5+8
./81+8
:/;<+7++8

 
 <38
    

       -9<
    

 







%/;-/:+=+8
1/=+;+8
:+;=35/6
.3
=3=35
%
7/;>:+5+8 =>;>8+8
:/;=+7+
.+;3
5/-/:+=+8
1/=+;
=3=35
%
=/;= =/;2+.+:
@+5=>

 

     -9<
    

  

    <38
    
     %/;<+7++8
1/697,+81
<=+<398/;
+53,+=
:/7+8=>6+8 :+.+
>4>81
=+63
,/,+<
+.+6+2




  -9<
  <38 
     

 ./81+8
 -9<
  +.+6+2
+7:63=>.9
1/697,+81  <>:/;:9<3<3 "/=+5
<37:>6
.+;3
>4>81
:/7+8=>6+8
:+.+
>4>81
=+63 ,/,+<
+.+6+2

 
   

"/=+5
:/;>=
./697,+81
.+;3
>4>81
:/7+8=>6+8
:+.+ >4>81
=+63
,/,+<
+.+6+2

      %/;<+7++8
1/697,+81
<=+<398/;
+53,+=
:/7+8=>6+8 :+.+
>4>81
=+63
=/;35+=
+.+6+2




   <38 
  -9<
      

./81+8
 <38 
  +.+6+2
+7:63=>.9
1/697,+81  "/=+5
<37:>6
.+;3
>4>81
:/7+8=>6+8
:+.+
>4>81
=+63 =/;35+=
+.+6+2

     

.+:>8
6/=+5
:/;>=
1/697,+81
.+;3
>4>81 :/7+8=>6+8
:+.+
>4>81
=+63
=/;35+=
+.+6+2

 
   

 /697,+81
.+:+=
7/81+6+73
:/7+8=>6+8
:/7,3+<+8 :96+;3<+<3
38=/;0/;/8<3
.+8
.30;+5<3

 >5>7
(8/663><
:+.+
:/7,3+<+8
1/697,+81 .38A+=+5+8
./81+8
:/;<+7++8   <38     <38 

 0/5
9::6/;
7/;>:+5+8
0/897/8+
:/;>,+2+8 0;/5>/8<3
1/697,+81
+53,+=
:/81+;>2
1/;+5
;/6+=30 +8=+;+
<>7,/;
1/697,+81
.+8
:/81+7+=
(/-+;+ >7>7
.38A+=+5+8
./81+8
:/;<+7++8         

Peta Konsep %/7+8=>6+8

%/7,3+<+8

/697,+81 #/5+835

%96+;3<+<3

8=/;0/;/8<3

.+:+=
7/81+6+73 ,/;.+<+;5+8 <>7,/;8A+ =/;.3;3
+=+<

/697,+81 6/5=;97+18/=35

$'%&

,/;.+<+;5+8 +;+2
1/=+;8A+ =/;.3;3
+=+<

,/;.+<+;5+8 +7:63=>.98A+ =/;.3;3
+=+<

/697,+81 (=+<398/;

30;+5<3

/697,+81 );+8
/697,+81 "9813=>.38+6

/697,+81 /;4+6+8

Refleksi Setelah mempelajari bab ini, Anda tentu telah mengetahui jenis-jenis gelombang serta sifat-sifatnya. Dari materi bab ini, bagian mana yang Anda anggap sulit? Coba diskusikan dengan teman atau guru Fisika Anda.

Selain itu, coba Anda sebutkan manfaat lainnya yang Anda peroleh setelah mempelajari materi bab ini.

Gejala Gelombang

27

Tes Kompetensi Bab 1

 !$! $
*$
*+,
"-&
.&
($!&
+(+
&
#)"#&$
(
,#,
$+! &









(/,>+2
1/697,+81
7/736353
:/;<+7++8
<37:+81+8    <38
     .+6+7
<+=>+8
(
;/5>/8<3 .+8
:+84+81
1/697,+81
=/;=
+.+6+2
 + 
B
.+8

7 . 
B
.+8

7 , 
B
.+8

7 / 
B
.+8

7 - 
B
.+8

7 *8=>5
,/8.+
A+81
7/81+6+73
1/=+;+8
2+;79835
:+.+ <++=
<37:+81+8
7+5<37>7
 + 5/-/:+=+8
.+8
:/;-/:+=+8
7+5<37>7 , 5/-/:+=+8
.+8
:/;-/:+=+8
73837>7 - 5/-/:+=+88A+
7+5<37>7
.+8
:/;-/:+=+88A+
896 . 5/-/:+=+88A+
896
.+8
:/;-/:+=+88A+
7+5<37>7 / <37:+81+8
7+5<37>7
/8/;138A+
7+5<37>7 /:+=
;+7,+=
1/697,+81
:+.+
=+63
+.+6+2

-7< ./81+8
0;/5>/8<3

B
+;+5
.>+
=3=35
:+.+
=+63
435+ ,/.+
0++
=3=35

+.+6+2


+ 
-7 . 
-7 , 
-7 / 
-7 - 
-7 /697,+81
(
7/;+7,+=
./81+8
:/;<+7++8


   
<+=>+8

.+8

.+6+7
7/=/;
.+8  

   <38
 






.+6+7
2

7
.+;3
(
+.+6+2
 +  
7 . 
7 , 
7 /  
7 -  
7 (/,>+2
1/697,+81
,/;4+6+8
7/6+6>3
=3=35

.+8

A+81 ,/;4+;+5

-7
.+6+7
+;+2
.+;3

5/

%+.+
<++=


 <37:+81+8
1/697,+81
.3

+.+6+2
896
35+
:+84+81 1/697,+81

-7
.+8
+7:63=>.9

-7
<37:+81+8
=3=35

:+.+
<++=
0++2
1/697,+81
=;+8

    
./81+8

.+8


.+6+7
-7
.+8

.+6+7
/8<38A+

B   +7:63=>.98A+

-7   -/:+=
;+7,+=8A+

-7< %/;8A+=++8
A+81
,/8+;
+.+6+2
 + 

.+8
.
<+4+ ,
.+8
/ +
,/8+; - 
.+8



28

Mudah dan Aktif Belajar Fisika untuk Kelas XII



%+.+
:/;-9,++8
#/6./
.31>8+5+8
=+<
.+@+3
A+81 :+84+818A+

-7
.+8
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1
35+
.+@+3 .3=/1+815+8
./81+8
1+A+



)+63
A+81
:+84+818A+

7
.3=/1+815+8
./81+8
1+A+ 
$
5/7>.3+8
.3;+7,+=3
1/697,+81
=;+8=

7<
7+<<+
=+63 +.+6+2
 + 
1;+7 . 
1;+7 , 
1;+7 / 
1;+7 - 
1;+7

+&*



 



%+.+
:/;-9,++8
 
.31>8+5+8
=+<
,/8+81
A+81 :+84+818A+

7/=/;
7+<<+
,/8+81

1;+7
35+
,/,+8 815+8
:+.+
,/8+81
=/;=
.+8 

7<
7+5+
5/-/:+=+8
:/;+7,+=+8
1/697,+81 :+.+
,/8+81
+.+6+2
 + 
7< . 
7< ,
7< /
7< - 
7<

 (+=>
:+84+81
1/697,+81
+.+6+2
4+;+5
+8=+;+
 + :>8-+5
5/
:>8-+5
,/;>;>=+8 , <37:>6
.+8
:/;>=
,/;>;>=+8 - :/;>=
.+8
6/7,+2
,/;>;>=+8 . :>8-+5
5/
6/7,+2
,/;>;>=+8 / <37:>6
5/
<37:>6
,/;>;>=+8

 (/,>+2
1/697,+81
6+>=
.+=+81
7/7/-+2
:+8=+3
1/697,+81
7/8>4>
:+8=+3

 +7,+;
,/;35>=
7/8>84>55+8
1/697,+81
=;+8
U R

S

T

V

+6+7
<+=>
:/;39./
.3=/7:>2
4+;+5
 + %D&D'D( . &D'D( , 'D(D)D* / 'D(D) - %D&D'D(D)

 35+
1/697,+81
+3;
7/;+7,+=
.+;3
=/7:+=
A+81
.+6+7 5/
=/7:+=
A+81
6/,32
.+815+6
7+5+
 + -/:+=
;+7,+=
.+8
:+84+81
1/697,+81
=/=+: , -/:+=
;+7,+=
7/84+.3
6/,32
,/<+;
.+8
:+84+81 1/697,+81
=/=+:

-

-/:+=
;+7,+=
7/81/-36
.+8
:+84+81
1/697,+81 7/7,/<+; . -/:+=
;+7,+=
.+8
:+84+81
1/697,+81
7/81/-36 / -/:+=
;+7,+=
7/81/-36
.+8
0;/5>/8<3
7/7,/<+;

 /697,+81
7/;+7,+=
7/6+6>3
.>+
B+=
7>6+7>6+ ./81+8
5/-/:+=+8

7<
<>.>=
.+=+81
C
.+8 .3,3+<5+8
:+.+
B+=
5/.>+
./81+8
<>.>=
 C
/:+= ;+7,+=
1/697,+81
:+.+
B+=
5/.>+
+.+6+2
 +
7< . 
7< , 
7< / 
7< - 
7<

 (/>=+<
=+63
:+84+818A+

7
.+8
,/;7+<<+

1;+7 .3=/1+815+8
./81+8
1+A+

$
35+
0;/5>/8<3 <>7,/;
1/=+;+88A+

B
-/:+=
;+7,+=
1/697,+81 =+63
=/;=
+.+6+2
 +
7< . 
7< , 
7< / 
7< - 
7<

 %/;3<=3@+
:/;:+.>+8
+8=+;+
.>+
1/697,+81
.3
<>+=> =3=35
.+6+7
@+5=>
A+81
,/;<+7++8
.3=
 + :/7,3+<+8 . .30;+5<3 , :/7+8=>6+8 / :96+;3<+<3 - <>:/;:9<3<3

 (/>=+<
=+63
A+81
:+84+818A+

7
5/.>+
>4>818A+ .335+=
./81+8
/;+=
!/7>.3+8
:+.+
=+63
.3=37,>65+8 1/697,+81
5

,>+2
:/;>=
"/=+5 :/;>=
5/637+
.+;3
>4>81
=/;4+>2
+.+6+2
 + 
7 . 
7 , 
7 / 
7 - 
7

 35+
+2
1/697,+81
7/6/@+=3
:/812+6+81
A+81 7/736353
-/6+2

 35+
:+.+
:/;-9,++8
#/6./
=/1+81+8
.3,>+=
6+
-/:+=
;+7,+=
1/697,+81
:+.+
=+63 =/;=
7/84+.3


+ 5+63
6+ . 
5+63
6+

,
5+63
6+ / 
5+63
6+  -
5+63
6+   (/,>+2
<>7,/;
,>8A3
./81+8
0;/5>/8<3
 
B
,/; 1/;+5
7/8./5+=3
8A3
.3
>.+;+ +.+6+2
23
<>7,/;
,>8A3
./81+8
5/-/:+=+8

7< ,/<+;
0;/5>/8<3
,>8A3
A+81
.3./81+;
96/2
:/81+7+= +.+6+2
 + 
B . 
B ,  
B /  
B -  
B

  8.3
,/;.3;3
.3
<+7:381
<>7,/;
,>8A3
A+81
0;/5>/8 <38A+

B
(/,>+2
<>7,/;
,>8A3
6+38
./81+8 0;/5>/8<3

B
7/8./5+=3
8.3
./81+8
5/-/:+=+8 
7<
35+
5/-/:+=+8
7/;+7,+=
,>8A3
.3
>.+;+
+.+6+2
7<
8.3
+5+8
7/8./81+;
6+A+81+8
,>8A3 ./81+8
0;/5>/8<3
 + . 
B , 
B / 
B -
B  (/,>+2
<>7,/;
,>8A3
,/;1/;+5
./81+8
5/-/:+=+8


7<
7/8>4>
:/8./81+;
A+81
.3+7
/8<3
A+81
.3./81+;
96/2
:/8./81+;
+.+6+2
  35+
<>7,/;
,>8A3
=/;=
.3+7
7,/;
,>8A3
0;/5>/8<3
A+81
.3./81+; 96/2
:/8./81+;
+.+6+2

35+
-/:+=
;+7,+=
,>8A3
.3 >.+;+

7<
:/;,+8.381+8



+.+6+2
 +





,





  



  

.





/



-





  (/9;+81
:/8/;,+81
A+81
:/<+@+=
=/;,+818A+
7/8>4> 5/
7/8+;+
,+8.+;+
7/8./81+;
,>8A3
<3;38/
7/8+;+ ./81+8
0;/5>/8<3

B
35+
:+.+
<++=
3=>
<3;38/ 7/7+8-+;5+8
,>8A3
./81+8
0;/5>/8<3

B
.+8 -/:+=
;+7,+=
,>8A3
.3
>.+;+

7<
5/-/:+=+8 :/<+@+=
=/;,+81
=/;=
+.+6+2
 + 
574+7 . 
574+7 , 
574+7 /  
574+7 -  
574+7   (/,>+2
1/697,+81
,/;4+6+8
7/6+6>3
=3=35

.+8

A+81 ,/;4+;+5

-7
%+.+
<++=



<37:+81+8
1/697,+81 .3

+.+6+2

35+
:+84+81 1/697,+818A+

-7
.+8 +7:63=>.98A+

-7 <37:+81+8
=3=35

:+.+
<++=
0+


+.+6+2

-7 

+



.



,

 

/

-   (/>=+<
=+63
:+84+818A+

7
.31/=+;5+8
=;+8
;+7,+=
1/697,+81
=;+8= 
7<
35+
1+A+
=/1+81+8
:+.+
=+63
=/;=

$ 7+<<+
=+63
+.+6+2
 +  
51 . 
51 ,  
51 / 
51 -  
51

Gejala Gelombang

29



-$
()+&.&
)!#,+
!&!
&&
+(+



%+.+
<>+=>
<++=
,/8=>5
1/697,+81
=;+881<3
.+;3
=/7:+=
5/.>.>5+8

.3=>84>55+8
= B A

F C

D

G

E

I

H





)/8=>5+8
.>+
=3=35
A+81
0+.
./81+8
1/697,+81
<=+<398/; /6+<5+8
./81+8
7/7:/;1>8+5+8
1+7,+; (/>=+<
=+63
./81+8
:+84+81

.31/=+;5+8
2+;79835 4>81

/8<3
1/697,+81

B
.+8
+7:63=>.98A+ 
-7
)3=35

,/;+.+

7
.+;3

35+

7>6+3
.31/;+5 5+8
5/
+=+<
23=>816+2 + <37:+81+8
.3
 , 0++2
1/697,+81
=;+8


+.+6+2   <38 
    ./81+8

.+8

.3 

8A+=+5+8
.+6+7
<+=>+8
7
.+8

.38A+=+5+8
.+6+7 <+=>+8

3=>816+2 + +7:63=>.98A+ , :+84+81
1/697,+818A+ - -/:+=
;+7,+=8A+ . :/;39./8A+ %/;<+7++8
1/697,+81
69813=>.38+6
,/;4+6+8
.+6+7











 

,+=+81
+.+6+2
    <38 
 



30

 

.+8

 






.+6+7
-7
.+8

.+6+7
<
3=>816+2
,/<+; + +7:63=>.9
1/697,+81 , :+84+81
1/697,+81 - -/:+=
;+7,+=
1/697,+81 . :/;39./
1/697,+81 35/=+2>3
:/;<+7++8
1/697,+81
,/;4+6+8
   <38 
  
./81+8

.+8

.38A+=+5+8




 



 

Mudah dan Aktif Belajar Fisika untuk Kelas XII



.+6+7
<+=>+8
-7
.+8

.38A+=+5+8
.+6+7
<+=>+8
< )/8=>5+8
836+3
.+;3 + 0;/5>/8<3
1/697,+81 , :+84+81
1/697,+81 - -/:+=
;+7,+=
1/697,+81
=/;= (/,>+2
1/697,+81
,/;4+6+8
.+;3
=3=35

5/
=3=35
 ./81+8
+7:63=>.9

-7
.+8
:/;39./

=

7<
=/8=>5+8 + 3
5/.>.>5+8 +2
1/697,+81
<=+<398/;
.3,/8=>5
.+;3
38=/;0/;/8<3 .>+
,>+2
1/697,+81
A+81
7+<3817+<381
7/736353


+7:63=>.9

-7
,36+81+8
1/697,+81


  
-7  .+8
0;/5>/8<3
<>.>=
 

;+.< + 3=>81
4+;+5
+8=+;+
.>+
:/;>=
1/697,+81
A+81 ,/;>;>=+8 , /;+:+
+7:63=>.9
1/697,+81
<=+<398/;
.3 


-7 (/,>+2
<>7,/;
,>8A3
,/;0;/5>/8<3

B
,/;1/;+5 ./81+8
5/-/:+=+8

7<
7/8./5+=3
5+8
0;/5>/8<3
A+81 .3./81+;
96/2
:/81+7+=
435+ + =3.+5
+.+
+8138 , =/;.+:+=
+8138
A+81
,/;1/;+5
./81+8
5/-/:+=+8


7<
7,/;
,>8A3 - =/;.+:+=
+8138
A+81
,/;1/;+5
./81+8
5/-/:+=+8


7<
,/;6+@+8+8
./81+8
+;+2
1/;+5
<>7,/; ,>8A3 (/,>+2
79,36
A+81
,/;1/;+5
./81+8
5/-/:+=+8 
7<
%+.+
+;+2
,/;6+@+8+8
,/;1/;+5
79,36 +7,>6+8<
./81+8
5/-/:+=+8

7<
7,>6+8<
7/7 ,>8A35+8
<3;38/
8A3
<3;38/ ,/;0;/5>/8<3

B
)/8=>5+8
0;/5>/8<3
<3;38/
A+81 =/;./81+;
96/2
:/81/7>.3
79,36
=/;=
435+
+8138 ,/;=3>:
./81+8
5/-/:+=+8

7< + 6+8< ,
Bab

2 Sumber: Physics Today, 1995

Cahaya merupakan gelombang elektromagnetik yang banyak digunakan untuk kepentingan teknologi komunikasi.

Gelombang Cahaya Hasil yang harus Anda capai: menerapkan konsep dan prinsip gejala gelombang dalam menyelesaikan masalah. Setelah mempelajari bab ini, Anda harus mampu: • •

mendeskripsikan gejala dan ciri-ciri gelombang cahaya; menerapkan konsep dan prinsip gelombang cahaya dalam teknologi.

5>?=5>165>?=5>1
C5>C1>7
3181G1
B5A9>7
>41
:D=@19
41<1= ;5894D@1>
B581A981A9
%5C9;1
CDAD>
8D:1>
@141
B91>7
81A9
;141>7 ;141>7
;9C1
41@1C
=5>9;=1C9
9>418>G1
@5<1>79
G1>7
=5>7891B9
35A18>G1 <1>79C
@1;18
B525>1A>G1
@5<1>79
9CD 181G1
=5AD@1;1>
757
5<5;CA?=17>5C9;
*141
12

>41 C5<18
=5=@5<1:1A9
757
=5;1>9;
=9B1<>G1
757
19A
41> 757
C1<9
@1;18
@5A25411>
1>C1A1
757
5<5;CA?=17>5C9; 41>
757
=5;1>9; '1>DB91
C5<18
=5=1>611C;1>
757
5<5;CA?=17>5C9;
41<1= 2941>7
CA1>B@?AC1B9
1BCA?>?=9
=9<9C5A
41>
75?7A169
,18D;18
>41 21719=1>1
75:1<1
41>
39A939A9
757
5<5;CA?=17>5C9;
;8DBDB>G1 3181G1
B5AC1
@5>5A1@1>>G1
$1F121>
1C1B
@5AC1>G11>@5AC1>G11>
C5AB52DC 41@1C
>41
C5=D;1>
@141
212
9>9
)<58
;1A5>1
9CD
@5<1:1A9
212
9>9 45>71>
219;

A. Interferensi Cahaya B. Difraksi Cahaya C. Polarisasi Cahaya

Gelombang Cahaya

31

Tes Kompetensi Awal !(&(.6/
/(/2(.%,%3+
-104(2
(.1/&%0)
%*%8%
-(3,%-%0.%*
41%.41%.
&(3+-65
'%.%/
&6-6
.%5+*%0 @1;18
G1>7
49=1;BD4
45>71>
757
5<5;CA?  5A1@1;18
;535@1C1>
3181G1
41<1=
AD1>7
81=@1 =17>5C9;  +52DC;1>
B961CB961C
757
3181G1

@1;18
3181G1
C5A=1BD;
757
5<5;CA?=17>5C9;

Tokoh

A. Interferensi Cahaya  
141<18
 "
!
!

 !

# 

1AC9>G1
757
=5=9<9;9
!

41>
 #
 
+5@5AC9
G1>7
C5<18
492181B
@141
@5=2181B1>
757 218F1
757
3181G1
25AB961C
B5@5AC9
81<>G1
757
2D>G9
G19CD 41@1C
25A9>C5A65A5>B9
)<58
;1A5>1
9CD
D>CD;
=5>41@1C;1>
9>C5A65A5>B9 3181G1
@D>
49@5A
BD=25A
3181G1
G1>7
;?85A5>
G19CD
BD=25A
3181G1 G1>7
=5=9<9;9
6A5;D5>B9
B1=1
41>
2541
61B5
C5C1@
+D=25A
3181G1
G1>7 ;?85A5>
41@1C
491=1C9
=5<1
G1>7
49<1;D;1>
?<58
$160) 41>
3(40(..

Augustine Fresnell (1788–1827)

1. Percobaan Young dan Fresnell Augustine Fresnell (1788–1827) adalah fisikawan Prancis yang mengembangkan teori gelombang transversal cahaya berdasarkan hasil penemuannya tentang lensa dan interferensi. Fresnell memperlihatkan bahwa cahaya matahari terdiri atas bermacammacam warna cahaya, yang setiap warna memiliki sudut bias tertentu. Ia juga menemukan sebuah bentuk lensa yang pada kedua permukaannya berbentuk cembung. Bentuk lensa ini dikenal sebagai lensa cembung. Lensa ini memiliki sifat mengumpulkan cahaya.

a.

Percobaan Celah Ganda Young *5A3?211>
9>9
49<1;D;1>
?<58
$160)
45>71>
=5>77D>1;1>
4D1
@5>7 81<1>7
*5>781<1>7
@5AC1=1
=5=9<9;9
B1CD
7
;539<
41>
@5>781<1>7 ;54D1
49<5>7;1@9
45>71>
4D1
7
;539<
*5A81C9;1>
%/&%3
 terang gelap terang gelap terang gelap terang gelap terang gelap terang gelap terang gelap terang gelap terang gelap terang gelap terang gelap terang gelap

gelombang datang S2

S0

Sumber: Science Encyclopedia, 1998

S1

Gambar 2.1 Interferensi pada celah ganda Young

terang A S C1 S1 O

S2

C2

P

Gambar 2.2 Percobaan Fresnell untuk menunjukkan interferensi cahaya.

32

B

C

+9>1A
=?>?;A?=1C9B
49@5A?<58
41A9
<1=@D
B521719
BD=25A
3181G1
G1>7 =5=1>31A
=5<1
B9>1A
41A9
35<18
+
49@1>31A;1>
;5 @5>781<1>7
;54D1
D1
35<18
@141
@5>781<1>7
;54D1
G19CD
+
41>
+ G1>7
49@1B1>7
B5:1:1A
45>71>
+
1;1>
25A6D>7B9
B521719
@5=1>31A
B9>1A B9>1A
;?85A5>
%54D1
25A;1B
41A9
35<1835<18
+
41>
+
9>9
25A9>C5A65A5>B9 @141
<1G1A

"1B9<
9>C5A65A5>B9
25AD@1
71A9B
C5A1>7
41>
71A9B
75<1@ b. Percobaan Fresnell 5>71>
=5>77D>1;1>
B52D18
BD=25A
3181G1
+
3(40(..
=5=@5A?<58 4D1
BD=25A
3181G1
+
41>
+
G1>7
;?85A5>
41A9
81B9<
@5=1>CD<1>
4D1 35A=9>
*5A81C9;1>
%/&%3


Mudah dan Aktif Belajar Fisika untuk Kelas XII

*141
%/&%3

C5AB52DC
41@1C
49<981C
218F1
+
141<18
BD=25A
B9>1A =?>?;A?=1C9B
+
41>
+
141<18
21G1>71>
41A9
+
?<58
35A=9>

41>
 5>71>
45=9;91>
B9>1AB9>1A
G1>7
41C1>7
@141
<1G1A
B5?<18?<18
25A1B1< 41A9
+
41>
+
!57
3181G1
41A9
+
41>
+
9>9
1;1>
B1<9>7 25A9>C5A65A5>B9
@141
<1G1A
41>
81B9<>G1
25A71>CD>7
@141
B5<9B98
41A9
<9>C1B1> ;54D1
B9>1A
9CD
*5A41
;5C18D9
:9;1
;54D1
BD=25A
3181G1
=5=9<9;9 1=@<9CD4?
G1>7
B1=1
@141
C5=@1C
C5A:149>G1
9>C565A5>B9
=9>9=D=
1;1> C5A25>CD;
71A9B
75<1@
+521<9;>G1
:9;1
1=@<9CD4?
C941;
B1=1
9>C5A65A5>B9 =9>9=D=>G1
C941;
75<1@
B1=1
B5;1<9 r2

D P

S2

y r1

S1

d b

r1

Gambar 2.3 (a) Sinar gelombang dari celah S1 dan S2 berinterferensi di titik P. (b) Untuk D
d, r1 dan r 2 dianggap sejajar dan membentuk sudut  terhadap sumbu tengah.



r2 S1



S2



b  L

2

B

Pembahasan Soal

C

1

*5A81C9;1>
%/&%3
 
757
3181G1
41C1>7
=5>D:D
35<18
41>
35<18



G1>7
C5A<5C1;
@141
2941>7

181G1
C5AB52DC
C5A496A1;B9
?<58 ;54D1
35<18
C5AB52DC
41>
=5>781B9<;1>
@?<1
9>C5A65A5>B9
@141
<1G1A
 %9C1
41@1C
=5>5>CD;1>
49
=1>1
B5C91@
@9C1
C5A1>7
1C1D
@9C1
75<1@ C5A<5C1;
@141
<1G1A
45>71>
=5=25A9;1>
BD4DC

41A9
BD=2D
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Seberkas cahaya monokromatis dijatuhkan pada dua celah sempit vertikal berdekatan dengan jarak d = 0,01 mm. Pola interferensi yang terjadi ditangkap layar pada jarak 20 cm dari celah. Diketahui bahwa jarak antara garis gelap pertama sebelah kiri ke garis gelap sebelah kanan adalah 7,2 mm. Panjang gelombang cahaya tersebut adalah .... a. 180 nm d. 720 nm b. 270 nm e. 1.800 nm c. 360 nm SPMB 2003 Pembahasan: Jarak pola gelap ke-1 ke pusat adalah 7,2  10 3 m y  3,6  10 3 m 2 Dari soal diketahui: m = 1; d = 10–5 m; D = 0,2 m sehingga yd  m 1  D 2 yd  D m 1 2







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2. Menentukan Jarak Pita Terang ke-m atau Pita Gelap ke-m dari Terang Pusat

Tantangan untuk Anda Seberkas cahaya monokromatis dijatuhkan pada dua celah sempit vertikal berdekatan dengan jarak d = 0,01 mm. Pola interferensi yang terjadi ditangkap pada jarak 20 cm dari celah. Diketahui bahwa jarak antara garis gelap pertama di sebelah kiri ke garis gelap pertama di sebelah kanan adalah 7,2 mm. Hitunglah panjang gelombang berkas cahaya tersebut.

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34

Mudah dan Aktif Belajar Fisika untuk Kelas XII

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Gelombang Cahaya

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Sumber: www.designprodygzone.com

Gambar 2.5 Interferensi oleh busa sabun.


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Sumber: www.funsci.com

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Sumber: www.instckphoto.com

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36

Mudah dan Aktif Belajar Fisika untuk Kelas XII

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Tugas Anda 2.1 Coba Anda perhatikan kembali Gambar 2.5. Gelembung tersebut sebenarnya berwarna-warni. Mengapa demikian? Anda dapat mencari jawabannya dari buku referensi atau internet.


 
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Tes Kompetensi Subbab

Kata Kunci • • • •

interferensi sinar monokromatis interferensi maksimum interferensi minimum

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37

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1. Difraksi Celah Tunggal 5 4



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Gambar 2.8 Difraksi cahaya pada celah tunggal.

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Gambar 2.9 Maksimum utama terjadi untuk k = 0 atau  = 0.

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38

Mudah dan Aktif Belajar Fisika untuk Kelas XII

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Pembahasan Soal

garis gelap



Suatu berkas sinar sejajar mengenai celah yang lebarnya 0,4 mm secara tegak lurus. Di belakang celah diberi lensa positif dengan jarak titik api 40 cm. Garis terang pusat (orde nol) dengan garis gelap pertama pada layar di bidang titik api lensa berjarak 0,56 mm. Panjang gelombang sinar adalah .... a. 6,4 × 10–7 m b. 6,0 × 10–7 m c. 5,2 × 10–7 m d. 5,6 × 10–7 m e. 0,4 × 10–7 m PPI 1983 Pembahasan: Jarak titik api = jarak celah ke layar =  = 40 cm. Gelap pertama m = 1 dp  m 
0,4  10 3 m
0,5  10 3 m  1 0,4 m –7  = 5,6 × 10 m

Q

 terang

pusat

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T

B



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d C D E F G

M



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Gambar 2.10 Difraksi pada kisi

Gelombang Cahaya

39

intensitas m=1

m=2


1 31 
1 23 



2

d sin

Gambar 2.11 Difraksi minimum kedua untuk N = 2 celah. m = –1 m = 0 m = 1 m = 2

d sin 2



0

Gambar 2.12 Difraksi minimum kedua untuk N banyak celah.

me rah bir u

spektrum orde ke 2 ah er m u bir

cahaya putih kisi difraksi

u bir rah me

b m iru er ah

spektrum orde ke-1 spektrum orde ke-0 (putih) spektrum orde ke-1

spektrum orde ke-2

Gambar 2.13 Difraksi cahaya putih akan menghasilkan pola berupa pita-pita spektrum.

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Tantangan untuk Anda Tentukan daya urai sebuah celah dengan diameter 2 mm, jarak celah ke layar 1 meter dengan panjang gelombang cahayanya 590 nm.

40

Mari Mencari Tahu *5A>18;18
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Mudah dan Aktif Belajar Fisika untuk Kelas XII

3. Daya Urai Optik #
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Gambar 2.14 Bayangan dari optik fisis dua benda yang berdekatan karena (a) beririsan dan (b) terpisah dengan baik.

celah bulat

pola difraksi

r

 D  cahaya sumber datang

2

Gambar 2.15 Lukisan sinar dari sumber cahaya dari sebuah celah bulat.

Informasi untuk Anda

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==
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Merak jantan dengan bulu-bulu ekornya yang berwarna-warni dan berukuran lebar menyebabkan lebih kelihatan menarik dibanding merak betina. Keindahan bulu merak tersebut merupakan contoh efek difraksi gelombang cahaya oleh bulu merak.

Information for You Male with the largest or most colorful adornments are often the most attractive to females. The extraordinary feathers of a peacock’s tail are an example of diffraction effect of peacock’s tail light wakes. Sumber: Biology Concepts & Connections, 2006

Gelombang Cahaya

41

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Kata Kunci • • • • • • • •

difraksi gelombang kisi difraksi sudut simpang (deviasi) difraksi maksimum difraksi celah tunggal difraksi pada kisi daya urai optik apertur

!    
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Tes Kompetensi Subbab

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C. Polarisasi Cahaya 1. Polarisasi pada Kristal

cahaya alami yang datang getaran vertikal diserap sebagian getaran horizontal diserap sempurna oleh polaroid

cahaya diteruskan terpolarisasi linear

Gambar 2.16 Cahaya tak terpolarisasi dilewatkan pada sebuah kristal.

42

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B1CD
1A18
75C1A1> C5AC5>CD
1C1D
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C5A@?<1A9B1B9
;1A5>1
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%
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41>
1>1<9B1C?A
G1>7
49@1B1>7

Mudah dan Aktif Belajar Fisika untuk Kelas XII

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1C1D
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3181G1
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25A;1B
3181G1
1<1=918
45>71>
9>C5>B9C1B

49<5F1C;1>
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C571;

C1;
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3181G1
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16+4
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3?B  
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cahaya yang diteruskan terpolarisasi

sumber cahaya

polarisator

analisator

1 tidak ada cahaya

sumber cahaya

polarisator

analisator

2 Gambar 2.17 (a) Polarisator dan analisator dipasang sejajar sehingga cahaya yang diteruskan di belakang analisator akan terpolarisasi linear. (b) Polarisator dan analisator dipasang tegak lurus sehingga cahaya tidak diteruskan oleh analisator.

*5AB1=11>
9>9
49B52DC
:D71
6-6/
%.64
45>71>

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9>C5>B9C1B 3181G1
G1>7
<5F1C
1>1<9B1C?A
41>

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1>C1A1
@?<1A9B1C?A
41> 1>1<9B1C?A
+1CD1>
9>C5>B9C1B
3181G1
141<18
F1CC =

Contoh 2.9 +525A;1B
3181G1
1<1=918
49<5F1C;1>
@141
4D1
;5@9>7
;131
@?<1A?94
G1>7
1A18
@?<1A9B1B9 B1CD
B1=1
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=5=25>CD;
BD4DC
J
$9;1
9>C5>B9C1B
3181G1
1<1=918>G1
141<18 
. = 
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9>C5>B9C1B
3181G1
G1>7
C5<18
=5<5F1C9
;54D1
;131
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=5>77D>1;1>
(34%/%%0
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ip

 

 
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cermin terpolarisasi

1

2. Polarisasi pada Pemantulan dan Pembiasan *5A81C9;1>
%/&%3
&
+525A;1B
B9>1A
41C1>7
G1>7
49<5F1C;1> @141
@5A=D;11>
2941>7
21C1B
4D1
=549D=
G1>7
9>45;B
291B>G1
25A2541 =9B1<>G1

41>
 
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9CD
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49B52DC
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G19CD 
B9>



B9>

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B9>



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B9>   

 3?B  

ip n1 n2

r terpolarisasi

2 Gambar 2.18 Polarisasi pada (a) pemantulan dan (b) pembiasan.

Gelombang Cahaya

43

  C1>   

normal

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5>71>

141<18
BD4DC
@?<1A9B1B9
1C1D
BD4DC
" 
41>

B5AC1
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9>45;B
291B
=549D=
B1CD
41>
=549D=
4D1

n1 n2

3. Polarisasi pada Pembiasan Ganda (1)

(2)

Gambar 2.19 Polarisasi pembiasan ganda.

*141
;A9BC1<
;1
5B 3181G1
41@1C
=5>71<1=9
@5=291B1>
71>41
;1A5>1
=5=9<9;9
4D1
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291B
*141
%/&%3

C1=@1;
141
4D1
21791>
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G1>7 49291B;1>
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@5=291B1>
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+>5<<9DB 1C1D
49B52DC
B9>1A
9BC9=5F1
B9>1A
 
=5>79;DC9
8D;D=
@5=291B1>
+>5<<9DB 1C1D
49B52DC
B9>1A
291B

4. Polarisasi karena Hamburan

partikel-partikel gas gelombang datang tak terpolarisasi gelombang hamburan terpolarisasi

Gambar 2.20 Polarisasi karena hamburan.

181G1
G1>7
41C1>7
@141
H1C
71B
1;1>
=5>71<1=9
@?<1A9B1B9
B521791> <5;CA?>5<5;CA?>
41<1=
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1;1>
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41>
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#

 
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B9>1A
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41>
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;5=21<9
C5ADC1=1 B9>1A
29AD>G1
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81A9
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3181G1
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1C=?B65A B589>771
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7 G1>7
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71A9B
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5. Pemutaran Bidang Polarisasi cahaya alami tak terpolarisasi

cahaya terpolarisasi



E 0 cos

E0 sumbu polarisator sumbu analisator

Gambar 2.21 Pemutaran bidang polarisasi.

%/&%3
 
=5=@5A<981C;1>
B52D18
 
G1>7
C5A49A9
1C1B 4D1
2D18
@?<1A?94
G1>7
49@1B1>7
@141
B52D18
1<1C
G1>7
49<5>7;1@9
45>71> B;1<1
45A1:1C
41>
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<1ADC1>
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G1>7
45;1C
45>71>
BD=25A 3181G1
49B52DC
@?<1A9B1C?A
41>
G1>7
<19>>G1
141<18
1>1<9B1C?A 'D<1=D<1
=1C1
49
25<1;1>7
1>1<9B1C?A
C941;
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3181G1
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75<1@
*5>D>:D;
1>1<9B1C?A
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BD4DC
 
;5=D491> 49
1>C1A1
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7D<1
181G1
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<1ADC1>
9>9 B5251<9B1C?A
+5C5<18
491=1C9
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141>G1
3181G1
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41<1=
81<
9>9
25A6D>7B9 B521719
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2941>7
75C1A
71A
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1>1<9B1C?A 49@DC1A
B589>771
=5>D>:D;;1>
BD4DC

$149
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BD4DC
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75C1A
3181G1
G1>7
49<1;D;1>
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<1ADC1>
7D<1
141<18

  

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&1ADC1>
7D<1
C5AB52DC
49B52DC
<1ADC1>
 
 
&1ADC1>
C5AB52DC 141
G1>7
41@1C
=5=DC1A
2941>7
75C1A
@?<1A9B1B9
;5
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41>
141
:D71 G1>7
;5
;1>1>
5>71>
1<1C
B5=131=
9>9
?A1>7
41@1C
=5>5>CD;1>

44

Mudah dan Aktif Belajar Fisika untuk Kelas XII

;?>B5>CA1B9
<1ADC1>
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1;C96
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41>
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41@1C
49CD<9B;1> B521719
25A9;DC  K 

     %5C5A1>71>    
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@1>:1>7
<1ADC1>
C521<  
BD4DC
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:5>9B
<1ADC1>

Kata Kunci • • • • • • • • •

bidang polarisasi dichroic polarisator analisator sudut polarisasi/sudut Brewster hamburan polarimeter larutan optik aktif sacharimeter

Contoh 2.10 $9;1
9>45;B
291B
D41A1
141<18

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(34%/%%0
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49@5A?<58 C1>




 



C1>
 



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$149
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21
Contoh 2.11 +52D18
B1381A9=5C5A
G1>7
@1>:1>7>G1

3=
25A9B9
<1ADC1>
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49@5A?<58

      



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Tes Kompetensi Subbab

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291B
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Gelombang Cahaya

45

Rangkuman





181G1
C5A=1BD;
757
5<5;CA?=17>5C9; B589>771
@5A1=21C1>>G1
C941;
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757
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41@1C
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1AC9>G1
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B1=1
41>
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3181G1
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B5<9B98
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757
141<18    B9>     

45>71>






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757
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   B9>    45>71>






$1A1;
41A9
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Peta Konsep (.1/&%0)
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#>C5A65A5>B9

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96A1;B9

C5A49A9
1C1B

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=5=2181B

#>C5A65A5>B9 '9>9=D=

#>C5A65A5>B9 '1;B9=D=

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@141
*5=1>CD<1> 41>
*5=291B1>

*5=DC1A1> 941>7 *?<1A9B1B9

96A1;B9 5<18
,D>771<

96A1;B9 @141
%9B9

*?<1A9B1B9
;1A5>1 "1=2DA1>

Refleksi Setelah mempelajari bab ini, tentu Anda menjadi tahu bahwa cahaya merupakan gelombang elektromagnetik yang dapat mengalami proses interferensi, difraksi, dan polarisasi. Dari semua materi pada bab ini,

46

Mudah dan Aktif Belajar Fisika untuk Kelas XII

bagian mana yang menurut Anda sulit dipahami? Coba Anda diskusikan bersama teman atau guru Fisika Anda. Selain itu, coba Anda sebutkan manfaat yang Anda peroleh setelah mempelajari materi bab ini.

Tes Kompetensi Bab 2  +.+*.%*
4%.%*
4%56
,%7%&%0
8%0)
2%.+0)
5(2%5
'%0
-(3,%-%0.%*
2%'%
&6-6
.%5+*%0











+D4DC
@9C1
C5A1>7
C5>718
41A9
496A1;B9
?A45
;54D1
G1>7 4981B9<;1>
?<58
;9B9
45>71>
 
35<18 3=
B525B1A
J '1;1
@1>:1>7
757
3181G1
G1>7
49@5A7D>1;1> 141<18
1  
I 2  
 I 3  
I 4  
 I 5  
I $9;1
B9>1A
G1>7
:1CD8
C571;
=9>G1;
49
1C1B
19A
C521<
<1@9B1>
=9>G1;

9>45;B
291B>G1 
41>
@1>:1>7
757
41A9
3181G1
G1>7
=5>7 81B9<;1>
9>C5A65A5>B9
=1;B9=D=

=1;1
@5AB1=11> G1>7
=5=5>D89
141<18
1


  2


 

  3 

  4 

 

  5 

 ->CD;
=5>5>CD;1>
@1>:1>7
757
=?>?;A?=1C9B 497D>1;1>
@5A3?211>
0?D>7
G1>7
41C1>G1
B521719 25A9;DC $1A1;
1>C1A1
;54D1
35<18>G1


==
:1A1;
35<18
;5 <1G1A


3=
B5AC1
:1A1;
1>C1A1
71A9B
75<1@
;5
41> 71A9B
75<1@
;5
@141
<1G1A


== *1>:1>7
757
B9>1A
=?>?;A?=1C9B
C5AB52DC 141<18
1 
>= 4 
>= 2 
>= 5 
>= 3 
>= *141
@5A3?211>
0?D>7
4D1
35<18
25A:1A1;

==
49 <5C1;;1>
@141
:1A1;

=
41A9
B52D18
<1G1A
$9;1
:1A1; C5A45;1C
1>C1A1
@?<1
9>C5A65A5>B9
71A9B
C5A1>7
@5AC1=1 41>
71A9B
C5A1>7
;5B525<1B
141<18

==
@1>:1>7 757
3181G1
G1>7
=5>G9>1A9
141<18
1 
I 4  
I 2 
I 5  
I 3  
I *A9>B9@
41B1A
41A9
4D1
BD=25A
3181G1
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141<18 1 ;54D1>G1
B1>71C
25A45;1C1> 2 1=@<9CD4?>G1
B1=1 3 B9=@1>71>>G1
B5<1G1
141<18
C5C1@ 5 ;54D1>G1
=5=1>31A;1>
3181G1
G1>7
25A@1@1B1> 9>39>
(5FC?>
C5A:149
;1A5>1
75:1<1
1 496A1;B9 2 @?<1A9B1B9 3 49B@5AB9 4 9>C5A65A5>B9 5 A56A1;B9

 *141
@5A3?211>
0?D>7
35<18
71>41
:9;1
:1A1;
1>C1A1 4D1
35<18>G1
49:149;1>
4D1
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B5=D<1
=1;1
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4D1
71A9B
75<1@
G1>7
25ADADC1>
1 
;1<9
B5=D<1 2
;1<9
B5=D<1 3 M
;1<9
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;1<9
B5=D<1 5 C5C1@
C941;
25AD218  @129<1
9>C5>B9C1B
3181G1
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=5<5F1C9 4D1
@<1CA5C1A41B9
141<18
 
1>C1A1
;54D1
@<1C =5=25>CD;
BD4DC
J
9>C5>B9C1B
3181G1
G1>7 49<5F1C;1>
?<58
;54D1
@<1C
141<18
1





4

  

  5     3  +52D18
@?<1A?94
=5=9<9;9
@?<1A9B1C?A
41>
1>1<9B1C?A
G1>7 49@1B1>7
=5=25>CD;
BD4DC
J
9>C5>B9C1B
3181G1
G1>7 49C5ADB;1>
1;1>
B521>49>7
45>71>
1 4 

  2 5 

3  D1
35<18
B5=@9C
G1>7
C5A@9B18
@141
:1A1;

== 49B9>1A9
C571;
7
;5C971
C5A<5C1;

== 41A9
71A9B
C5A1>7
;5
>?<
@141
<1G1A
G1>7
:1A1;>G1

= 41A9
35<18
*1>:1>7
757
B9>1A
G1>7
49@1;19 141<18
1 
L
K
== 2 
L
K
== 3 
L
K
== 4 
L
K
== 5 
L
K
== +525A;1B
B9>1A
=?>?;A?=1C9B
45>71>
@1>:1>7 757

L
K
=
41C1>7
C571;
71> 71A9B
>?A=1<
@141
;9B9
=1;1
:D=<18
71A9B
@5A
3=
;9B9 141<18
4
L
 1
L
 2 
L
 5 
L
 3 
L
 1
=5>781B9<;1>
B9>1A
@1>CD<
C5A@?<1A9B1B9 <9>51A
:9;1
B9>1A
@1>CD<
41>
B9>1A
291B
=5=25>CD; BD4DC
1 J 4 J 2 J 5 J 3 J 2









Gelombang Cahaya

47

 +525A;1B
3181G1
=?>?;A?=1C9B
49:1CD8;1>
@141 4D1
35<18
B5=@9C
E5AC9;1<
25A45;1C1>
45>71>
:1A1; 


==
*?<1
9>C5A65A5>B9
G1>7
C5A:149
49C1>7;1@ @141
:1A1;

3=
41A9
35<18
9;5C18D9
218F1
:1A1; 1>C1A1
71A9B
75<1@
@5AC1=1
49
B525<18
;9A9
;5
71A9B
75<1@ @5AC1=1
49
B525<18
;1>1>
141<18

==
*1>:1>7 757
25A;1B
3181G1
141<18
1 
>= 2 
>= 3 
>= 4  
>= 5 
>= !

()+10%.
  +D1CD
3181G1
=5>5A1>79
35<18
71>41
G1>7
=5=9<9;9 :1A1;
1>C1A35<18
 
3=
B545=9;91>
89>771
C5A25>CD; @?<1
75<1@C5A1>7
@141
<1G1A
G1>7
25A:1A1;

3=
%5C9;1



%7%&.%*
2(35%08%%0
&(3+-65
+0+
'(0)%0
5(2%5



D1
35<18
45>71>
:1A1;

==
49B9>1A9
C571;
7
;5C971
C5A<5C1;

==
41A9
71A9B
C5A1>7
;5 >?<
@141
<1G1A
G1>7
25A:1A1;

=
41A9
35<18
5A1@1 @1>:1>7
757
B9>1A
G1>7
49@1;19 181G1
BD1CD
BD=25A
=5<17 C5A@9B18

==
$9;1
:1A1;
1>C1A1
4D1
35<18
B5=@9C C5A8141@
<1G1A

3=
41>
:1A1;
1>C1A1
71A9B
75<1@ @5AC1=1
41>
71A9B
C5A1>7
@5AC1=1
141<18

== 25A1@1;18
@1>:1>7
757
3181G1
G1>7 497D>1;1>



48

Mudah dan Aktif Belajar Fisika untuk Kelas XII

@5=9B181>
1>C1A@?<1
C5A1>7
141<18

3=
=1;1 @1>:1>7
757
3181G1
G1>7
497D>1;1>
C5AB52DC 141<18
1 
>= 4 
>= 2 
>= 5 
>= 3 
>= #"

%810
  .1A>1
D>7D
41A9
B@5;CAD=
?A45
;5C971
25A9=@9C
45>71> F1A>1
=5A18
?A45
;54D1
41A9
BD1CD
@5A9BC9F1
496A1;B9 G1>7
=5=@5A7D>1;1>
;9B9
9>9
25A1AC9
@5A21>49>71> 1>C1A1
B9>1A
@1>:1>7
757
B9>1A
D>7D
41>
B9>1A =5A18
141<18
1 

2

 3 

 4 

 5 B5=D1
:1F121>
B1<18

 

$9;1
@5A3?211>
35<18
71>41
0?D>7
4935
;5
41<1= 19A
21719=1>1;18
@5AD2181>
@141
9>C5A65A5>B9
G1>7 C5A:149 +52D18
B1381A9=5C5A
=5=9<9;9
C12D>7
G1>7
499B9
<1ADC1> 7D<1
@1>:1>7>G1

3=
%?>B5>CA1B9
7D<1
G1>7 497D>1;1>

41>
BD4DC
@DC1A
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<1ADC1>
141<18 


@5A

3=
$9;1
497D>1;1>
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@5=DC1A1>
2941>7
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Bab

3 Sumber: Fundamental of Physics, 2001

Kelelawar ladam kuda (horseshoe bat) dapat memancarkan gelombang bunyi ultrasonik yang berfungsi sebagai pengindra ketika terbang pada malam hari.

Gelombang Bunyi Hasil yang harus Anda capai: menerapkan konsep dan prinsip gejala gelombang dalam menyelesaikan masalah. Setelah mempelajari bab ini, Anda harus mampu: • •

mendeskripsikan gejala dan ciri-ciri gelombang bunyi; menerapkan konsep dan prinsip gelombang bunyi dalam teknologi.

2A2D<29
?52
>6>32H2?8<2?
5E?:2
D2?A2
252?H2
3E?H:
E?:2 2<2?
D6B2C2
CE?H:
D2?A2
252?H2
3E?H:
6B282>
;6?:C
3E?H:
>6?;25:<2? 5E?:2
:?:
=63:9
O9:5EAP
?52
52A2D
>6?56?82B<2?
>EC:<
<2B6?2
252?H2 A6B2>32D2?
86=@>32?8
3E?H:
H2?8
C2>A2:
<6
D6=:?82
?52
+29E<29 ?52
C:72DC:72D
52?
<=2C:7:<2C:
86=@>32?8
3E?H:
,=DB2C@?:<
>6BEA2<2? C2=29
C2DE
;6?:C
3E?H:
H2?8
52A2D
5:A2?42B<2?
@=69
<6=6=2G2B
6=@>32?8 3E?H:
D6BC63ED
36BA6B2?
C63282:
OA6?8:?5B2P
<6D:<2
<6=6=2G2B
D6B32?8 A252
>2=2>
92B: *6=2:?
E=DB2C@?:<
D6B52A2D
;6?:C
86=@>32?8
3E?H:
=2:??H2
,?DE< >6?86D29E:
=63:9
=2?;ED
D6?D2?8
;6?:C;6?:C
86=@>32?8
3E?H:
C:72DC:72D 86=@>32?8
3E?H:
52?
76?@>6?276?@>6?2
=2:??H2
H2?8
36B<2:D2?
56?82? 86=@>32?8
3E?H:
C6BD2
A6?6B2A2??H2
52=2>
D62D6B: A252
323
:?:
C642B2
32:<

A. Sifat Dasar Gelombang Bunyi B. Cepat Rambat Bunyi C. Unsur Bunyi dan Pemanfaatan Gelombang Bunyi D. Sifat-Sifat Gelombang Bunyi E. Dawai dan Pipa Organa sebagai Sumber Bunyi F. Energi Gelombang Bunyi

Gelombang Bunyi

49

Tes Kompetensi Awal '$'.6/
/'/2'.#,#3+
-104'2
'.1/$#0)
608+
-'3,#-#0.#*
41#.41#.
$'3+-65
&#.#/
$6-6
.#5+*#0  6B52C2B<2?
CE>36B?H2
3E?H:
D6B>2CE<
;6?:C  A2
H2?8
5:>22
52?
82E?8 86=@>32?8
2A2  A2<29
86=@>32?8
H2?8
5:A2?42B<2?
A6>2?42B
B25:@  A2<29
86=@>32?8
3E?H:
52A2D
>6?82=2>:
A@=2B:C2C: D6B>2CE<
86=@>32?8
=@?8:DE5:?2=

 A2
H2?8
5:>2
A. Sifat Dasar Gelombang Bunyi

Gambar 3.1 Gelombang bunyi dari lonceng yang bergetar merambat melalui media udara sampai ke telinga.

#$'.
 6A2D
)2>32D
E?H:
636B2A2
292? A252
*E9E

L #/#
#*#0 '&+6/

'2#5
#/$#5 608+
/4

:B :B
=2ED =E>:?:E> 6C:
52?
2;2 6=2C 6=:E> :5B@86? #2HE
<6B2C ,52B2













 




 Sumber: Physics, 1995

A2<29
?52
>6?86D29E:
AB@C6C
D6B;25:?H2
3E?H:
(252
52C2B?H2 C6CE2DE
H2?8
36B86D2B
252=29
CE>36B
3E?H:
E?H:
C66<@B
?H2>E< D6B56?82B
<2B6?2
<6A2<2?
3E=E3E=E
<65E2
C2H2A?H2
H2?8
>6>:=:<: 7B6<E6?C:
C6<:D2B



I
E?H:
C66<@B
;2?86?6BEC 6=@>32?8
:?7B2C@?:<
5:A2?42B<2?
@=69
C66<@B
A2EC
56?82?
7B6<E6?C: C2?82D
B6?529
5:
32G29

I
D6D2A:
>2>AE
5:D2?8<2A
@=69
A2EC
H2?8 =2:?
G2=2EAE?
;2?8<2E2??H2
B2DEC2?
<:=@>6D6B 6B3652
56?82?
CE>36BCE>36B
3E?H:
D6BC63ED
3E?H:
A252
2=2D2=2D >EC:<
H2?8
52A2D
<:D2
56?82B
D6B>2CE<
86=@>32?8
2E5:@C@?:<
E?H: H2?8
<6=E2B
52B:
2=2D
>EC:<
36BC:72D
D6B2DEB
C69:?882
5:?2>2<2?
56?82? ?252
52AE?
3E?H:
H2?8
D:52<
D6B2DEB
5:?2>2<2?
56C29 (6B92D:<2?
#/$#3
 
(252
82>32B
D6BC63ED
D6B=:92D
86D2B2? =@?46?8
>6?892C:=<2?
86=@>32?8
3E?H:
H2?8
>6B2>32D
>6=2=E:
>65:2 E52B2
>6?E;E
D6=:?82
H2?8
36BEA2
86=@>32?8
=@?8:DE5:?2= (252
52C2B?H2
86=@>32?8
3E?H:
52A2D
>6B2>32D
5:
>65:E>
2A2AE? (6B92D:<2?
#$'.
 
<6>2>AE2?
>65:E>
E?DE<
>6?886D2B<2?
A2BD:<6= 36B36523652
329<2?
252
>65:E>
H2?8
52A2D
>6B652>
3E?H:
>:C2=?H2
2:B 6=@>32?8
3E?H:
52A2D
>6?E?;E<<2?
C:72DC:72D
E>E>
86=@>32?8 C6A6BD:
52A2D
>6?82=2>:
:?D6B76B6?C:
5:7B26>A6=2;2B:
A6B:CD:G2
CEA6BA@C:C:
C6A6BD:
CEA6BA@C:C:
5E2 86=@>32?8
3E?H:
H2?8
<64:=
A6B36522?
7B6<E6?C:?H2
C69:?882
>6?8 92C:=<2?
3E?H:
<E2D=6>29
C642B2
A6B:@5:<
H2?8
5:C63ED
A6=2H2?82?
3E?H: 6B52C2B<2?
2B29
B2>32D?H2
D6B>2CE<
86=@>32?8
2A2<29
3E?H:
:DE 82B
?52
>6>292>:?H2
A6B92D:<2?
#/$#3
  tekanan udara

Gambar 3.2

rapatan

Getaran dari membran bas drum membuat partikel udara merapat dan merenggang.

x renggangan

#/$#3

>6?E?;E<<2?
CE>36B
3E?H:
36BEA2
86D2B2?
>6>3B2? 32C
5BE>
6D2B2?
C6=2AED
H2?8
>6BEA2<2?
56?HED
36B5:>6?C:
5E2 5:A2?DE=<2?
C642B2
36BE=2?8E=2?8
@=69
32D2C
C6=2AED
C69:?882
C6=2AED
2<2? 36B86D2B
C642B2
A6B:@5:<
#6D:<2
C6=2AED
36B86B2<
<6=E2B
2<2?
D6B;25: A6>2>A2D2?
E52B2
5:
56A2??H2
(6>2>A2D2?
D6BC63ED
>6?H6323<2? D6<2?2?
E52B2
36BD2>329
H2?8
5:C63ED
526B29
B2A2D2?
)2A2D2?
36B86B2< >6?;2E9:
C6=2AED
52=2>
32D2C
<646A2D2?
3E?H:
)2A2D2?
52?
B6?882?82? H2?8
D6B36?DE<
52B:
A2BD:<6=A2BD:<6=
E52B2
>:B:A
56?82?
B2A2D2?
52? B6?882?82?
86=@>32?8
=@?8:DE5:?2=
A252
C=:?<:
"25:
86=@>32?8
=@?8:DE 5:?2=
>6BEA2<2?
C:72D
52C2B
86=@>32?8
3E?H:

50

Mudah dan Aktif Belajar Fisika untuk Kelas XII

E?H:
52A2D
D6B56?82B
;:<2
86D2B2?
52B:
CE>36B?H2
52A2D
>6B2>32D 9:?882
<6
D6=:?82
%65:2
B2>32D2??H2
52A2D
36BEA2
I2D
A252D
I2D
42:B 52?
I2D
82C
"25:
5:A6B=E<2?
>65:E>
E?DE<
>6?;2=2B<2?
3E?H:
%2C:9 :?82D<29
?52
A6B:CD:G2
D6?D2?8
3E?H:
H2?8
52A2D
>6B2>32D
>6=2=E:
C6ED2C 36?2?8
32C29
H2?8
<65E2
E;E?8?H2
D6B9E3E?8
<2=6?8
CECE
56?82?
32?DE2? CE2B2
D6>2?
?52
&29
>6?EBED
?52
52A2D<29
3E?H:
>6B2>32D
5:
52=2> F2<E>
,?DE<
>6?86D29E:
;2G232??H2
=2<E<2?=29
2
Ingatlah Gelombang longitudinal adalah gelombang yang arah getarnya searah dengan arah rambatannya

Aktivitas Fisika 3.1 Bunyi dalam Ruang Vakum Tujuan Percobaan Menyelidiki apakah bunyi dapat merambat melalui ruang hampa. Alat-Alat Percobaan 1. Jam weker 2. Tabung hampa udara (vakum) Langkah-Langkah Percobaan 1. Bunyikan jam weker tanpa ditutup oleh tabung vakum. 2. Tutup jam weker dengan tabung vakum. Amati yang terdengar dari jam weker tersebut. 3. Gunakan sebuah pengisap untuk mengisap udara dalam tabung vakum secara perlahan-lahan. 4. Amati perubahan intensitas bunyi jam weker tersebut.

Tugas Anda 3.1 Apakah bunyi termasuk gelombang elektromagnetik? Sebutkan alasanalasan yang mendasari jawaban Anda tersebut.

Kata Kunci

A2
H2?8
52A2D
?52
C:>AE=<2?
52B:
2
D23E?8
5::C2A
=2>2<6=2>22?
3E?H:
;2>
G6<6B D6BC63ED
D:52<
D6B56?82B
2=
D6BC63ED
>6>3E32?8 3E?H:
>6>6B=E<2?
>65:E>
E?DE<
>6B2>32D

Tes Kompetensi Subbab

• • • • • • • •

gelombang bunyi gelombang longitudinal gelombang infrasonik gelombang audiosonik nada desah pelayangan bunyi ruang vakum

A

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.#5+*#0  



A2<29
H2?8
5:>236B
3E?H: *63ED<2?
C:72DC:72D
H2?8
5:>:=:<:
86=@>32?8
3E?H: A2<29
3E?H:
52A2D
>6B2>32D
5:
BE2?8
92>A2
E52B2 %6?82A2

 

6B52C2B<2?
2B29
86D2B?H2
D6B>2CE<
86=@>32?8 2A2<29
3E?H:
:DE A2<29
A6B36522?
2?D2B2
?252
52?
56C29

B. Cepat Rambat Bunyi 6A2D
B2>32D
3E?H:
D:52<
36B82?DE?8
A252
D6<2?2?
E52B2
"25:
;:<2 D6B;25:
A6BE3292?
D6<2?2?
E52B2
46A2D
B2>32D
3E?H:
D:52<
36BE329
6A2D B2>32D
3E?H:
36B82?DE?8
A252
CE9E
*6>2<:?
D:?88:
CE9E
E52B2
C6>2<:? 36C2B
46A2D
A6B2>32D2?
3E?H:
:
526B29
A68E?E?82?
46A2D
B2>32D
3E?H: =63:9
=2>32D
52B:A252
46A2D
B2>32D
3E?H:
5:
<2G2C2?
A2?D2:
:
E52B2 46A2D
B2>32D
3E?H:
36B82?DE?8
A252
;6?:C
A2BD:<6=
H2?8
>6>36?DE<
E52B2 D6BC63ED
(6BC2>22??H2
52A2D
5:DE=:C<2?
C63282:
36B:<ED $



 

 N

Gelombang Bunyi

51

Ingatlah Cepat rambat bunyi di udara bergantung pada suhu udara.

#6D6B2?82?  
<@?CD2?D2
$2A=246  
D6D2A2?
E>E>
82C
 
" >@=
#  
CE9E
>ED=2<
82C
<6=F:?


>2CC2
>@=6<E=
82C
8B2> >@= 6A2D
B2>32D
3E?H:
52=2>
I2D
A252D
5:D6?DE<2?
@=69
>@5E=EC
.@E?8
52? >2CC2
;6?:C
I2D
D6BC63ED
(6BC2>22??H2
52A2D
5:DE=:C<2?
C63282:
36B:<ED

$



Tugas Anda 3.2 Bandingkan nilai cepat rambat bunyi di udara yang Anda dapatkan dari percobaan dan dari buku referensi. Jika ada perbedaan, berikan analisis Anda mengenai hal tersebut.

 N 

#6D6B2?82?


>@5E=EC
.@E?8
I2D
A252D
& >   
>2CC2
;6?:C
I2D
A252D
<8 >  :
52=2>
I2D
42:B
46A2D
B2>32D
3E?H:
5:D6?DE<2?
@=69
>@5E=EC
E=< 52?
<6B2A2D2?
>2CC2
;6?:C
42:B2?
D6BC63ED
(6BC2>22??H2
52A2D 5:DE=:C<2?
C63282:
36B:<ED $



 N 

#6D6B2?82? 
>@5E=EC
E=<
& >   
>2CC2
;6?:C
I2D
42:B
<8 > 

Contoh 3.1

Pembahasan Soal Iskandar mendengarkan kembali suaranya sebagai gema dari sebuah tebing setelah waktu 4 detik. Jika  adalah perbandingan panas jenis udara pada tekanan dan suhu konstan, serta Iskandar mengetahui bahwa suhu saat itu T kelvin dan massa molekul relatif udara M, Iskandar dapat menentukan jarak tebing menurut persamaan ....

 RT M  RT b. 2 M

a.

c. 4

d. 6

 RT M

e. 8

 RT M

UMPTN 1996

d = vt v =

2

 RT M

 RT t M = 2

 RT 4 M 2

= 2  RT M Jawaban: b

52

$  

     
& > 




<8 > 


 
> C

$  
     
& > 



<8 > 

> 4

#646A2D2?
3E?H:
5:
2:B
D6B?H2D2
=63:9
<64:=
52B:A252
<646A2D2?
3E?H:
52=2> 2=E>:?:E>
2=
D6BC63ED
5:C6323<2?
42:B2?
=63:9
<@>AB6C:36=
=63:9
>E529 36BE329
36?DE<
5:32?5:?8<2?
56?82?
I2D
A252D
>@5E=EC
E=<
42:B2?
=63:9 <64:=
5:32?5:?8<2?
56?82?
>@5E=EC
.@E?8
E?DE<
I2D
A252D

 RT M

Pembahasan:

d=

2?5:?8<2?
<646A2D2?
3E?H:
52=2>
2=E>:?:E>
52?
2:B
;:<2
>@5E=EC
.@E?8
E?DE< 2=E>:?:E>


Q
 
& > 
>@5E=EC
E=<
2:B


Q
 
& > 
52?
<646A2D2? 2:B




<8 >  #7#$ :<6D29E:


Q
 
& >   





<8 > 


Q
 
& >   




<8 > 

Contoh 3.2 6A2D
B2>32D
86=@>32?8
3E?H:
5:
E52B2
A252
CE9E
L
252=29

> C
":<2
CE9E E52B2
?2:<
>6?;25:
L
D6?DE<2?
46A2D
B2>32D
86=@>32?8
3E?H:
D6BC63ED #7#$ #6D:<2
CE9E
E52B2
?2:<
<646A2D2?
3E?H:
2<2?
C6>2<:?
36BD2>329
2=
D6BC63ED 5:C6323<2?
86B2<2?86B2<2?
>@=6<E=
E52B2
H2?8
>6B2>32D<2?
3E?H:
36BD2>329
46A2D :<6D29E:   





#  





# $  

> C

Mudah dan Aktif Belajar Fisika untuk Kelas XII

$      $



Kata Kunci • • • •



  $ 
 $






> C  

cepat rambat bunyi konstanta Laplace modulus Young modulus Bulk

"25:
46A2D
B2>32D
3E?H:
36BD2>329
>6?;25:

> C

Tes Kompetensi Subbab

B

'3,#-#0.#*
&#.#/
$6-6
.#5+*#0 





+6?DE<2?
<646A2D2?
3E?H:
52=2>
82C
<2B3@?
5:@
L

;:<2
36B2D
>@=6<E=
'

52?  
  6A2D
B2>32D
3E?H:
A252
CE9E
L
C2>2
56?82?

> C +6?DE<2?
46A2D
B2>32D
3E?H:
A252
CE9E


L
 6A2D
B2>32D
3E?H:
52=2>
82C
'
252=29



> C    
6B2A2<29
46A2D
B2>32D?H2
52=2>
82C
'   

A252
CE9E
H2?8
C2>2

 

:DE?8
46A2D
B2>32D
3E?H:
A252
C632D2?8
36C: 


Q
 
& >
52?



8B2> 4>  6B2A2
<2=:
=:A2D<29
<646A2D2?
3E?H:
A252
32;2
5: 32?5:?8<2?
56?82?
<646A2D2?
3E?H:
A252
2=E>:?:E> ;:<2
>@5E=EC
.@E?8
E?DE<
32;2
52?
2=E>:?:E>
>2C:?8 >2C:?8

Q
 
& >
52?

Q
 
& > 
C6BD2
>2CC2 ;6?:C
32;2
52?
2=E>:?:E>
>2C:?8>2C:?8

8B2> 4> 52?

8B2> 4> 

C. Unsur Bunyi dan Pemanfaatan Gelombang Bunyi ,?CEBE?CEB
3E?H:
2?D2B2
=2:?
C63282:
36B:<ED

1. Tinggi Nada Bunyi (252
A@<@<
3292C2?
C636=E>?H2
?52
D6=29
>6?86D29E:
329G2
3E?H: 5:D:>3E=<2?
@=69
CE2DE
36?52
H2?8
36B86D2B
*6>2<:?
32?H2<
;E>=29 86D2B2?
H2?8
5:92C:=<2?
52=2>
C2DE
C6=2?8
G22<:?
?H2B:?8 (252
82>32B
#/$#3

C6@B2?8
2?2<
D6B=:92D
C652?8
>6?6>A6=<2? 32D2?8
=:5:
5:
;2B:;2B:
B@52
C6A652
H2?8
C652?8
36BAED2B
":<2
B@52
D6BC63ED 5:AED2B
C6>2<:?
46A2D
2A2<29
3E?H:
H2?8
5:92C:=<2?
2<2?
C6>2<:?
?H2B:?8 +6B?H2D2
A252
AED2B2?
H2?8
=2>32D
3E?H:
H2?8
D6B56?82B
36B?252 B6?529
(252
C22D
AED2B2?
B@52
C6A652
5:A6B46A2D
3E?H:
H2?8
D6B56?82B 36B?252
D:?88:
2=
D6BC63ED
>6>3E36B
3E?H:

Gambar 3.3 Lidi dipegang oleh seorang anak dan ditempelkan pada jeruji roda yang berputar.

2. Kuat Bunyi #E2D
3E?H:
H2?8
5:92C:=<2?
36B82?DE?8
A252
2>A=:DE5@
86D2B2? (6B92D:<2?
#/$#3

H2?8
>6?6B2?8<2?
A6B4@322?
A252
C6?2B
8:D2B (6D:<=29
C2=29
C2DE
C6?2B
8:D2B
56?82?
C:>A2?82?
H2?8
D:52<
D6B=2=E 36C2B
6?82B<2?
32:<32:<
<E2D
3E?H:
H2?8
5:92C:=<2?
=2=E
32?5:?8<2? 56?82?
<E2D
3E?H:
H2?8
5:92C:=<2?
@=69
C6?2B
H2?8
C2>2
?2>E?
56?82? C:>A2?82?
H2?8
=63:9
36C2B
52B:A252
C6>E=2
(6D:<2?
>2?2<29
H2?8 >6?892C:=<2?
3E?H:
=63:9
<E2D

2

3. Warna Bunyi

3

-2B?2
3E?H:
>6BEA2<2?
3E?H:
<92C
H2?8
5:D:>3E=<2?
@=69
CE2DE CE>36B
3E?H:
E?H:
8:D2B
36B3652
56?82?
3E?H:
3:@=2
:DE
<2B6?2
252 G2B?2
3E?H:
(6B36522?
D6BC63ED
D6B;25:
<2B6?2
823E?82?
?252
2D2C
52? ?252
52C2B
52B:
CE>36B
3E?H:
36B36523652
G2=2EAE?
7B6<E6?C:?H2
C2>2

Gambar 3.4 Percobaan kuat bunyi yang dihasilkan oleh getaran senar gitar dengan (a) simpangan yang kecil dan (b) simpangan yang besar.

Gelombang Bunyi

53

4. Batas Pendengaran ?52
D6?DE
D6=29
>6?86D29E:
329G2
>2?EC:2
52?
96G2?
5:=6?8<2A: 56?82?
2=2D
A6?56?82B2?
H2?8
C6>AEB?2
&2>E?
<6>2>AE2? A6?56?82B2??H2
36B36523652
2D2C
<6>2>AE2?
A6?56?82B2?
5:E<EB 36B52C2B<2?
7B6<E6?C:
3E?H: %2?EC:2
?@B>2=
>6>:=:<:
32D2C
A6?56?82B2?
2?D2B2

96BDI
C2>A2: 56?82?



96BDI
26B29
7B6<E6?C:
D6BC63ED
5:?2>2<2?

 #!
# 
C652?8<2?
526B29
7B6<E6?C:
5:
32G29

96BDI
5:C63ED
 ! 52?
526B29
5:
2D2C
7B6<E6?C:
56?82B
2D2E
5:
2D2C



96BDI
5:C63ED #" ! 636B2A2
96G2?
>2>AE
>6?56?82B
3E?H:
E=DB2C@?:<
329<2?
96G2? C6A6BD:
<6=6=2G2B
H2?8
>6>:=:<:
2=2D
A6?8=:92D2?
D:52<
C632:<
2=2D A6?56?82B2??H2
>6?88E?2<2?
3E?H:
E=DB2C@?:<
E?DE<
>6?86D29E: 36?52
H2?8
252
5:
56A2??H2

5. Pemanfaatan Gelombang Bunyi penerima

pemancar

gelombang ultrasonik

Gambar 3.5 Sonar dapat digunakan untuk mendeteksi kedalaman laut.

6=@>32?8
3E?H:
32?H2<
5:>2?722D<2?
52=2>
<69:5EA2?
C692B:92B: >:C2=?H2
E?DE<
>6?6?DE<2?
<652=2>2?
=2ED
>6?42B:
CE>36B
>:?H2<
52? >6?56D6
,?DE<
>6?6?DE<2?
<652=2>2?
=2ED
2=2D A6?56D66>2?42B<2?
86=@>32?8
<6>E5:2?
A2?DE=2?
86=@>32?8 D6BC63ED
2<2?
5:D6B:>2
@=69
$ 
6?82?
>6?86D29E:
<646A2D2? 86=@>32?8
H2?8
5:E<EB
52?
G2AE9
86=@>32?8
D6BC63ED
#652=2>2? =2ED
52A2D
5:D6?DE<2?
56?82?
>6?88E?2<2?
A6BC2>22?
36B:<ED

  !  $ "    4 1

2

3

5

Sumber: www.mediajakartaselatan.com

Gambar 3.6 Teknologi 3D USG terbaru sekarang dapat melihat organ-organ tubuh bayi dalam kandungan.

 N

26?H2D2<2?
C6D6?829 G2AE9
52B:
C@?2B
<6
D6>A2D
A6>2?DE=2? B6<E6?C:
H2?8
5:8E?2<2?
@=69
C@?2B
36B252
A252
526B29
E=DB2C@?:< H2:DE
5:
2D2C



96BDI
B6<E6?C:
D6BC63ED
D:52<
52A2D
5:56?82B
52? A2?;2?8
86=@>32?8
A252
526B29
E=DB2C@?:<
C2?82D
<64:=
C69:?882
5:7B22<:?
<64:=
52?
86=@>32?8
D:52<
2<2?
>6?H632B #64:=?H2
A2?;2?8
86=@>32?8
H2?8
5:8E?2<2?
52A2D
;E82
5:8E?2<2? E?DE<
>6?56D632?8 E=DB2C@?:<
D6B52A2D
;E82
A252
2=2D
#" ! 
,*
=2D
D6BC63ED E>E>?H2
5:8E?2<2?
E?DE<
>6?82?2=:C:C
<692>:=2?
2D2E
>6?56D6
DE3E9

Contoh 3.3

Kata Kunci • • • • • • • •

tinggi nada kuat bunyi audiosonik infrasonik ultrasonik resonansi sonar ultrasonografi

54

*63E29
2=2D
C@?2B
5:8E?2<2?
E?DE<
>6?8E<EB
<652=2>2?
=2ED
*6=2?8
G232?8
A6B2>32D2?
C2>A2:
<6>32=:
=28:
<6
C@?2B
252=29
 C6<@?
":<2
46A2D
B2>32D
86=@>32?8
5:
52=2>
2:B
=2ED
252=29



> C
D6?DE<2?=29 <652=2>2?
=2ED
D6BC63ED #7#$ :<6D29E:
$




> C
"


C

  !

$
 "




> C
Q

Q

C


>6D6B "25:
=2ED
D6BC63ED
>6>:=:<:
<652=2>2?

>6D6B

Mudah dan Aktif Belajar Fisika untuk Kelas XII

Tes Kompetensi Subbab

C

'3,#-#0.#*
&#.#/
$6-6
.#5+*#0  



?52
>6?56?82B
92=:=:?D2B

C6<@?
C6D6=29
<:=2D D6B=:92D
":<2
46A2D
B2>32D
3E?H:
5:
E52B2


> C D6?DE<2?
;2B2<
2C2=
<:=2D
D6BC63ED
52B:
D6>A2D
?52 *63E29
2=2D
C@?2B
5:8E?2<2?
E?DE<
>6?8E<EB <652=2>2?
=2ED
*6=2?8
G232?8
>6B2>32D
C2>A2:
<6>32=:
<6
C@?2B 252=29

C6<@?
":<2
<652=2>2?
=2ED



>
D6?DE<2? 46A2D
B2>32D
3E?H:
5:
52=2>
2:B
=2ED %6?82A2
<6=6=2G2B
H2?8
D6B32?8
A252
>2=2>
92B:
D:52< A6B?29
>6?23B2<
36?5236?52
H2?8
252
5:
56<2D?H2





":<2
D:?88:
?252
5:?2:<<2?
3282:>2?2<29
A6BE3292? H2?8
D6B;25:
A252 2 7B6<E6?C: 3 A2?;2?8
86=@>32?8 4 46A2D
B2>32D
86=@>32?8 5 2>A=:DE5@
86=@>32?8 ":<2
<E2D
3E?H:
5:?2:<<2?
3282:>2?2<29
A6BE3292? H2?8
D6B;25:
A252 2 7B6<E6?C: 3 A2?;2?8
86=@>32?8 4 46A2D
B2>32D
86=@>32?8 5 2>A=:DE5@
86=@>32?8

D. Sifat-Sifat Gelombang Bunyi ?52
D6=29
>6>A6=2;2B:
C:72DC:72D
E>E>
D6?D2?8
86=@>32?8
H2:DE A6>3:2C2?
B67B22?DE=2?
B67=66BEA2<2?
C2=29
C2DE
36?DE<
86=@>32?8 '=69
<2B6?2
:DE
86=@>32?8
3E?H:
;E82
>6?82=2>:
A6B:CD:G2A6B:CD:G2 D6BC63ED

1. Pemantulan Gelombang Bunyi %6?82A2
C22D
?52
36BD6B:2<
5:
C6<:D2B
D63:?8
C6=2=E
252
3E?H:
H2?8 >6?:BE<2?
CE2B2
?52
D6BC63ED
%6?82A2
CE2B2
?52
D6B56?82B
=63:9 <6B2C
<6D:<2
36B252
5:
52=2>
865E?8
#65E2
A6B:CD:G2
D6BC63ED >6?E?;E<<2?
329G2
3E?H:
52A2D
5:A2?DE=<2?
,?DE<
>6>292>:
D6?D2?8 A6>2?DE=2?
3E?H:
=2<E<2?=29
2
Aktivitas Fisika 3.2 Pemantulan Bunyi Tujuan percobaan Membuktikan peristiwa pemantulan bunyi. Alat-Alat Percobaan 1. Weker 4. Dua buah pipa berdiameter 1 inci 2. Papan pemantul 5. Busur derajat 3. Meja 6. Kertas karton putih Langkah-Langkah Percobaan 1. Susunlah alat seperti tampak pada gambar. 2. Letakkan jam weker di salah satu ujung pipa paralon. 3. Atur pipa paralon yang lain agar Anda dapat mendengar bunyi jam weker yang paling jelas. 4. Gambarlah susunan pipa tersebut pada kertas karton putih yang Anda letakkan di bawahnya. Ukurlah besar sudut datang dan sudut pantulnya. 5. Ulangi langkah percobaan ke-3 dan ke-4 dengan sudut yang berbeda.

Tugas Anda 3.3 Gelombang bunyi merupakan gelombang longitudinal sehingga tidak dapat mengalami dispersi dan polarisasi. Diskusikanlah bersama teman Anda, mengapa hal tersebut bisa terjadi? Berikanlah penjelasan ilmiah mengenai hal tersebut.

Gelombang Bunyi

55

i r

A

C

N

Gambar 3.7 Proses pemantulan bunyi.

$E<:C2?
82B:C
H2?8
?52
3E2D
D6?D2?8
CE5ED
52D2?8
CE5ED
A2?DE= 52?
82B:C
?@B>2=
A252
A6>2?DE=2?
3E?H:
2<2?
C2>2
56?82?
#/$#3
 (252
82>32B
D6BC63ED
D6B=:92D
329G2
CE5ED
52D2?8

CE5ED
A2?DE=
  52?
82B:C
?@B>2=
D6B=6D2<
A252
C2DE
3:52?8
52D2B
52?


 %6?82A2
3E?H:
52=2>
BE2?82?
D6BDEDEA
D6B56?82B
=63:9
<6B2C
$6D2< 5:?5:?8
BE2?82?
H2?8
D6B=2=E
56<2D
>6?H6323<2?
3E?H:
A2?DE=
D:52<
4E<EA G26B2>32D
C69:?882
3E?H:
52D2?8
52?
3E?H:
A2?DE=
D6B56?82B C642B2
36BC2>22?
2=
D6BC63ED
>6>3E6>A6B<E2D
3E?H:
2C=:?H2
!DE=29
C6323?H2
CE2B2
>EC:<
2<2?
D6B56?82B
=63:9 <6B2C
5:
52=2>
BE2?82?
52B:A252
5:
=2A2?82?
D6B3E<2
6B3652
56?82?
86>2 H2?8
D6B;25:
;:<2
?52
36BD6B:2<
5:
526B29
36BD63:?8
"2B2<
D63:?8
52? CE>36B
3E?H:
H2?8
4E<EA
;2E9
>6?H6323<2?
3E?H:
A2?DE=
>6>6B=E<2? G22
E?DE<
>6B2>32D
C2>A2:
A6?56?82B2?
<:32D?H2 3E?H:
A2?DE=
2<2?
D6B56?82B
C6D6=29
3E?H:
2C=:
636B2A2
A6B:CD:G2
D6BC63ED >6BEA2<2?
A6?82=2>2?
H2?8
C6B:?8
?52
D6>E<2?

2.

Pembiasan Gelombang Bunyi

?52
D6=29
>6?86D29E:
86=@>32?8
3E?H: >6B2>32D
<6
C6>E2
2B29
52?
86=@>32?8
3E?H:
D:52< 52A2D
D6BA@=2B:C2C:
%6?82A2
A252
>2=2>
92B:
CE2B2 25I2?
2D2E
CE2B2
D6B:2<2?
H2?8
4E<EA
;2E9
D6B56?82B 2 =63:9
;6=2C
52B:A252
C:2?8
92B:
(252
C:2?8
92B:
CE9E E52B2
5:
A6B>E<22?
=63:9
D:?88:
52B:A252
5:
2D2C?H2 2=
D6BC63ED
>6?H6323<2?
=2A:C2?
E52B2
A252
328:2? 2D2C
=63:9
B2A2D
52B:A252
5:
32G29?H2
*632=:2=2>
92B:
=2A:C2?
E52B2
5:
A6B>E<22?
=63:9
B2A2D 52B:A252
5:
2D2C?H2
<:32D?H2
A252
C:2?8
92B:
;:<2 3 CE>36B
3E?H:
52D2?8
C642B2
9@B:I@?D2=
D6B9252A Gambar 3.8 A6?56?82B
52A2D
5:A2CD:<2?
2B29
B2>32D
3E?H:
2<2?
>6=6?8<E?8
<6
2D2C (a) Pada siang hari, arah rambat C652?8<2?
A252
>2=2>
92B:
2B29
B2>32D
3E?H:
2<2?
>6=6?8<E?8
<6 bunyi melengkung ke atas. (b) Pada malam hari, arah rambat 32G29
(6B92D:<2?
#/$#3
 bunyi melengkung ke bawah. A2
H2?8
2<2?
D6B;25:
;:<2
3E?H:
52D2?8?H2
>6B2>32D
F6BD:<2=
<6 32G29
(252
>2=2>
92B:
2B29
B2>32D
3E?H:
5:3:2C<2?
>6?56<2D:
82B:C ?@B>2=
*632=:32D
3E?H:
5:3:2C<2?
>6?;2E9: 82B:C
?@B>2= *6CE2:
56?82?
9E<E>
A6>3:2C2?
86=@>32?8
329G2
86=@>32?8 52D2?8
52B:
>65:E>
<EB2?8
B2A2D
<6
>65:E>
=63:9
B2A2D
2<2?
5:3:2C<2? >6?56<2D:
82B:C
?@B>2=
2D2E
C632=:
3. Difraksi Gelombang Bunyi

Ingatlah Pelayangan bunyi terjadi karena adanya interferensi dua gelombang yang frekuensinya berbeda sedikit.

56

A2<29
86=@>32?8
3E?H:
52A2D
>6?82=2>:
5:7B22D:
<6D:<2
?52
>6?56?82B
56BE 3E?H:
>@3:=
>6=:?D2C:
526B29
A6B3E<:D2?
H2?8
36B<6=@<
CE2B2
A6B42<2A2? D6>2?D6>2?
?52
5:
BE2?82?
H2?8
36B3652
2D2E
<242
A6>32D2C
=@<6D A6>32H2B2?
5:
C63E29
32?<
H2?8
C6?82;2
5:3E2D
56?82?
3636B2A2
=E32?8 <64:=
282B
86=@>32?8
3E?H:
D:52<
>6>2?DE=
G2=2EAE?
2B29
B2>32D
3E?H: D:52<
36BEA2
82B:C
=EBEC
+:82
A6B:CD:G2
D6BC63ED
>6BEA2<2?
4@?D@9 A6?6B2A2?
5:7B2232?8
<6D:<2
>6=6G2D: 46=29
H2?8
E<EB2?
46=29?H2
C6@B56
56?82?
A2?;2?8
86=@>32?8?H2

Mudah dan Aktif Belajar Fisika untuk Kelas XII

$2=E
>6?82A2
86=@>32?8
3E?H:
>E529
>6?82=2>:
5:7B232?8
3E?H:
5:
E52B2
>6>:=:<:
A2?;2?8
86=@>32?8
C6<:D2B
3636B2A2 C6?D:>6D6B
C2>A2:
3636B2A2
>6D6B
2?5:?8<2?
56?82?
4292H2
H2?8 >6>:=:<:
A2?;2?8
86=@>32?8
36B<:C2B



>>
!DE=29
C6323?H2
86=@>32?8 3E?H:
=63:9
>E529
E?DE<
D6B5:7B2
4. Interferensi Gelombang Bunyi #6D:<2
?52
36B5:B:
5:
56A2?
5E2
#!
A6?86B2C
CE2B2
C63282: CE>36B
3E?H:
<6>E5:2?
36B;2=2?
=EBEC
C6;2;2B
56?82?
82B:C
A6?89E3E?8
<65E2 #!
D6BC63ED
+2<
=2>2
C6D6=29
:DE
?52
2<2?
>6?56?82B
3E?H:
=6>29 52?
C6C22D
<6>E5:2?
D6B56?82B
3E?H:
<E2D
E?H:
<E2D
D6B;25:
<6D:<2
CEA6BA@C:C: 52B:
5E2
86=@>32?8
3E?H:
>6?892C:=<2?
:?D6B76B6?C:
56CDBE6BEA2<2?
<6=:A2D2?
3:=2?82?
3E=2D
52B: C6D6?829
A2?;2?8
86=@>32?8
3E?H:
C642B2
>2D6>2D:C
5:DE=:C<2?
C63282:
36B:<ED      !



56?82?






 





52?



36BDEBEDDEBED
E?DE<
3E?H:
<E2D
A6BD2>2
3E?H: <E2D
<65E2
52?
3E?H:
<E2D
<6D:82

Kata Kunci • • • • • • •

dispersi refraksi refleksi difraksi interferensi polarisasi superposisi

5. Pelayangan Bunyi !?D6B76B6?C:
H2?8
5:D:>3E=<2?
52B:
5E2
86=@>32?8
3E?H:
52A2D >6?H6323<2?
A6B:CD:G2
A6=2H2?82?
3E?H:
H2:DE
A6?8E2D2?
52?
A6=6>292? 3E?H:
2=
D6BC63ED
D6B;25:
2<:32D
CEA6BA@C:C:
5E2
86=@>32?8
H2?8
>6>:=:<: 7B6<E6?C:
H2?8
C65:<:D
36B3652
52?
>6B2>32D
52=2>
2B29
H2?8
C2>2 "25:
C2DE
A6=2H2?82?
5:567:?:C:<2?
C63282:
5E2
3E?H:
<6B2C
2D2E
5E2 3E?H:
=6>29
H2?8
D6B;25:
C642B2
36BEBED2?
=2H2?82?

<E2D
N
=6>29
N <E2D
2D2E
=6>29
N
<E2D
N
=6>29 ":<2
<65E2
86=@>32?8
3E?H:
D6BC63ED
>6B2>32D
36BC2>22?
2<2? >6?892C:=<2?
3E?H:
A2=:?8
<E2D
C22D
72C6
<65E2?H2
C2>2
":<2
<65E2 86D2B2?
36B=2G2?2?
72C6
2<2?
5:92C:=<2?
3E?H:
A2=:?8
=6>29 %:C2=<2?
5E2
CE>36B
3E?H:
>6>:=:<:
7B6<E6?C:

52?

C65:<:D 36B3652
C:>A2?82?
C6D:2A
86=@>32?8
A252
A@C:C:
D6BD6?DE
252=29 &


C:?  "
52?
&


C:?   " 6?82?
56>:<:2?
CEA6BA@C:C:
C:>A2?82?
<65E2
86=@>32?8
D6BC63ED
252=29 &  &  &   C:?   "   C:?   " &

  C:?   "  C:?   "  56?82?
>6?88E?2<2?
2DEB2?
C:?EC
2<2?
5:A6B@=69

Ingatlah sin A + sin B = 2 cos 21 (A–B) sin 21 (A+B)

  &   4@C   "    "C:?   "    "        

  4@C  " 
C:?  "         ":<2
    
52?
    
>2<2      f           
52?
          (6BC2>22?
CEA6BA@C:C:
52A2D
5:DE=:C<2?
>6?;25:    &


4@C
 "
 
C:?
 "
A
C:?
 "  

 N

Gelombang Bunyi

57

       4@C   
"  

Pembahasan Soal Dua buah dawai baja yang identik memberikan nada dasar dengan frekuensi 400 Hz. Bila tegangan dalam salah satu dawai ditambah 2%, frekuensi pelayangan yang terjadi adalah .... a. 0 Hz d. 6 Hz b. 2 Hz e. 8 Hz c. 4 Hz Tes ITB 1975 Pembahasan: Diketahui: f0 = 400 Hz Tegangan ditambah 2% p’ = F + 2% F = 1,02 F maka

 N

6B52C2B<2?
'34#/##0
9
:?D6?C:D2C
3E?H:
2<2?
>22=
;:<2 &

 
'=69
<2B6?2
:DE
=2H2?82?
3E?H:
D6B;25:
;:<2
 4@C  

 "

36B?:=2:

52?
N
C69:?882
7B6<E6?C:
=2H2?82??H2
252=29

f p  ( f1  f 2 )

 N

#6D6B2?82?  
7B6<E6?C:
A6=2H2?82?
 I  
7B6<E6?C:
86=@>32?8
&
 I  
7B6<E6?C:
86=@>32?8
&
 I

Contoh 3.4

f0 1,02F  F’   1,01 f F F sehingga frekuensi layangannya adalah fp = (f2 – f1) = (1,01 f0 – f0) = 0,01 f0 = 0,01 · 400 Hz = 4 Hz

E2
ED2C
C6?2B
H2?8
:56?D:<
>6>36B:<2?
?252
52C2B
56?82?
7B6<E6?C:



I
":<2 D682?82?
C2=29
C2DE
C6?2B
5:D2>329

36B2A2
A6=2H2?82?
H2?8
D6B;25: #7#$ :<6D29E:







I








 


  
52A2D
5:9:DE?8
56?82?
A6BC2>22?
 

Jawaban: c

  

 



    

 

      

Hz  
I

B6<E6?C:
A6=2H2?82?
5:A6B@=69
56?82?
>6?88E?2<2?
'34#/##0
9
252=29 


N
 




N

I

N
I "25:
7B6<E6?C:
A6=2H2?82??H2
252=29
N
I

Tes Kompetensi Subbab

D

'3,#-#0.#*
&#.#/
$6-6
.#5+*#0  "6=2C<2?
A6?86BD:2?
5:7B232?8
3E?H:  ":<2
252
D:82
CE>36B
3E?H:
H2?8
:56?D:<
5:=6D2<<2? >2C:?8>2C:?8
36B;2B2<

>
56?82?
H2?8
=2:??H2 <2A2?<29
D6B;25:
:?D6B76B6?C:
>2E>
52B:
<6D:82 CE>36B
3E?H:
D6BC63ED



E2
3E29
C6?2B
H2?8
:56?D:<
>6>36B:<2?
?252
52C2B 56?82?
7B6<E6?C:



I
":<2
D682?82?
C2=29
C2DE 52G2:
5:D2>329

36B2A2
7B6<E6?C:
A6=2H2?82?
H2?8 D6B;25:

E. Dawai dan Pipa Organa sebagai Sumber Bunyi 1. Frekuensi Gelombang pada Dawai ?52
D6=29
>6>A6=2;2B:
329G2
36?5236?52
H2?8
36B86D2B
>6BEA2<2? CE>36B
3E?H:
D6D2A:
D:52<
C6>E2
3E?H:
52A2D
5:56?82B
@=69
D6=:?82
>2?EC:2 2=
D6BC63ED
5:C6323<2?
D6=:?82
>2?EC:2
?@B>2=
>6>:=:<:
7B6<E6?C: 2>32?8
>:?:>2=
H2:DE
7B6<E6?C:
D6B6?529
52B:
CE>36B
3E?H:
H2?8
>2C:9 52A2D
5:D6B:>2
@=69
D6=:?82

58

Mudah dan Aktif Belajar Fisika untuk Kelas XII

*2=29
C2DE
CE>36B
3E?H:
252=29
52G2:
8:D2B
&252
H2?8
5:92C:=<2? 52G2:
8:D2B
52A2D
5:E329E329
56?82?
42B2
>6?6<2?
C6?2B
A252
D:D:< D:D:<
D6BD6?DE
(@=2
86=@>32?8
H2?8
52A2D
D6B;25:
A252
52G2:
8:D2B 5:DE?;E<<2?
A252
#/$#3

56?82?


36BE329E329 *2DE
52G2:
52A2D
>6>36?DE<
36B3282:
7B6<E6?C:
2D2E
A@=2
86D2B2? C6A6BD:
D2>A2<
A252
#/$#3


1
 0 2


 1

3
 2 2


 2 2

2

3

4

5

(@=2
86D2B2?
52G2:
A252
#/$#3
#
>6?892C:=<2?
?252
52C2B  $ $  
56?82?

 
2D2E
 

>2<2
    
(@=2
86D2B2?
52G2:
A252
#/$#3
$
>6?892C:=<2?
?252
2D2C $ $ A6BD2>2

56?82?

 
2D2E
 

>2<2
      
(@=2
86D2B2?
52G2:
A252
#/$#3
%
>6?892C:=<2?
?252
2D2C
<65E2 $ $

$

      
56?82?

 
2D2E
 

>2<2
  



(@=2
86D2B2?
52G2:
A252
#/$#3
&
>6?892C:=<2?
?252
2D2C
<6D:82 $ $ $   
56?82?

 
2D2E
 

>2<2
         

(6B32?5:?82?
7B6<E6?C:
?252?252
A252
52G2:
H2?8
5:92C:=<2? 36BEA2
A@=2
86=@>32?8
C6A6BD:
A252
#/$#3

252=29
C63282:
36B:<ED 















Gambar 3.9 (a) Nada dasar f0; (b) Nada atas pertama f1; (c) Nada atas kedua f2; (d) Nada atas ketiga f3.

Informasi untuk Anda

 N

'=69
<2B6?2
46A2D
B2>32D
86=@>32?8
A252
52G2:
C6CE2:
56?82?
92C:= Echolocation adalah bentuk sensor yang digunakan oleh binatang seperti kelelawar, paus, dan lumbalumba. Binatang tersebut memancarkan gelombang suara (gelombang longitudinal), kemudian dipantulkan oleh objek. Gelombang pantul kemudian dideteksi oleh binatang tersebut. Gelombang echolocation yang dipancarkan oleh kelelawar memiliki frekuensi sekitar 175.000 Hz.


2'3%1$##0
'.&'
H2:DE
$

>2<2
5:A6B@=69  2 7B6<E6?C:
?252
52C2B    


 3 7B6<E6?C:
?252
2D2C
A6BD2>2  

   


 4 7B6<E6?C:
?252
2D2C
<65E2   

Information for You




  

 5 7B6<E6?C:
?252
2D2C
<6              


 

 N

Echolocation is a form of sensory perception used by animals such as bats, toothed whales, and porpoises. The animal emits a pulse of sound (a longitudinal wave) which is reflected from objects. The reflected pulse is detected by the animal. Echolocation wave emitted by bats have frequencies of about 175.000 Hz.

Gelombang Bunyi

59

(6BC2>22?
7B6<E6?C:
?252?252
D6BC63ED
<2=:
A6BD2>2
5:E?8<2A<2?
@=69 '34'00'
C69:?882
'34#/##0
9
D6BC63ED
5:<6?2=
56?82?
6-6/ '34'00'

Contoh 3.5 *6ED2C
52G2:
H2?8
<65E2
E;E?8?H2
D6B:<2D
>6>:=:<:
A2?;2?8

>
>2CC2
;6?:C?H2   


8 4>
>6?892C:=<2?
7B6<E6?C:
?252
52C2B
 


I 2 :DE?8=29
82H2
D682?82?
52G2: 3 6B2A2<29
36C2B
7B6<E6?C:
52?
A2?;2?8
86=@>32?8
A252
?252
2D2C
A6BD2>2 4 +6?DE<2?=29
7B6<E6?C:
52?
A2?;2?8
86=@>32?8
A252
52G2:
E?DE<
?252
2D2C <65E2 #7#$ :<6D29E:




>

 

<8 4> 





I 2 6?82?
>6?88E?2<2?
'34#/##0
9
E?DE<
?252
52C2B
5:A6B@=69

3

4

   


2D2E


   

 


Q
 N
<8 4>   
I 



& "25:
36C2B
82H2
D682?82?
52G2:
252=29

& B6<E6?C:
?252
2D2C
A6BD2>2

252=29 





2D2E






I


I (2?;2?8
86=@>32?8
A252
?252
2D2C
A6BD2>2
252=29






> B6<E6?C:
A252
?252
2D2C
<65E2



252=29 





2D2E






I


I (2?;2?8
86=@>32?8
A252
?252
2D2C
<65E2
252=29
 


2D2E


 
      


>





2. Pipa Organa dan Pola Gelombangnya

Gambar 3.10 Seruling merupakan pipa organa terbuka.

+6=29
?52
<6D29E:
329G2
36?52
H2?8
36B86D2B
52A2D
>6?892C:=<2? 3E?H:
+:52<
92?H2
36?52
A252D
C6A6BD:
52G2:
D2=:
>2EAE?
A6=2D
=@82> H2?8
52A2D
36B86D2B
86D2B2?
AE?
52A2D
D6B;25:
A252
82C
>:C2=?H2
E52B2 6D2B2?
<@=@>
E52B2
A252
C6BE=:?8
2<2?
>6?892C:=<2?
3E?H:
+:?88: ?252
H2?8
5:A6B@=69
52B:
C63E29
C6BE=:?8
52A2D
5:2DEB
7B6<E6?C:?H2
: 2?D2B2
A6B2?8<2D
>EC:<
C656B92?2
H2?8
52A2D
>6?892C:=<2?
86D2B2? <@=@>
E52B2
C6=2:?
C6BE=:?8
252=29
A:A2
@B82?2
+6B52A2D
5E2
>242> A:A2
@B82?2
H2:DE
A:A2
@B82?2
D6B3E<2
52?
A:A2
@B82?2
D6BDEDEA a. Pipa Organa Terbuka (:A2
@B82?2
D6B3E<2
252=29
2=2D
D:EA
36BEA2
D23E?8
H2?8
<65E2
E;E?8 A6?2>A2?8?H2
D6B3E<2
#65E2
E;E?8
A:A2
@B82?2
D6B3E<2
>6?;25:
A6BED 86=@>32?8
A252
<@=@>
E52B2
(@=2
86=@>32?8
H2?8
5:92C:=<2?
C6A6BD: D2>A2<
A252
#/$#3
 
B6<E6?C:
H2?8
5:92C:=<2?
@=69
A:A2
@B82?2 D6B3E<2
252=29
C63282:
36B:<ED

  &252
52C2B
 
 
2D2E  
 B6<E6?C:
H2?8
5:92C:=<2?
 

60

Mudah dan Aktif Belajar Fisika untuk Kelas XII

$ $   



 &252
2D2C
A6BD2>2

  2D2E
 
 B6<E6?C:
H2?8
5:92C:=<2?
 

nada dasar atau harmonik pertama

$ $   





 &252
2D2C
<65E2
 
 
2D2E
  



1
 0 2

$ $   B6<E6?C:
H2?8
5:92C:=<2?
  


nada atas pertama atau harmonik kedua

   &252
2D2C
<6

?
          B6<E6?C:
H2?8
5:92C:=<2?       



 

 

$


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(6B32?5:?82?
7B6<E6?C:
86=@>32?8
H2?8
5:>:=:<:
<@=@>
E52B2
A252 A:A2
@B82?2
D6B3E<2
C2>2
56?82?
A6B32?5:?82?
7B6<E6?C:
86=@>32?8 A252
52G2:
(6B32?5:?82?
7B6<E6?C:
H2?8
5:92C:=<2?
@=69
C6D:2A
A@=2 86=@>32?8
E?DE<
A:A2
@B82?2
D6B3E<2
252=29





 





$ $ $



 









 1 nada atas ketiga atau harmonik ketiga

 N 

'34#/##0
9 
H2?8
>6BEA2<2?
A6B32?5:?82?
7B6<E6?C:
5:<6?2= C63282:
6-6/

'3016..+
52?
>6BEA2<2?
A6B32?5:?82?
3:=2?82? 3:=2?82?
3E=2D

3
  2

Gambar 3.11 Pola gelombang resonansi suatu pipa organa terbuka.

Contoh 3.6 *63E29
A:A2
@B82?2
D6B3E<2
A2?;2?8?H2

4>
+6?DE<2? ?252
52C2B
H2?8
5:92C:=<2?
A:A2
D6BC63ED
;:<2
46A2D
B2>32D 3E?H:
5:
E52B2


> C #7#$

1
   40cm 2

:<6D29E:
A2?;2?8
A:A2

 

4>


> 46A2D
B2>32D
3E?H:
$


> C 6?82?
>6?88E?2<2?
'34#/##0
9
7B6<E6?C:
?252
52C2B
5:A6B@=69 
 
2D2E
 


>2<2
      >    > $  > C *H2B2D
D6B;25:?H2
?252
52C2B
H2:DE       I



  >

Contoh 3.7 *63E29
A:A2
@B82?2
D6B3E<2
H2?8
A2?;2?8?H2

>
52?
7B6<E6?C:
92B>@?:32D
3E?H:
A252
A:A2
@B82?2 D6BC63ED

Gelombang Bunyi

61

#7#$ %:C2=<2?
?


I
52?
?



 
I 6?82?
>6?88E?2<2?
'34#/##0
9
5:A6B@=69 $
?


N
?

J

N

K
     I
        
I 

$

$




   

$


> C

harmonik pertama

1
 0 4

harmonik kedua

b. Pipa Organa Tertutup (:A2
@B82?2
D6BDEDEA
252=29
2=2D
D:EA
36BEA2
D23E?8
H2?8
C2=29
C2DE E;E?8
A6?2>A2?8?H2
D6BDEDEA
*2=29
C2DE
E;E?8
A:A2
@B82?2
D6BDEDEA >6?;25:
C:>AE=
86=@>32?8
A252
<@=@>
E52B2
52?
E;E?8
=2:??H2
H2?8 D6B3E<2
>6?;25:
A6BED
86=@>32?8
(@=2
86=@>32?8
H2?8
5:92C:=<2?
D2>A2< C6A6BD:
#/$#3
 
B6<E6?C:
H2?8
5:92C:=<2?
@=69
A:A2
@B82?2
D6BDEDEA 252=29
C63282:
36B:<ED

  &252
52C2B
 

 
2D2E
  
 

$ $    


7B6<E6?C:
H2?8
5:92C:=<2?
  3
 1 4

harmonik ketiga

  &252
2D2C
A6BD2>2


 
2D2E
 
 

$  $  
     


7B6<E6?C:
H2?8
5:92C:=<2?
 

 

 &252
2D2C
<65E2
 

 
2D2E
 
   5
 2 4

Gambar 3.12 Pola gelombang resonansi pada suatu pipa organa tertutup.

7B6<E6?C:
H2?8
5:92C:=<2?
 

$  $   
     



  &252
2D2C
<6N
?

     
2D2E
        B6<E6?C:
H2?8
5:92C:=<2?            



$ 


 N

(6B32?5:?82?
7B6<E6?C:
H2?8
5:92C:=<2?
@=69
C6D:2A
A@=2
86=@>32?8 E?DE<
A:A2
@B82?2
D6BDEDEA
252=29





 





$  $   $ 


 


   


 
 
  


 N 

'34#/##0
9 
H2?8
>6BEA2<2?
A6B32?5:?82?
7B6<E6?C:
5:<6?2= C63282:
6-6/

'3016..+
52?
>6BEA2<2?
A6B32?5:?82?
3:=2?82? 3:=2?82?
82?;:=

62

Mudah dan Aktif Belajar Fisika untuk Kelas XII

Contoh 3.8 *63E29
A:A2
@B82?2
D6BDEDEA
>6>:=:<:
A2?;2?8

4>
A23:=2
46A2D
B2>32D
3E?H: C22D
:DE

> C
D6?DE<2?
7B6<E6?C:
?252
52C2B
 
?252
2D2C
A6BD2>2

52?
?252
2D2C <65E2

  #7#$ :<6D29E:




4>

 
>

$


> C 6B52C2B<2?
'34#/##0
9             

M ?252
52C2B



 

$ 





> C   
>

$





5:A6B@=69

 
I

M ?252
2D2C
A6BD2>2


 0 

1


   


I M ?252
2D2C
<65E2



 0  

1


  




I

Contoh 3.9 *63E29
A:A2
@B82?2
D6B3E<2

56?82?
A2?;2?8

4>
D6B;25:

3E29
C:>AE=
&252 A:A2
@B82?2
D6BC63ED
36B6C@?2?C:
56?82?
A:A2
@B82?2
D6BDEDEA
 
52?
>6>36?DE<
3E29
C:>AE=
6B2A2
A2?;2?8
A:A2
@B82?2
H2?8
D6BDEDEA #7#$ )6C@?2?C:
D6B;25:
;:<2

C2>2
56?82?



  (2?;2?8
A:A2
@B82?2
D6B3E<2



4>
D6B;25:

C:>AE=
52?

A6BED
     C69:?882
 




2D2E
 







>2<2
   4>   4>

Pembahasan Soal Pipa organa terbuka A dan pipa organa tertutup sebelah B memiliki panjang yang sama. Perbandingan frekuensi nada atas pertama antara pipa organa A dengan pipa organa B adalah .... a. 1 : 1 d. 3 : 2 b. 2 : 1 e. 4 : 3 c. 2 : 3 UMPTN 1995 Pembahasan: Pipa organa terbuka:
A  1  ,  , 3  , .... 2 2 Pada nada atas pertama:
A    fA  v  v x
A Pipa organa tertutup:
B  1  , 3  , 5  , .... 4 4 4 Pada nada atas pertama:
B  3     4
B 4 3 fB  v  v  3 v  4
B 4
3 B



Untuk
A 
B , diperoleh v fA
 A 4 fB 3 v 3 4
B Jawaban: e

  (252
A:A2
@B82?2
D6BDEDEA
D6B;25:

C:>AE=
52?

A6BED
  
C69:?882  





2D2E
 





>2<2
 

$



56?82?
>6?88E?2<2?
CH2B2D
B6C@?2?C:
A2?;2?8
A:A2
@B82?2
D6BDEDEA


52A2D
5:9:DE?8  



$









$





 








 
4>





4>

3. Resonansi (6B:CD:G2
:<ED
36B86D2B?H2
CE2DE
36?52
<2B6?2
252
36?52
=2:?
H2?8 36B86D2B
5:C63ED
!!
*H2B2D
D6B;25:?H2
B6C@?2?C:
252=29
7B6<E6?C:
86D2B <65E2
36?52
92BEC
C2>2
2D2E
7B6<E6?C:
36?52
H2?8
:<ED
36B86D2B
C2>2 56?82?
<6=:A2D2?
3:=2?82?
3E=2D
52B:
7B6<E6?C:
36?52
H2?8
36B86D2B (6B92D:<2?
5E2
4@?D@9
A6B:CD:G2
B6C@?2?C:
36B:<ED
:?: a. Resonansi pada Garputala ":<2
C63E29
82BAED2=2
5:AE<E=
82BAED2=2
D6BC63ED
36B86D2B
C6A6BD: A252
#/$#3
 
B6<E6?C:
86=@>32?8
3E?H:
H2?8
5:92C:=<2? 36B82?DE?8
A252
36?DE<
36C2B
52?
3292?
=@82>
82BAED2=2
D6BC63ED #6D:<2
5:56<2D<2?
A252
82BAED2=2
H2?8
=2:?
52?
>6>:=:<:
7B6<E6?C:
3E?H: C2>2
82BAED2=2
<65E2
2<2?
:<ED
36B86D2B
(6B:CD:G2
:<ED
36B86D2B?H2 82BAED2=2
H2?8
<65E2
5:C63ED
B6C@?2?C:

Gambar 3.13 Dua garputala yang memiliki frekuensi yang sama akan beresonansi.

Gelombang Bunyi

63

Gambar 3.14 Sebuah kolom udara di atas permukaan air digetarkan oleh sebuah garputala.

b. Resonansi pada Kolom Udara (6B92D:<2?
#/$#3

H2?8
>6>A6B=:92D<2?
A6B:CD:G2
B6C@?2?C: A252
<@=@>
E52B2
":<2
<@=@>
E52B2
H2?8
D6B=6D2<
5:
2D2C
A6B>E<22?
2:B 5:86D2B<2?
56?82?
>6?88E?2<2?
C63E29
82BAED2=2
>@=6<E=>@=6<E= E52B2
5:
52=2>
<@=@>
E52B2
2<2?
:<ED
36B86D2B
*H2B2D
D6B;25:?H2
B6C@?2?C: A252
<@=@>
E52B2
252=29  D6A2D
A252
A6B>E<22?
2:B
92BEC
D6B36?DE<
B2A2D2? C:>AE=
86=@>32?8  A252
E;E?8
A6B>E<22?
<@=@>
E52B2
>6BEA2<2?
A6BED B6?882?82? 86=@>32?8 ,?DE<
>6?52A2D<2?
B6C@?2?C:
A2?;2?8
<@=@>
E52B2
5:
2D2C
A6B>E<22? 92BEC
5:E329E329
52AE?
7B6<E6?C:
52?
A2?;2?8
86=@>32?8
3E?H:?H2
D6D2A )6C@?2?C:
A6BD2>2
2<2?
D6B;25:
;:<2
A2?;2?8
<@=@>
E52B2
5:
2D2C
A6B>E<22? 

 2:B
 
B6C@?2?C:
<65E2
 
B6C@?2?C:
<6D:82
 
52?
C6D6BEC?H2    *2>2
92=?H2
56?82?
86D2B2?
E52B2
A252
A:A2
@B82?2
D6BDEDEA 9E3E?82?
A2?;2?8
<@=@>
E52B2


D6B9252A
A2?;2?8
86=@>32?8
3E?H:   
A252
A6B:CD:G2
B6C@?2?C:
<@=2>
E52B2
52A2D
5:DE=:C
C63282:
36B:<ED


       

 N 

#6D6B2?82?  





 
A2?;2?8
<@=@>
E52B2
<6N  
A2?;2?8
86=@>32?8
3E?H: ":<2
7B6<E6?C:
82BAED2=2
5:<6D29E:
46A2D
B2>32D
86=@>32?8
3E?H:
82BAED2=2 52A2D
5:D6?DE<2?
>6=2=E:
A6BC2>22?
$

  
82B
?52
52A2D >6?82A=:<2C:<2?
AB@C6C
A6?8E<EB2?
46A2D
B2>32D
3E?H:
5:
E52B2
36B52C2B<2? <@?C6A
B6C@?2?C:
3E?H:
A252
<@=@>
E52B2
=2<E<2?=29
2
Aktivitas Fisika 3.3 Cepat Rambat Bunyi Tujuan Percobaan Mengukur cepat rambat bunyi di udara.

gelas air mineral

Alat-Alat Percobaan garputala 1. Garputala 2. Lampu neon yang sudah tidak terpakai 3. Botol air mineral 100 m 4. Selang plastik berdiameter 1,5 cm dan panjang 2 m 5. Pipa paralon berdiameter 3 cm dan panjang 1 m 6. Penjepit 7. Penggaris neon
paralon 8. Gabus sebagai sumbat selang plastik Langkah-Langkah Percobaan 1. Rangkailah alat-alat tersebut seperti pada gambar. gabus 2. Isilah lampu neon dengan air sampai ketinggian kolom udara 5 cm. 3. Getarkanlah garputala di atas lampu neon yang terbuka. 4. Lakukan pemompaan terhadap air dari lampu neon agar air keluar melalui selang dengan botol air mineral secara perlahanlahan.

64

Mudah dan Aktif Belajar Fisika untuk Kelas XII

5.

Berhentilah memompa air pada saat terdengar bunyi mendengung. Ukur panjang kolom saat itu dan catat sebagai l1 (resonansi pertama) pada tabel pengamatan seperti berikut. Data Pengamatan Resonansi Udara untuk Percobaan Cepat Rambat Udara Resonansi ke–n

Panjang Kolom Udara (

1. 2. 3. 6.

)

.... .... ....

Turunkan permukaan air dalam lampu neon sampai terdengar bunyi mendengung lagi. Ukur panjang kolom saat itu dan catat sebagai l2 (resonansi kedua) pada tabel pengamatan. Ulangi langkah tersebut sampai didapatkan l3, l4, dan seterusnya.

7.

6B52C2B<2?
26?52A2D<2?
52D2 A2?;2?8
<@=@>
E52B2
#6>E5:2?
3282:>2?2<29
42B2
>6?89:DE?8
46A2D B2>32D
3E?H:
5:
E52B2
(6B92D:<2?
<6>32=:
'34#/##0
9  
? 
 


  

 

Ingatlah


 
 

Nilai frekuensi setiap garputala sudah diketahui. Analoginya seperti nilai fokus dari lensa.

":<2
?52
>6?89:DE?8
C6=:C:9

1 52?

0 
2<2?
5:52A2D<2?

N




N
 

   

 



N

 '=69
<2B6?2
A2?;2?8
86=@>32?8
CE529
5:<6D29E:
?52
52A2D
>6?89:DE?8 46A2D
B2>32D?H2
36B52C2B<2?
A6BC2>22? $

 

 N

#6D6B2?82? $ 
46A2D
B2>32D
3E?H:
> C  
A2?;2?8
86=@>32?8
82BAED2=2
>  
7B6<E6?C:
82BAED2=2
 I

Contoh 3.10




(252
82>32B
D2>A2<
C63E29
<@=@>
E52B2
H2?8
A2?;2?8?H2


E?DE< >6?892C:=<2?
B6C@?2?C:
<@=@>
E52B2
C642B2
F6BD:<2=
":<2
<@=@>
E52B2 D6BA6?56<
>6?892C:=<2?
B6C@?2?C:
C22D

 
4> 
D6?DE<2? 7B6<E6?C:
82BAED2=2

 +6?DE<2?
AE=2
A2?;2?8
<@=@>
E52B2
A252
B6C@?2?C:
C6=2?;ED?H2 :<6D29E:
46A2D
B2>32D
3E?H:
5:
E52B2
$


> C

#7#$ :<6D29E:




4>


>

$


> C (252
A6B:CD:G2
B6C@?2?C:
A6BD2>2
52=2>
<@=@>
E52B2
36B=2<E
9E3E?82?
36B:<ED


 


2D2E
  
  4>  
4>   
>

Gelombang Bunyi

65

Kata Kunci • • • • •



resonansi Hukum Mersenne pipa organa Hukum I Bernoulli Hukum II Bernoulli

$






> C

 
>

  


I

"25:
7B6<E6?C:
82BAED2=2
252=29
 


I (2?;2?8
<@=@>
E52B2
A252
B6C@?2?C:
<65E2



52?
<6D:82



52A2D
5:D6?DE<2? 

56?82?
'34#/##0
9 
H2:DE

     
52?
E?DE<
B6C@?2?C:
<65E2


 

252=29




         >   
>  

 "25:
E?DE<
B6C@?2?C:
<6D:82




        m    
>  

Tes Kompetensi Subbab

E

'3,#-#0.#*
&#.#/
$6-6
.#5+*#0 







(252
C6ED2C
52G2:
H2?8
A2?;2?8?H2

>
56?82?
>2CC2
8
A6B2>32D2?
86=@>32?8
DB2?F6BC2=
H2?8
5:92C:=<2? >6>:=:<:
<646A2D2?

> C
+6?DE<2?
D682?82?
52G2: D6BC63ED +2=:
H2?8
A2?;2?8?H2

>
5:D682?8<2?
56?82?
82H2
&
(6B2>32D2?
86=@>32?8
DB2?CF6BC2=
H2?8
5:92C:=<2? >6>:=:<:
<646A2D2?

> C
+6?DE<2?
>2CC2
D2=: D6BC63ED E2
ED2C
D2=:
H2?8
C2>2
A2?;2?8
5:D2B:<
@=69
82H2
H2?8 C2>2
%2CC2
D2=:
A6BD2>2

<2=:
>2CC2
D2=:
<65E2
+2=: A6BD2>2
5:86D2B<2?
56?82?
7B6<E6?C:



I C652?8<2?
D2=:
<65E2



I
":<2
A2?;2?8
86=@>32?8 A252
D2=:
A6BD2>2

4>
D6?DE<2?
A2?;2?8
86=@>32?8 A252
D2=:
<65E2 *63E29
86=@>32?8
C:?EC
>6B2>32D
A252
D2=:
H2?8 A2?;2?8?H2



4>
36B86B2<
52B:
C:>A2?82?
>:?:>E>

  

<6
>2E>
*E2DE
D:D:<
>6>6B=E<2?
G2
86B2<
D6BC63ED
+6?DE<2? 2 A6B:@56
86=@>32?8 3 7B6<E6?C:
86=@>32?8 4 82H2
D682?8
D2=:
;:<2
5:<6D29E:
A2?;2?8
86=@>32?8 


>
52?
>2CC2
D2=:

8B2> *63E29
A:A2
@B82?2
D6B3E<2
A2?;2?8?H2

4> +6?DE<2?
7B6<E6?C:
?252
52C2B
H2?8
5:92C:=<2?
A:A2 D6BC63ED
;:<2
46A2D
B2>32D
3E?H:
5:
E52B2




> C *63E29
A:A2
@B82?2
D6BDEDEA
A2?;2?8?H2

4> +6?DE<2?
?252
52C2B
H2?8
5:92C:=<2?
A:A2
:DE
;:<2
46A2D B2>32D
3E?H:
5:
E52B2




> C +6?DE<2?
A2?;2?8
>:?:>E>
A:A2
@B82?2
D6B3E<2
H2?8 36B6C@?2?C:
56?82?
82BAED2=2
H2?8
36B86D2B
56?82? 7B6<E6?C:

I
;:<2
46A2D
B2>32D
3E?H:
5:
E52B2



> C

F. Energi Gelombang Bunyi +6=29
?52
<6D29E:
329G2
86=@>32?8
252=29
86D2B2?
H2?8
>6B2>32D >6=2=E:
I2D
A6B2?D2B2
>65:E>
D6D2A:
>65:E>?H2
D:52<
:<ED
>6B2>32D 2=2>
A6B2>32D2??H2
86=@>32?8
>6>:?529<2?
6?6B8:
52B:
C2DE
D6>A2D <6
D6>A2D
=2:?
?6B8:
86=@>32?8
3E?H:
H2?8
5:D:>3E=<2?
52B:
CE2B2 A6C2G2D
;6D
!# !
52A2D
>6>6429<2?
<242
;6?56=2
CE2DE
32?8E?2? 2=
D6BC63ED
>6?E?;E<<2?
329G2
6?6B8:
86=@>32?8
3E?H:
>6=2<E<2? A6B2>32D2?
?6B8:
86=@>32?8
3E?H:
52A2D
5:D6?DE<2?
52B:
6?6B8: A@D6?C:2=
>2E>
86D2B2?
<2B6?2
3E?H:
>6BEA2<2?
86=@>32?8
=@?8: DE5:?2=
92C:=
A6B2>32D2?
86D2B2?
'=69
<2B6?2
:DE
6?6B8:
86=@>32?8 3E?H:
52A2D
5:?H2D2<2?
56?82?
H2:DE
C63282:
36B:<ED 

  <2B6?2     

66

Mudah dan Aktif Belajar Fisika untuk Kelas XII

2D2E 

 


>2<2
5:A6B@=69









 N

#6D6B2?82?  
>2CC2
A2BD:<6=
<8  
2>A=:DE5@
86=@>32?8
>  
7B6<E6?C:
86D2B
 I  
<@?CD2?D2
A682C
& > 6B52C2B<2?
'34#/##0
9 
6?6B8:
H2?8
5:A:?529<2?
@=69
CE2DE 86=@>32?8
C632?5:?8
56?82?
<E25B2D
2>A=:DE5@?H2

1. Intensitas Gelombang Bunyi !?D6?C:D2C
86=@>32?8
3E?H:
5:567:?:C:<2?
C63282:
;E>=29
6?6B8:
3E?H: A6B
C2DE2?
G26?6>3EC
D682<
=EBEC
CE2DE
3:52?8
A6B
C2DE2? =E2C
E3E?82?
2?D2B2
52H2

=E2C

52?
:?D6?C:D2C
3E?H:
  >6>6?E9:
A6BC2>22?



 

 N

#6D6B2?82?  :?D6?C:D2C
86=@>32?8
3E?H:
- >   =E2C
A6?2>A2?8
>    52H2
86=@>32?8
3E?H:
- *E>36B
3E?H:
36BEA2
C63E29
D:D:<
C6A6BD:
A252
#/$#3

2<2? >6>2?42B<2?
3E?H:
<6
C682=2
2B29
BD:?H2
86=@>32?8
3E?H:
>6B2>32D 52=2>
D:82
5:>6?C:
2=
H2?8
C2>2
D6B;25:
AE=2
A252
86=@>32?8
4292H2 !?D6?C:D2C
3E?H:
A252
3:52?8
A6B>E<22?
52=2>
3@=2
H2?8
>6>:=:<:
;2B: ;2B:

2<2?
>6>:=:<:
=E2C


 
2?882A
D:52<
252
6?6B8:
3E?H:
H2?8 9:=2?8
2D2E
D6BC6B2A
C6=2>2
A6B2>32D2? !?D6?C:D2C
3E?H:
2<2?
C6>2<:?
<64:=
;:<2
;2B2<
CE>36B
3E?H:
<6
CE2DE 3:52?8
A6B>E<22?
C6>2<:?
;2E9
(6?8EB2?82?
:?D6?C:D2C
3E?H:
2<:32D A6BD2>3292?
;2B2<
52B:
CE>36B
3E?H:
>6>6?E9:
9E3E?82?
36B:<ED     

       

 

r2 P

r1

Gambar 3.15 Sumber bunyi berupa titik. Semakin jauh dari sumber bunyi, luas permukaan semakin besar.

Ingatlah 1 bel (B) = 10 desibel (dB)









 N

#6D6B2?82? 
;2B2<
CE>36B
3E?H:
<6
A@C:C:
3:52?8

>  
;2B2<
CE>36B
3E?H:
<6
A@C:C:
3:52?8

> 6B52C2B<2?
'34#/##0
9 
52A2D
5:?H2D2<2?
329G2
:?D6?C:D2C 3E?H:
36B32?5:?8
D6B32=:<
56?82?
<E25B2D
;2B2<
CE>36B
3E?H:
D6BC63ED <6
3:52?8
A6?56?82B2?

Tugas Anda 3.4 Berapakah rasio intensitas bunyi 90 dB terhadap intensitas bunyi 60 dB?

2. Taraf Intensitas (TI) *6D:2A
>2<9=E<
5:
2=2>
C6>6CD2
:?:
>6>:=:<:
32D2C
:?D6?C:D2C A6?56?82B2?
2D2C
:?D6?C:D2C
3E?H:
H2?8
52A2D
5:56?82B
D6=:?82
>2?EC:2 252=29
C63282:
36B:<ED

Gelombang Bunyi

67

2

!?D6?C:D2C
D6B<64:=
H2?8
>2C:9
52A2D
>6?:>3E=<2?
B2?8C2?82? A6?56?82B2?
A252
D6=:?82
252=29
C636C2B
 N
G2DD >
H2?8
5:C63ED C63282:
"!"!

  3 !?D6?C:D2C
D6B36C2B
H2?8
>2C:9
>2>AE
5:D6B:>2
D6=:?82
>2?EC:2
D2?A2
B2C2 C2<:D
252=29
C636C2B

G2DD >
H2?8
5:C63ED
"!

  "25:
32D2C
A6?56?82B2?
D6B6?529
252=29
 N
G2DD >
52?
32D2C :?D6?C:D2C
D6BD:?88:
252=29

G2DD > 
+2B27
:?D6?C:D2C
 
C632?5:?8
56?82? =@82B:D>2
A6B32?5:?82?
2?D2B2
:?D6?C:D2C
3E?H:
52?
92B82
2>32?8
:?D6?C:D2C 3E?H:
H2?8
D6B56?82B
52?
>6>:=:<:
C2DE2?
56C:36=
5
*642B2
>2D6>2D:C 9E3E?82?
2?D2B2

52?

5:DE=:C

   =@8

Pembahasan Soal sumber gempa

Malang

100 km

Surabaya Suatu gelombang gempa terasa di Malang dengan intensitas 6 × 105 W/m2. Sumber gempa berasal dari suatu tempat yang berjarak 300 km dari Malang. Jika jarak antara Malang dan Surabaya sebesar 100 km dan ketiga tempat itu membentuk segitiga siku-siku dengan sudut siku-siku di Malang, maka intensitas gempa yang terasa di Surabaya adalah .... a. 2 × 105 W/m2 b. 3 × 105 W/m2 c. 4,5 × 105 W/m2 d. 5,4 × 105 W/m2 e. 7,5 × 105 W/m2 UMPTN 1996

Pembahasan: 2

I2  r1  r    I2  I1  1  I1  r2   r2   300  I2  6  10 5   5   10  5

= 5,4 × 10 W/m

2

2

Jawaban: d

68

#$'.


!?D6?C:D2C
52?
+2B27
!?D6?C:D2C
52B:
636B2A2
*E>36B
E?H:

>32?8
A6?56?82B2? >32?8
B2C2
C2<:D *:B:?6
A252
;2B2<

> $2=E
=:?D2C
B2>2: (6B42<2A2?
3:2C2
A252
;2B2<

4> )25:@
A6=2? :C:<3:C:<

#3#(
05'04+5#4
& 05'04+5#4
"/ 


















Q
 N  
Q
 N 
Q
 N


Q
 N 
Q
 N 
Q
 N

Sumber: College Physics, 1980

Contoh 3.11 +2B27
:?D6?C:D2C
3E?H:
C63E29
A6C2G2D
A252
;2B2<

>
252=29

5
6B2A2
D2B27 :?D6?C:D2C
3E?H:
A252
;2B2<



> #7#$ :<6D29E:




>






>
 


5 !?D6?C:D2C
3E?H:
36B32?5:?8
D6B32=:<
56?82?
<E25B2D
;2B2<                 



     

56?82?
>6?88E?2<2?
'34#/##0
9 
E?DE<

52?
 
5:A6B@=69    =@8

    =@8

2

 N

#6D6B2?82?  
D2B27
:?D6?C:D2C
5 
:?D6?C:D2C
3E?H:
- >  
 N
- >

:?D6?C:D2C
2>32?8
A6?56?82B2? 6B:<ED
:?:
52D2
:?D6?C:D2C
52?
D2B27
:?D6?C:D2C
52B:
3636B2A2
CE>36B 3E?H:

6/$'3
608+

300 km







      =@8  =@8   



Mudah dan Aktif Belajar Fisika untuk Kelas XII



       =@8          =@8





Kata Kunci

   

• • • •

energi gelombang bunyi intensitas gelombang taraf intensitas batas ambang pendengaran


     =@8 

5


=@8
 N




5  



"25:
D2B27
:?D6?C:D2C
3E?H:
A252
;2B2<



>
252=29



5

":<2
D6B52A2D
CE>36B
3E?H:
H2?8
=63:9
52B:
C2DE
D2B27
:?D6?C:D2C D@D2=?H2
52A2D
5:<6D29E:
56?82?
>6?88E?2<2?
A6BC2>22?
36B:<ED  ?

 


=@8


 N

#6D6B2?82?   
D2B27
:?D6?C:D2C
@=69

3E29
CE>36B  
32?H236B   
D2B27
:?D6?C:D2C
@=69

3E29
CE>36B

Contoh 3.12 *63E29
CE>36B
3E?H:
>6?892C:=<2?
D2B27
:?D6?C:D2C
C636C2B

5
6B2A2
D2B27 :?D6?C:D2C
H2?8
5:92C:=<2?
;:<2
D6B52A2D



CE>36B
3E?H:
H2?8
C2>2
52?
36B3E?H: A252
C22D
36BC2>22? #7#$ :<6D29E:




5





 ,?DE<
C63E29
CE>36B
3E?H: 


=@8






=@8




 




,?DE<



3E29
CE>36B
3E?H:





Q
 


 


>2<2
+! 

=@8

 




=@8


=@8
 


5



Tes Kompetensi Subbab

F

'3,#-#0.#*
&#.#/
$6-6
.#5+*#0  



*63E29
CE>36B
3E?H:
>6>:=:<:
D2B27
:?D6?C:D2C
C636C2B 
5
":<2

3E29
CE>36B
3E?H:
H2?8
C2>2
36B3E?H: C6B6?D2<
36B2A2
D2B27
:?D6?C:D2C
3E?H:
H2?8
5:92C:=<2? 6B2A2
A6B32?5:?82?
:?D6?C:D2C
5E2
CE>36B
3E?H:
;:<2 D2B27
:?D6?C:D2C?H2

5 *63E29
CE>36B
3E?H:
D:D:<
56?82?
52H2
 
G2DD >6>2?42B<2?
86=@>32?8
3E?H:
36BEA2
86=@>32?8 C76B:C
3@=2
+6?DE<2?
D2B27
:?D6?C:D2C
3E?H:
H2?8 5:56?82B
@=69
C6C6@B2?8
A252
;2B2<



>6D6B
52B: CE>36B
3E?H:
D6BC63ED


 N
G2DD > 





*63E29
CE>36B
3E?H:
>6>:=:<:
D2B27
:?D6?C:D2C

36== ":<2

3E29
CE>36B
3E?H:
H2?8
C2>2
36B3E?H:
C642B2 C6B6?D2<
36B2A2<29
D2B27
:?D6?C:D2C
3E?H:
H2?8 5:92C:=<2? $5#0#4
  *63E29
CE>36B
3E?H:
>6>2?42B<2?
6?6B8:
<6
C682=2 2B29
A23:=2
;2B2<
A6?56?82B
D6B9252A
CE>36B
3E?H: 5:3E2D
>6?;25:

<2=:
C6>E=2
36B2A2
A6?8EB2?82?
D2B27 :?D6?C:D2C
3E?H:
H2?8
D6B56?82B

Gelombang Bunyi

69

Rangkuman  

6=@>32?8
3E?H:
252=29
86=@>32?8
=@?8:DE5:?2=
H2?8 52A2D
>6B2>32D
52=2>
>65:E>
I2D
A252D
42:B
52?
82C 6A2D
B2>32D
3E?H:
5:
E52B2
36B82?DE?8
A252
;6?:C A2BD:<6=
A6>36?DE<
E52B2
H2?8
5:?H2D2<2?
56?82?

 

C652?8<2?
5:
52=2>
I2D
A252D
36B82?DE?8
A252 >@5E=EC
.@E?8
52?
>2CC2
;6?:C
I2D $

$



A6BC2>22?

 
 C652?8<2?
7B6<E6?C:
?252
2D2C
<6
A252
52G2: 5:?H2D2<2?
56?82?
A6BC2>22?  

               
   (252
A:A2
@B82?2
D6B3E<2
7B6<E6?C:
?252
2D2C
<6 5:?H2D2<2?
56?82?
A6BC2>22?

52AE?
5:
52=2>
I2D
42:B
36B82?DE?8
A252
>@5E=EC E=<
52?
>2CC2
;6?:C
I2D
42:B $

 

 

+:?88:
B6?529
?252
3E?H:
36B82?DE?8
A252
7B6<E6?C: 3E?H:
#E2D
3E?H:
36B82?DE?8
A252
2>A=:DE5@
86D2B2? E5:@C@?:<
252=29
3E?H:
H2?8
36B7B6<E6?C:
2?D2B2
I
C2>A2:
56?82?



I
!?7B2C@?:<
252=29 3E?H:
H2?8
36B7B6<E6?C:
5:
32G29

I
C652?8<2? E=DB2C@?:<
252=29
3E?H:
H2?8
36B7B6<E6?C:
5:
2D2C 


I 6=@>32?8
3E?H:
52A2D
>6?82=2>:
AB@C6C
5:CA6BC: B67B232?8
3E?H:
D6B;25:
;:<2
3652
=:?D2C2? 86=@>32?8?H2
>6BEA2<2?
<6=:A2D2?
3:=2?82?
3E=2D 52B:
C6D6?829
A2?;2?8
86=@>32?8
3E?H: ! 





    


     (6BC2>22?
CEA6BA@C:C:
86=@>32?8
3E?H:

   &   4@C  "   C:?  "    B6<E6?C:
52?
?252
52C2B
A252
52G2:
5:?H2D2<2?
56?82?

70

Mudah dan Aktif Belajar Fisika untuk Kelas XII

$
C652?8<2?
A252
A:A2
@B82?2
D6BDEDEA
7B6<E6?C:
?252 2D2C
<6
5:?H2D2<2?
56?82? $        
E3E?82?
A2?;2?8
<@=@>
E52B2


D6B9252A A2?;2?8
86=@>32?8
3E?H:
  
A252
A6B:CD:G2 B6C@?2?C:
<@=@>
E52B2
252=29 
        2=2>
A6B2>32D2??H2
86=@>32?8
3E?H:
>6>:?529 <2?
6?6B8:
C636C2B
   &
56?82?
&
252=29 2>A=:DE5@
86=@>32?8 *6>2<:?
;2E9
;2B2<
CE>36B
3E?H:
D6B9252A
A6?56?82B :?D6?C:D2C
3E?H:
2<2?
C6>2<:?
<64:= !?D6?C:D2C
3E?H:
5:?H2D2<2?
56?82?
A6BC2>22?    =@8

":<2
D6B52A2D

3E29
CE>36B
3E?H:
D2B27
:?D6?C:D2C 3E?H:
D@D2=?H2
5:?H2D2<2?
56?82?
A6BC2>22?  ?

 


=@8
      

 



   



Peta Konsep '.1/$#0)
608+ >6>3292C

6A2D
)2>32D

#=2C:7:<2C: 6=@>32?8
E?H:

*:72D*:72D 6=@>32?8
E?H:

>6>3292C

>6>3292C

!?D6?C:D2C
E?H:

52=2>

/2D
(252D $

/2D
2:B $



+2B27
!?D6?C:D2C +:?88:
E?H:

!?D6B76B6?C:

#E2D
E?H:

(6>2?DE=2?

-2B?2
E?H:

:7B2
2C

(6=2H2?82?

;:<2
CE>36B 3E?H:?H2
32?H2<






=@8


$



Refleksi Setelah mempelajari bab ini, Anda tentu dapat mengetahui sifat-sifat dasar gelombang bunyi dan klasifikasi gelombang bunyi serta dapat menjelaskan peristiwa resonansi pada pipa organa. Dari materi bab ini, bagian manakah yang Anda anggap sulit? Coba diskusikan dengan teman atau guru Fisika Anda.

Pada bab ini pula, Anda telah mempelajari manfaat dari gelombang bunyi, seperti untuk mengukur kedalaman laut. Coba Anda sebutkan manfaat lain yang Anda rasakan setelah mempelajari bab ini.

Gelombang Bunyi

71

Tes Kompetensi Bab 3  +.+*.#*
4#.#*
4#56
,#7#$#0
8#0)
2#.+0)
5'2#5
�
-'3,#-#0.#*
2#&#
$6-6
.#5+*#0 





(252
CE2DE
C22D
D6B=:92D
<:=2D
52?

56D:<
<6>E5:2? D6B56?82B
CE2B2
8E?DEB?H2
":<2
36C2B
<646A2D2? 4292H2


Q
 
> C
52?
<646A2D2?
3E?H:


> C ;2B2<
2?D2B2
D6>A2D
2C2=
<:=2D
52?
A6?82>2D
 2 
> 5
Q
 
> 3 


> 6
Q
 
> 4  


> *63E29
2=2D
C@?2B
5:8E?2<2?
E?DE<
>6?8E<EB <652=2>2?
=2ED
*6=2?8
G232?8
>6B2>32D
9:?882
<6>32=:
=28:
<6 C@?2B
252=29

C6<@?
":<2
46A2D
B2>32D
86=@>32?8
5: 52=2>
2:B
=2ED
252=29



> C
<652=2>2?
=2ED D6BC63ED
252=29
 2 


> 5  
> 3 
> 6 


> 4 


> E2
3E29
CE>36B
3E?H:
C6A6BD:
D2>A2<
A252
82>32B 36B:<ED
36B86D2B
C642B2
<@96B6?
E?H:
H2?8
?H2B:?8 5:92C:=<2?
5:

;:<2



6?82?
>6?2:<<2?
C642B2 A6B=292?=292?

3E?H:
D6B=6>29
5:56?82B
<6D:<2

N
252=29

4>

4>
52?



4>
":<2
46A2D
B2>32D 3E?H:
5:
E52B2

> C
36C2B
7B6<E6?C:
3E?H:
252=29
 S



r1 P

S2





r2

2  
I 5 
I 3  
I 6 


I 4 
I E2
3E29
#!

52?

5:=6D2<<2?
C6A6BD:
A252 82>32B
36B:<ED
+6B?H2D2
A252
56D62E>
@B56
A6BD2>2
":<2
<646A2D2? 3E?H:
5:
E52B2




> C
7B6<E6?C:
86=@>32?8
3E?H: D6BC63ED
D6BC63ED
252=29
 40 m

A





3


N


2 
I 5 
I 3


I 6 
I 4 
I %@5E=EC
E=<
CE2DE
I2D
42:B
H2?8
>6>:=:<:
>2CC2
;6?:C


8B2> 4>
252=29

Q
 
& > 
6A2D
B2>32D 86=@>32?8
52=2>
I2D
42:B
D6BC63ED
252=29
 2 


> C 5 


> C 3 


> C 6 


> C 4 


> C

Mudah dan Aktif Belajar Fisika untuk Kelas XII



 $

 $

5


N


6





 $

 $ *6@B2?8
A6?6B32?8
H2?8
A6C2G2D
D6B32?8?H2
>6?E;E <6
>6?2B2
32?52B2
>6?56?82B
3E?H:
C:B:?6
>6?2B2 56?82?
7B6<E6?C:



I
":<2
C22D
:DE
C:B:?6
D6BC63ED >6>2?42B<2?
3E?H:
56?82?
7B6<E6?C:



I
52? 46A2D
B2>32D
3E?H:
5:
E52B2


> C
<646A2D2? A6C2G2D
D6B32?8
D6BC63ED
252=29
 2  
<> ;2> 5
<> ;2> 3


<> ;2> 6 
<> ;2> 4 
<> ;2>

4

B

72

2 D

9m



*63E29
CE>36B
3E?H:
36B7B6<E6?C:

< I
36B86B2< =2?8CE?8
<6
2B29
C6@B2?8
A6?56?82B
56?82?
<6=2;E2?


<2=:
<6=2;E2?
3E?H:
B6<E6?C:
3E?H:
H2?8 5:D6B:>2?H2
252=29
 2  

< I 5 
< I 3 
< I 6 
< I 4 
< I 1#.
!
9
!
 ?52
C652?8
36B86B2<
>6?56<2D:
C63E29
CE>36B
3E?H: *
D:52<
36B86B2<
H2?8
36B252
;2E9
5:
56A2?
?52
52? C652?8
>6>2?42B<2?
3E?H:
36B7B6<E6?C:


<6
C682=2 2B29
82<
;2E9
5:
36=2<2?8
?52
D6B52A2D
C63E29 5:?5:?8
A6>2?DE=

E?H:
H2?8
?52
56?82B =2?8CE?8
52B:
*
56?82?
7B6<E6?C:
C
52?
H2?8
36B2C2= 52B:
A2?DE=2?
5:

56?82?
7B6<E6?C:
 5
<6D:<2 36B:?D6B26?:>3E=<2?
=2H2?82?
<2B6?2 C

5

  3 D:52<
2<2?
>6?:>3E=<2?
=2H2?82?
<2B6?2 C

5


  4 2<2?
>6?:>3E=<2?
=2H2?82?
52?
>6>6?E9: C

5

  5 2<2?
>6?:>3E=<2?
=2H2?82?
52?
>6>6?E9: C



5 6 2<2?
>6?:>3E=<2?
=2H2?82?
52?
>6>6?E9: C


5
  *63E29
CE>36B
3E?H:
36B86B2<
56?82?
<646A2D2?

> C >6?E;E
<6
A6?56?82B
H2?8
5:2>
C69:?882
7B6<E6?C: H2?8
5:56?82B
@=69
A6?56?82B
252=29

":<2
CE>36B 3E?H:
D6BC63ED
5:2>
C652?8<2?
A6?56?82B
36B86B2< 56?82?
<646A2D2?

> C
>6?56<2D:
CE>36B
3E?H: 7B6<E6?C:
H2?8
5:56?82B
@=69
A6?56?82B
252=29
 
":<2 46A2D
B2>32D
3E?H:
5:
E52B2
$
> C
A6B32?5:?82?



 252=29






  (6=2H2?82?
3E?H:
D6B;25:
<2B6?2
252?H2
:?D6B76B6?C:
 2 =63:9
52B:

86=@>32?8
H2?8
C2>2
7B6<E6?C:?H2 3
86=@>32?8
H2?8
36B=2G2?2?
2B29
86D2B?H2 4
86=@>32?8
H2?8
C2>2
7B6<E6?C:?H2 5
86=@>32?8
H2?8
36C2B
3652
7B6<E6?C:?H2 6
86=@>32?8
H2?8
<64:=
3652
7B6<E6?C:?H2  ":<2
5E2
3E29
CE>36B
3E?H:
>2C:?8>2C:?8
56?82? 7B6<E6?C:



I
52?



I
36B3E?H:
C6B6?D2< 2<2?
D:>3E=
A6=2H2?82?
3E?H:
56?82?
7B6<E6?C:
 2
I 5 


I 3 
I 6 


I 4 
I   *6ED2C
52G2:
>6?892C:=<2?
B6C@?2?C:
A6BD2>2
A252



4>
":<2
A2?;2?8
<2G2D

4>
52?
36B86D2B A252
?252
2D2C
<65E2
46A2D
B2>32D
86=@>32?8
DB2?C F6BC2=
52=2>
<2G2D
252=29

46A2D
B2>32D
3E?H:
5: E52B2


> C 2  
> C 3 
> C 4 
> C 5 
> C 6 
> C   E2
3E29
52G2:
32;2
H2?8
:56?D:<
>6>36B:<2?
?252 52C2B
56?82?
7B6<E6?C:



I
":<2
D682?82?
52=2> C2=29
C2DE
52G2:
5:D2>329

7B6<E6?C:
A6=2H2?82? H2?8
D6B;25:
252=29
 2
I 5 
I 3
I 6 
I 4 
I  *63E29
D23E?8
86=2C
B6C@?2?C:
H2?8
36B:C:
2:B
>6>:=:<: =E2C
A6?2>A2?8

4> 
(252
A@C:C:
D6BC63ED
D6B;25: B6C@?2?C:
A6BD2>2
*6;E>=29



4>
2:B
5:3E2?8
<6=E2B C69:?882
D6B;25:
B6C@?2?C:
<65E2
":<2
<6=2;E2?
3E?H: 5:
E52B2




> C
7B6<E6?C:
82BAED2=2
H2?8
5:A2<2: 252=29
 2 
I 5 
I 3


I 6 


4 
I  (252
A6B4@322?
%6=56
5:8E?2<2?
C6ED2C
36?2?8
H2?8 A2?;2?8?H2

>6D6B
52?
36B>2CC2

8B2>
":<2
3632? C636C2B

<8
5:82?DE?8<2?
A252
36?2?8
D6BC63ED <646A2D2?
A6B2>32D2?
86=@>32?8
A252
36?2?8
252=29 



> C  2 
> C 5 
> C 3 
> C 6 
> C 4
> C  *6ED2C
52G2:
>6?892C:=<2?
?252
52C2B

":<2
52G2: D6BC63ED
5:A6?56<<2?

4>
D2?A2
>6?8E329
D682?82? 5:92C:=<2?
7B6<E6?C:
 

":<2
52G2:
5:A6?56<<2?
4>
=28:
7B6<E6?C:
H2?8
5:92C:=<2?
252=29
 2  5 
 3  
 6 
 4 




 *6ED2C
52G2:
H2?8
A2?;2?8?H2

4>
52?
36B>2CC2
 8B2>
5:;6A:D
<65E2
E;E?8?H2
52?
5:B6?D2?8<2?
56?82? D682?82?



&
B6<E6?C:
?252
2D2C
A6BD2>2
H2?8 5:92C:=<2?
252=29
 2  
I 5


I 3 
I 6 
I 4 
I  1,2 m

f = 340 Hz W









#6D:<2
82BAED2=2
5:86D2B<2?
A252
52G2:
D6B;25: 86=@>32?8
CD2C:@?6B
C6A6BD:
D2>A2<
A252
82>32B #646A2D2?
B2>32D
86=@>32?8
A252
52G2:
252=29
 2 
> C 5  
> C 3 
> C 6 
> C 4 
> C *63E29
A:A2
@B82?2
D6BDEDEA
>6>:=:<:
A2?;2?8

4> ":<2
46A2D
B2>32D
3E?H:
5:
E52B2

> C
36C2B
7B6<E6?C: ?252
2D2C
<65E2
252=29
 2 
I 5 
I 3  
I 6 
I 4 
I &252
2D2C
A6BD2>2
A:A2
@B82?2
D6B3E<2
H2?8 A2?;2?8?H2

4>
36B6C@?2?C:
56?82?
A:A2
@B82?2 D6BDEDEA
":<2
A252
C22D
36B6C@?2?C:
;E>=29
C:>AE=
A252 <65E2
A:A2
C2>2
A2?;2?8
A:A2
@B82?2
D6BDEDEA
D6BC63ED 52=2>
4>
252=29
 2

5 

3

6 

4 

! 

#810
 (252
A:A2
@B82?2
D6B3E<2
?252
2D2C
<65E2
5:92C:=<2? A2?;2?8
86=@>32?8
C636C2B
%
52?
A252
A:A2
@B82?2 D6BDEDEA
?252
2D2C
<65E2
5:92C:=<2?
A2?;2?8
86=@> 32?8
C636C2B
&
":<2
A2?;2?8
<65E2
A:A2
C2>2
?:=2: A6B32?5:?82?
&
52?
%
252=29
 2

 3

 4 

5 

 6 

 +2B27
:?D6?C:D2C
3E?H:
C63E29
>6C:?
B2D2B2D2

5 ":<2



>6C:?
5:9:5EA<2?
36BC2>2
D2B27
:?D6?C:D2C?H2 252=29
 2
5 3 
5 4 
5 5 
5 6 
5

Gelombang Bunyi

73

 ":<2
C63E29
C6A652
>@D@B
>6=6G2D:
C6C6@B2?8
>@D@B D6BC63ED
2<2?
>6?:>3E=<2?
D2B27
:?D6?C:D2C
+!
C636C2B 
5
":<2
@B2?8
D6BC63ED
5:=6G2D:

C6A652
>@D@B
;6?:C H2?8
C2>2
C6<2=:8EC
D2B27
:?D6?C:D2C?H2
252=29
 2 
5 5 
5 3 
5 6 


5 4 
5  +2B27
:?D6?C:D2C
3E?H:
C63E29
>6C:?
252=29

5 56?82?
24E2?
:?D6?C:D2C
2>32?8
A6?56?82B2?
C636C2B  N
- >
":<2
D2B27
:?D6?C:D2C
5:
52=2>
BE2?8
A23B:< H2?8
>6?88E?2<2?
C6;E>=29
>6C:?
D6BC63ED
252=29 
5
;E>=29
>6C:?
H2?8
5:8E?2<2?
252=29
 

#7#$.#*
2'35#08##0
$'3+-65
&'0)#0
5'2#5



*63E29
82BAED2=2
36B7B6<E6?C:

I
5:86D2B<2?
5: A6B>E<22?
D23E?8
B6C@?2?C:
(252
C22D
<@=@>
E52B2 >6?42A2:
<6D:?88:2?

4>
D6B;25:
B6C@?2?C:
H2?8 A6BD2>2
6B2A2<29
46A2D
B2>32D
3E?H:
C22D
:DE *6ED2C
<2G2D
H2?8
A2?;2?8?H2

4>
5::<2D
<65E2 E;E?8?H2
#6>E5:2?
<2G2D
5:86D2B<2?
=@?8:DE5:?2= 6A2D
B2>32D
86=@>32?8
H2?8
5:92C:=<2?



> C %@5E=EC
.@E?8
<2G2D


Q
 
& > 
:DE?8=29 2 7B6<E6?C:
86D2B?H2 3 >2CC2
;6?:C
<2G2D %@3:=

>6?56<2D:
A6?82>2D

5:2>
56?82? <646A2D2?

> C
C2>3:=
>6>3E?H:<2?
C:B6?6 36B7B6<E6?C:
 
I
(252
C22D
:DE
;E82
>@3:=
>6?56<2D:

52B:
2B29
H2?8
36B=2G2?2?
56?82?
 A252
<646A2D2?

> C
C2>3:=
>6>3E?H:<2?
C:B:?6 36B7B6<E6?C:

I
":<2
46A2D
B2>32D
3E?H:
5:
E52B2 C22D
:DE



> C
36B2A2<29
7B6<E6?C:
=2H2?82?
H2?8 5:56?82B
@=69
 *6ED2C
<2G2D
32;2
H2?8
>2CC2?H2

8
52?
A2?;2?8

> 5:36B:
D682?82?

&
+6?DE<2? 2 46A2D
B2>32D
A2?;2?8
86=@>32?8
DB2?CF6BC2= C6A2?;2?8
<2G2D 3 A2?;2?8
86=@>32?8
52?
7B6<E6?C:
?252
52C2B?H2 4 7B6<E6?C:
?252
2D2C
<6C2DE
52?
<65E2







74

Mudah dan Aktif Belajar Fisika untuk Kelas XII

2

5

3 

6 

4 

 +2B27
:?D6?C:D2C
3E?H:
A252
C63E29
;6?56=2
D6B3E<2 H2?8
=E2C?H2

>
252=29

5
":<2
?:=2:
2>32?8 3E?H:
 N
- 4> 
52H2
2<ECD:<
H2?8
>2CE<
>6=2=E: ;6?56=2
D6BC63ED
252=29
 2  N
3  N
4  N
5  N
6  N
-



B6<E6?C:
?252
2D2C
A6BD2>2
52B:
A:A2
@B82?2
D6B3E<2 


I
":<2
46A2D
B2>32D
3E?H:
5:
E52B2

> C 9:DE?8=29 2 7B6<E6?C:
?252
52C2B?H2 3 7B6<E6?C:
?252
2D2C
<6D:82  6A2D
B2>32D
3E?H:
52=2>
C63E29
A:A2
@B82?2
252=29


> C
":<2
7B6<E6?C:
?252
52C2B
A:A2
@B82?2
252=29 
I
36B2A2
A2?;2?8
>:?:>E>
A:A2
@B82?2
E?DE< <2CEC 2 A:A2
@B82?2
D6B3E<2 3 A:A2
@B82?2
D6BDEDEA  E2
ED2C
52G2:
32;2
C6;6?:C
>6>36B:<2?
?252
52C2B 
I
":<2
A2?;2?8
C2=29
C2DE
52G2:
5:D2>329
 36B2A2<29
7B6<E6?C:
=2H2?82?
H2?8
D6B;25:  E2
ED2C
52G2:
32;2
H2?8
:56?D:<
>6>36B:<2?
?252 52C2B
56?82?
7B6<E6?C:




I
":<2
D682?82?
52=2> C2=29
C2DE
52G2:
5:D2>329

36B2A2
7B6<E6?C: =2H2?82?
52G2:
H2?8
>E?8<:?  *66<@B
D2G@?
H2?8
>6?56?8E?8
A252
;2B2<

> >6>:=:<:
D2B27
:?D6?C:D2C

5
":<2
252



D2G@? >6?56?8E?8
A252
;2B2<
%
D2B27
:?D6?C:D2C?H2

5 D6?DE<2?
;2B2<
%   (252
C63E29
D23E?8
B6C@?2?C:
D6B52A2D
<@=@>
E52B2 A252
"
*6=:C:9
A2?;2?8
<@=@>
B6C@?2?C:


4> #646A2D2?
3E?H:
A252
CE9E
L
252=29

> C :DE?8
"
;:<2
7B6<E6?C:
B6C@?2?C:


I

Bab

4 Sumber: Physics for Scientists & Engineers with Modern Physics, 2000

Berbagai jenis kapasitor yang berfungsi untuk menyimpan muatan listrik.

Medan Listrik, Potensial Listrik, dan Kapasitor Hasil yang harus Anda capai: menerapkan konsep kelistrikan dan kemagnetan dalam berbagai penyelesaian masalah dan produk teknologi. Setelah mempelajari bab ini, Anda harus mampu: memformulasikan gaya listrik, kuat medan listrik, fluks, potensial listrik, energi potensial listrik, serta penerapannya pada keping sejajar.

:
-(+
?52
D6?DE
D6=29
>6>A6=2;2B:
>6?86?2:
=:CDB:<
':CDB:<
CE529 252
C6;2<
;282D
B2I2
:?:
=29:B
'652<2?
A6D:B
>6BEA2<2?
4@?D@9
<636B2522? =:CDB:<
<2B6?2
A6D:B
252=29
92C:=
A6=6A2C2?
>E2D2?
=:CDB:<
5:
2G2?

2?I2<
C6<2=:
2A=:<2C:
=:CDB:<
52=2>
<69:5EA2?
C692B: 92B:
.6=6F:C: B25:@
<@>AED6B
52?
2=2D 2=2D
6=66?88E?2<2?
=:CDB:<
C63282: CE>36B
6?6B8:?I2
-6D6=29
5:D6>E<2??I2
=:CDB:<
32?I2<
5:D6>E<2? D6
C63E29 B25:@
:
52=2>?I2
?52
52A2D
>6=:92D
36B>242> >242>
<@>A@?6? =:CDB:<
2A2D<29
?52
>6?I63ED<2?
?2>2 ?2>2
<@>A@?6?
=:CDB:<
I2?8 D6B52A2D
5:
52=2>
B25:@ :
2?D2B2
C6<:2?
32?I2<
<@>A@?6?
I2?8
D6B52A2D
5:
52=2>
B25:@ ?52
A2CD:
>6?6>E<2?
<2A2C:D@B
A2<29
<2A2C:D@B
:DE
A2
<68E?22? <2A2C:D@B
282:>2?2
42B2
<6B;2
C63E29
<2A2C:D@B ?52
A2CD:
:?8:?
>6?86D29E:
;2G232?
52B:
A6BD2?I22? A6BD2?I22? D6BC63ED
/?DE<
:DE
D6>E<2?
;2G232??I2
56?82?
>6>A6=2;2B:
323
:?:

A. Muatan Listrik B. Medan Listrik C. Energi Potensial Listrik dan Potensial Listrik D. Kapasitor

Medan Listrik, Potensial Listrik, dan Kapasitor

75

Tes Kompetensi Awal "*'*192
2*25*1&/&6.
0437*5
*)&3
.786.0
!48*37.&1
.786.0
)&3
&5&7.846
0*6/&0&31&-
74&174&1
'*6.098
)&1&2 '909
1&8.-&3

A2<29
C6D:2A
36?52
>6>:=:<:
>E2D2?
=:CDB:<  A2
I2?8
?52
<6D29E:
D6?D2?8
<2A2C:D@B
A2  A2
C2;2<29
A6?IECE?
52B:
C63E29
2D@>
%6=2C<2? C2;2<29
7E?8C:?I2
282:>2?2
AB:?C:A
52C2B
<6B;2?I2 A6?IECE? A6?IECE??I2
D6BC63ED

 A2
I2?8
5:>2652?
=:CDB:<

A. Muatan Listrik

Sumber: Conceptual Physics, 1998

Gambar 4.1 Mesin fotokopi

batang karet

batang kaca

Gambar 4.2 Setelah batang karet digosok kain wol dan batang kaca digosok kain sutra kedua batang tersebut tarikmenarik.

batang karet

batang karet

Gambar 4.3 Setelah kedua batang karet digosok kain wol dan didekatkan, kedua batang karet tolak-menolak.

+B@C6C
5EA=:<2C:
?2C<29
@=69
>6C:?
7@D@<@A:
>6BEA2<2?
C2=29
C2DE 4@?D@9
2A=:<2C:
<@?C6A
=:CDB:<
CD2D:C
52=2>
<69:5EA2?
C692B: 92B:

+6>36?DE<2?
C2=:?2?
92C:=
7@D@<@A:
<2B6?2
A6?6>A6=2?
C6B3E<
2D2E
D@?6B I2?8
36B>E2D2?
?682D:7
A252
A6B>E<22?
<6BD2C
@?D@9
=2:?
86;2=2
=:CDB:< CD2D:C
I2?8
=63:9
C656B92?2
252=29
>6?6>A6=?I2
B2>3ED
2D2E
A@D@?82? <6BD2C
A252
A6?882B:C
A=2CD:<
I2?8
D6=29
5:8@C@<
<2:?
G@=

&65E2
86;2=2
!:C:<2
D6BC63ED
>6?E?;E<<2?
329G2
5E2
3E29
36?52
52A2D C2=:?8
>6?2B:<
282:>2?2
A2B2
29=:
7:C:<2
>6?;6=2C<2?
86;2=2
D6BC63ED +6B92D:<2?
<65E2
A6B:CD:G2
A252
&2'&6

52?
&2'&6


&6D:<2
C63E29
32D2?8
<2B6D
5:8@C@<
56?82?
<2:?
G@=
<6>E5:2?
5:56<2D<2? A252
32D2?8
<242
I2?8
5:8@C@<
<2:?
CEDB2
D6B?I2D2
<65E2
32D2?8
D6BC63ED D2B:< >6?2B:<
-632=:6?@=2<
/?DE<
>6?;6=2C<2?
86;2=2
D6BC63ED
52A2D
5:2?882A 329G2
A6?88@C@<2?
<2:?
G@=
52?
<2:?
CEDB2
A252
32D2?8
>6>36B:<2? >E2D2?
=:CDB:<
A252
32D2?8
D6BC63ED

6B52C2B<2?
86;2=2
7:C:<2
D6C63ED
;6=2C
329G2
>E2D2?
A252
<242 52?
>E2D2?
A252
32D2?8
<2B6D
36B3652
;6?:C?I2
*3/&2.3
6&301.3   O  
>6?I63ED<2?
329G2
>E2D2?
=:CDB:<
A252
32D2?8
<242
C63282: >E2D2?
A@C:D:7
52AE?
>E2D2?
A252
32D2?8
<2B6D
5:C63ED
>E2D2?
?682D:7

6?52
D6BCECE?
2D2C
B:3E2?
329<2?
;ED22?
2D@>
D@>
D6BCECE?
2D2C AB@D@?
?6EDB@?
52?
6=6
:?D: 2D@>
C652?8<2?
6=6
%E>=29
AB@D@?
52?
6=6
C63E29
2D@>
252=29
C2>2
$DE=29
C6323?I2
2D@>
36BC:72D
?6DB2=

-63282:
4@?D@9
2D@>
9:5B@86?
>6?82?5E?8
C2DE
AB@D@?
52?
C2DE
6=6
D@>
36B>E2D2?
A@C:D:7
252=29
2D@>
I2?8
>6>:=:<:
;E>=29
AB@D@?
=63:9 32?I2<
52B:
;E>=29
6=6
36B>E2D2?
?682D:7
252=29 2D@>
I2?8
>6>:=:<:
;E>=29
AB@D@?
=63:9
C65:<:D
52B:A252
;E>=29
6=6
6?52
36B>E2D2?
?6DB2=
D6BCECE?
2D2C
2D@> 2D@>
I2?8
D:52<
36B >E2D2?
?6DB2=
D@>
?6DB2=
252=29
2D@>
I2?8
>6>:=:<:
;E>=29
AB@D@? C2>2
56?82?
;E>=29
6=6:C2=?I2
2D@>
9:5B@86?
C6A6BD:
D6B=:92D A252
&2'&6


1. Interaksi Elektrostatis antara Dua Muatan Listrik

Gambar 4.4 Pada atom hidrogen terdapat satu proton dan satu elektron.

76

%6?:C
:?D6B2242>
I2:DE 2 D2B:< >6?2B:<
2?D2B2
>E2D2? >E2D2?
D:52<
C6;6?:C 3 D@=2< >6?@=2<
2?D2B2
>E2D2? >E2D2?
C6;6?:C

-&61*7
9,978.3
49142'
 O  
>6?8E<EB
36C2B?I2
D2B:< 2?
52?
D@=2<2?
=:CDB:<
C642B2
<E2?D:D2D:7
$2
;E82
>6?I:>AE=<2?
9E<E> I2?8
>6?82DEB
D2B:<2?
52?
D@=2<2?
=:CDB:<
D6BC63ED
#E<E>
D6BC63ED 5:<6?2=
56?82?
1'1)
+1(+)
#E<E>
@E=@>3
>6?I2D2<2?
329G2

Mudah dan Aktif Belajar Fisika untuk Kelas XII

5
0.%')!*.%'
01
0+(')!*+('
*0.
1
1$
)10*
(%/0.%' !/.*5
!.* %*#
(1.1/
!*#*
$/%(
'(%
!/.
'! 1
)10*
0!./!10
* !.* %*#
0!.(%'
!*#*
'1 .0
&.'
*0.
'! 1
)10*
0!./!10  +6B92D:<2?
&2'&6

52?
&2'&6

-642B2
>2D6>2D:C
82I2 D2B:< >6?2B:<
2D2E
D@=2< >6?@=2<
82I2
@E=@>3
52=2>
F2<E>
52A2D 5:DE=:C
C63282:
36B:<ED



-
-


2D2E


F2<E>  '   . 

O 

,  

F q2 r

Gambar 4.5 Interaksi muatan listrik berlainan jenis akan tarik-menarik F

F q1

q2

F

"2I2
@E=@>3
I2?8
D6B;25:
52=2>
CE2DE
>65:E>
2D2E
3292?
>6>:=:<: A6BC2>22?
82I2
@E=@>3
C63282:
36B:<ED

3292?

F q1

-
-  
 . 

O

F q1

r

q2

Gambar 4.6 Interaksi muatan listrik sejenis akan tolak-menolak

%:<2
+6BC2>22?
O 2
52?
+6BC2>22?
O 3
5:823E?8<2?
5:52A2D

F2<E>    . 3292? 

O2

-69:?882
5:52A2D
9E3E?82?
82I2
@E=@>3
A252
3292?
52?
82I2
@E=@>3 A252
F2<E>
C63282:
36B:<ED

3292? 

F2<E> .


O3

&6D6B2?82?   82I2
@E=@>3
) -  >E2D2?
=:CDB:<



<@?CD2?D2
5:6=6    A6B>:D:F:D2C
BE2?8
92>A2
  

P
O 
)>  r  A6B>:D:F:D2C
B6=2D:7
3292? .  ;2B2<
2?D2B>E2D2?
>

' 

Tugas Anda 4.1

Contoh 4.1 E2
<6A:?8
=@82>
I2?8
D6B3E2D
52B:
3292?
C2>2
5:36B:
>E2D2?
C2>2
36C2B

2 6B2A2<29
>E2D2?
5:
C6D:2A
<6A:?8
;:<2
5:<6D29E:
82I2
@E=@>3
C636C2B

)
52? ;2B2<
2?D2B<6A:?8
=@82>

> 3 %:<2
<65E2
<6A:?8
36B252
5:
52=2>
3292?>65:E>
56?82?  . 

36B2A2<29 36C2B
82I2
@E=@>3?I2 &:&' :<6D29E: 


) .



> - -  2 

'

56?82?
-

-

.  . .  - -

' ' - 
.

Anda masih ingat alat yang dinamakan dengan elektroskop? Apakah fungsi alat tersebut? Bagaimanakah cara kerjanya? Coba Anda cari informasi dari internet atau buku-buku referensi.


) 


>


P
O


   '    )>   

Medan Listrik, Potensial Listrik, dan Kapasitor

77

-

-

-

   %25:
36C2B
>E2D2?
=:CDB:<
5:
C6D:2A
<6A:?8
252=29

 

3

3292?   F2<E> 

)   ) B  

%25:
36C2B
82I2
@E=@>3?I2
252=29
)






2. Resultan Gaya Coulomb F12

q1

q2

r1

F23 F21

F32

q3

r2

Gambar 4.7 Resultan gaya Coulomb

+6B92D:<2?
&2'&6

"2>32B
D6BC63ED
52A2D
5:56C3
I2?8
5:2=2>:
>E2D2?
-
2<:32D A6?82BE9
>E2D2?
-

252=29
82I2
@E=@>3
I2?8
5:2=2>:
>E2D2?
- 2<:32D
A6?82BE9
>E2D2?
-

252=29
82I2
@E=@>3
I2?8
5:2=2>:
>E2D2? -
2<:32D
A6?82BE9
>E2D2?
-

252=29
82I2
@E=@>3
I2?8
5:2=2>: >E2D2?
-
2<:32D
A6?82BE9
>E2D2?
-

,6CE=D2?
82I2
@E=@>3
5:
CE2DE
D:D:<
5:BE>EC<2?
C63282:
36B:<ED

       

  *

O

6B2BD:
B6CE=D2?
82I2
@E=@>3
I2?8
5:2=2>:
>E2D2?
-
A252
&2'&6


252=29

- 


 2=2>
92=
:?:
  '

- -  -
52?
  '   

 . .

Contoh 4.2

Tokoh Charles Coulomb (1736– 1806)

.:82
3E29
A2BD:<6=
36B252
A252
C2DE
82B:C
=EBEC
C6D:2A
A2BD:<6=
36B>E2D2?
     
52?
 
?D2BA2BD:<6=
36B;2B2<
C2>2
I2:DE

4>

2 .6?DE<2?
36C2B
2B29
82I2
A252
A2BD:<6=
36B>E2D2?
 

3 :
>2?2
A2BD:<6=
 
5:=6D2<<2?
282B
82I2
@E=@>3
A252
A2BD:<6=
D6BC63ED ?@= &:&' :<6D29E:
-
   
-
   
-
   2 '6D2<
>E2D2?
- 
-
52?
-
C6A6BD:
82>32B
36B:<ED

%2B2<
2?D2B2
<65E2
A2BD:<6=
C2>2
36C2B
I2:DE
. 

.


4>
52AE?
  >6BEA2<2?
82I2
@E=@>3
92C:=
:?D6B26?2B:<
-
52?
-
C652?8<2?
 >6BEA2<2?
82I2
@E=@>3
92C:=
:?D6B26?2B:<
-
52?
-
6?82? >6>AB65:2<2
B6CE=D2?
82I2
@E=@>3?I2
>6>6?E9: A6BC2>22?
36B:<ED

r12 = 10 cm

q1 Coulomb adalah ahli Fisika kelahiran Angouleme, Prancis. Dia merupakan orang pertama yang menunjukkan perilaku bahwa muatan listrik saling tolak satu sama lain. Namanya kemudian digunakan sebagai satuan muatan listrik. Sumber: Jendela IPTEK, 1995

78

r23 = 10 cm

F12

q2



 
O


'  



F23

q3

- -  - '    .  .

' -
-  -   .


  
)>     
    


Mudah dan Aktif Belajar Fisika untuk Kelas XII



> 

           
)

3

82B
?:=2:
B6CE=D2?
82I2
@E=@>3
5:
-
C2>2
56?82?
?@=
>2<2
36C2B
 
92BEC C2>2
56?82?

C69:?882
A6BC2>22??I2
>6?;25: '- '- .               .  . .  -

.       .    .   .  (:C2=<2?
>E2D2?
-
5:=6D2<<2?
4
>
52B:
>E2D2?

- 
>2<2 . 

4
> 52? .

 
O
4
> .

 
. 



O
4

 
4

4

 
>

Kata Kunci • • • • •

proton elektron neutron atom netral gaya Coulomb

%25:
282B
82I2
@=@E>3
A252
-
C2>2
56?82?
?@=
>E2D2?
-
5:=6D2<<2?  
>
52B:
>E2D2?
-

Contoh 4.3 E2
3E29
A2BD:<6=
>2C:?8 >2C:?8
36B>E2D2?
 
5:82?DE?8
A252
D2=:
C6A6BD:
A252 82>32B
%:<2
D2=:
5:2?882A
D:52<
36B>2CC2
D6?DE<2?
36C2B?I2 2 D682?82?
D2=:
 T 60° T 3 >2CC2
C6D:2A
A2BD:<6=
;:<2
#


>C

&:&' :<6D29E:
- 
-
-

P
O
 q q #


>C
'


P

)>

30 cm %:<2
82I2 82I2
I2?8
36<6B;2
A252
A2BD:<6= A2BD:<6=
D6BC63ED 5:82>32B<2?
2<2?
D6B=:92D
C6A6BD:
82>32B
36B:<ED

30° T

T cos 30°

F T sin 30° w

?52
52A2D
>6?6?DE<2?
36C2B
D682?82?
D2=:

52?
>2CC2
)
92?I2
56?82?
>6?:?;2E C6D6?829
328:2?
52B:
C:CD6>

2 -:CD6>
5:<2D2<2?
52=2>
<62522?
C6D:>32?8
;:<2 4


2D2E



C:?
 L 2B:
!*67&2&&3
 < 
5:<6D29E:
329G2


' - 2<2?
5:A6B@=69

C:?
 L

 .







   Nm2 / C 2   O C

    m





' - >2<2 .

Tantangan untuk Anda Perhatikan gambar berikut. q

r








) 3

r

 5


52?

4@C
 L

)
#

 
)

)

>C

)


<8 

%25:
>2CC2
C6D:2A
A2BD:<6=
252=29

<8

–q

+q


Hitunglah resultan gaya F yang bekerja pada muatan q.

Medan Listrik, Potensial Listrik, dan Kapasitor

79

Tes Kompetensi Subbab

A

*6/&0&31&-
)&1&2
'909
1&8.-&3

6B2A2<29
36C2B
82I2
D2B:< >6?2B:<
2?D2B2
AB@D@?
52? 6=6
C63E29
2D@>
9:5B@86?
I2?8
36B;2B: ;2B: 2D@>

K
 
P O
>
  +6B92D:<2?
82>32B
36B:<ED

q1

q2 a





(E2D2?

I2?8
36C2B?I2
  
5:=6D2<<2?
A252
82B:C 9E3E?8
2?D2B2
>E2D2?

I2?8
36C2B?I2
O  
52? >E2D2?


I2?8
36C2B?I2
  
(E2D2?

D6B=6D2<
 4>
52B:

52?

4>
52B:

%:<2
;2B2<



4> D6?DE<2?
36C2B?I2
B6CE=D2?
82I2
I2?8
36<6B;2
A252 2 >E2D2?


B 3 >E2D2?




q3 a

%:<2
5:<6D29E:
- 
-
  
-
O  
52?



> D6?DE<2?
B6CE=D2?
82I2
@E=@>3
A252
-

.:82
3E29
A2BD:<6=
36B252
52=2>
C2DE
82B:C
=EBEC
>2C:?8 >2C:?8
36B>E2D2?
O    
   
52?
O  

?D2BA2BD:<6=
>6>:=:<:
;2B2<
I2?8
C2>2

>

2 .6?DE<2?
36C2B
52?
2B29
82I2
A252
A2BD:<6= 36B>E2D2?
  

3 :
>2?2
A2BD:<6=
I2?8
36B>E2D2?
   5:=6D2<<2?
282B
82I2
@E=@>3
A252
A2BD:<6= D6BC63ED
36B?:=2:
?@= E2
3E29
A2BD:<6=
>2C:?8 >2C:?8
36B>E2D2?
   5:82?DE?8
A252
D2=:
:C@=2D@B
C6A6BD:
82>32B
36B:<ED

.2=:
5:2?882A
D:52<
36B>2CC2

.6?DE<2?
36C2B?I2 T 74° 2 D682?82?
D2=:
 3 >2CC2
C6D:2A
A2BD:<6=
;:<2 #


>C

+q +q

A



-63E29
36?52
36B>2CC2

8
52?
36B>E2D2?
-

   5:82?DE?8<2?
A252
C6ED2C
D2=:
B:?82?
I2?8
>2CC2?I2 52A2D
5:232:<2?
.6A2D
5:
C636=29
<2?2?
36?52
A252 ;2B2<

4>
5:=6D2<<2?
-

O  
I2?8
>6?I6323<2? A@C:C:
36?52
>6?;25:
C6A6BD:
A252
82>32B
36B:<ED

T

q

q' 15 cm

%:<2


P 
)>
52?
#


>C

90 cm



C



D6?DE<2?=29 2




36C2B
CE5ED

52? 3




36C2B
D682?82?
A252
D2=:

282:>2?2<29
;2B2<
2?D2B2
5E2
A2BD:<6=
36B>E2D2? 92BEC
5:E329
282B
82I2
D@=2<
<65E2?I2
>6?;25:

2




<2=:
C6>E=2 3





<2=:
C6>E=2



B. Medan Listrik ?52
>E?8<:?
A6B?29
>6?56?82B
329G2
C6D:2A
>E2D2?
I2?8 5:=6D2<<2?
5:
CE2DE
526B29
2<2?
>6>:=:<:
>652?
=:CDB:<
5:
C6<:D2B?I2

%:<2
C63E29
>E2D2?
E;:
-
5:=6D2<<2?
A252
526B29
D6BC63ED
>E2D2? D6BC63ED
2<2?
>6?82=2>:
82I2
@E=@>3
&E2D
>652?
=:CDB:<
5:
CE2DE
D:D:< 5:567:?:C:<2?
C63282:
82I2
@E=@>3
A6B
C2DE2?
>E2D2?
I2?8
5:2=2>:
@=69 C63E29
>E2D2?
5:
D:D:<
D6BC63ED
-642B2
>2D6>2D:C
5:DE=:C<2?
C63282: 36B:<ED

 -

O


2D2E
   .  . 

O



56?82?
  >2<2

'--  . '

80

Mudah dan Aktif Belajar Fisika untuk Kelas XII

&6D6B2?82?  
<E2D
>652?
=:CDB:<
) - 
>E2D2?
=:CDB:<
 . 
;2B2<
2?D2B>E2D2?
> ' 

P

)>

1. Garis-Garis Gaya

2

/?DE<
>6?882>32B<2?
>652?
=:CDB:<
52A2D
;E82
5:=E<:C<2?
52=2> 36?DE<
82B:C 82B:C
82I2
(%*!/
+"
"+.!
#E3E?82?
2?D2B2
82B:C 82B:C
82I2 52?
F6652?
=:CDB:<
252=29
C63282:
36B:<ED

2 06652?
5:
CE2DE
D:D:<
A252
82B:C
82I2
>6?I:?88E?8
82B:C 82I2
5:
D:D:<
D6BC63ED

3 2?I2A2?8
I2?8
D682<
=EBEC
56?82? 82B:C 82B:C
D6BC63ED
252=29
C632?5:?8
56?82?
36C2B?I2
>652?
=:CDB:<


3

Contoh 4.4 +2BD:<6=
36B>E2D2?
   52?
36B>2CC2

>8
D6B2AE?8
3632C
52=2>
>652?
=:CDB:< C6A6BD:
A252
82>32B
36B:<ED

q

4

%:<2
#


>C
D6?DE<2?=29
36C2B?I2
<E2D
>652?
=:CDB:<
I2?8
>6>6?82BE9:
A2BD:<6= D6BC63ED

&:&' :<6D29E: )


>8

O
<8 #



>C -


   

P
O
 +6B92D:<2?
82>32B
36B:<ED

2=2>
<62522?
C6D:>32?8 





3 F -


)# E w



)#
 <8 

>C    
)    

5 Gambar 4.8 (a) Garis-garis gaya listrik untuk partikel bermuatan positif; (b) Garis-garis gaya listrik untuk partikel bermuatan negatif; (c) Garis-garis gaya untuk dua muatan yang sejenis; (d) Garis-garis gaya untuk dua muatan yang berbeda jenis.

Contoh 4.5 +6B92D:<2?
82>32B
36B:<ED

v elektron

E

-63E29
6=6E2D2?
?682D:7 >6?E;E
A6=2D
A@C:D:7
I2?8
36B;2B2<

4>
52=2>
>652? =:CDB:<
9@>@86?



)
#:DE?8=29 2 A6B46A2D2?
I2?8
5:>:=:<:
6=66?42A2:
A6=2D
A@C:D:7 4 =2;E
6=6
Medan Listrik, Potensial Listrik, dan Kapasitor

81

&:&' *=69
<2B6?2
>2CC2
6=63


-


:<6D29E: - 
!

O 
P
O 
 ) 

P
O
<8 

4>


>  

) 2 2B:
#E<E>
$$
)6GD@?
5:A6B@=69   )  -  ) -
      
 
)      >C )    <8 D2?52
?682D:7
>6?E?;E<<2?
A6B46A2D2?
I2?8
5:2=2>:
6=6
3

4

12AE9
6=622?
"'
I2:DE
/

2 0
 0  56?82?
2


/



 0  
>


P

>C
0   0 

P
O
C6<@?

'2;E
6=66?I6?DE9
A6=2D
A@C:D:7
5:A6B@=69
56?82?
A6BC2>22? 20 
2 


/ 




P

>C
 
> 20 

P

>C

2. Resultan Kuat Medan Listrik q1

E2 r1

p

q2

E1 r2

Gambar 4.9 Titik p dipengaruhi oleh muatan q1 dan q2.

(652?
=:CDB:<
>6BEA2<2?
36C2B2?
F6=292??I2
>6?8:<ED:
2DEB2?
A6?;E>=292?
F66?882>32B
F6652?
=:CDB:<
5:
C6<:D2B
>E2D2?
CD2D:C
I2?8 >6?E?;E<<2?
36C2B
52?
2B29
>652?
=:CDB:<
A252
D:D:< D:D:<
5:
C6<:D2B >E2D2?
D6BC63ED
,6CE=D2?
36C2B
<E2D
>652?
5:
D:D:<
,
252=29
C63282: 36B:<ED

     

 *

O

&6D6B2?82?  
<E2D
>652?
=:CDB:<
5:
D:D:<
,
2<:32D
>E2D2?
- 

<E2D
>652?
=:CDB:<
5:
D:D:<
,
2<:32D
>E2D2?
- (652?
=:CDB:<
>6BEA2<2?
36C2B2?
F66?89:DE?8
B6CE=D2?
52B:
>652?
=:CDB:<
52A2D
5:=2<E<2?
56?82?
42B2 >6D@56
2?2=:C:C
>6?88E?2<2?
F66?88E?2<2? >6D@56
8B27:<
(6D@56
8B27:<
52A2D
5:=2<E<2?
56?82?
CI2B2D
C6D:2A
>652? =:CDB:<
5:<6D29E:
2B29
F6
Contoh 4.6 +6B92D:<2?
82>32B
36B:<ED

5 cm

0,2  C

P

5 cm

0,05 C

.6?DE<2? 2 <E2D
>652?
=:CDB:<
5:
D:D:<
 3 82I2
A252
>E2D2?
O
P O

5:
D:D:<


82

Mudah dan Aktif Belajar Fisika untuk Kelas XII

&:&' 2 +6B92D:<2?
82>32B
36B:<ED

P E1

q1

E2

0,2  C

Pembahasan Soal

q2

0,05 C r1 = 5 cm

r1 = 5 cm

(E2D2?
E;:

3:2C2?I2
5:2?882A
36B>E2D2?
A@C:D:7



<E2D
>652?
=:CDB:<
2<:32D
- 

<E2D
>652?
=:CDB:<
2<:32D
-   ' 
52?
  '  . .

-    ,   '   '  
*=69
<2B6?2
. 
.

.
>2<2 .   .    )>  
'  , 

-  -               .
   >  


3

   )>   


   > 

Pada titik-titik sudut B dan D sebuah bujur sangkar ABCD masing-masing diletakkan sebuah partikel bermuatan +q. Agar kuat medan listrik di titik A = nol, maka di titik C harus diletakkan sebuah partikel bermuatan sebesar .... a. – q b. + q c. q 2 d.

q 2

e.

2q 2 UMPTN 1991

Pembahasan: EBD

EAD

EAB

A

         )

%25:
<E2D
>652?
=:CDB:<
5:
D:D:<

252=29

P

)

(E2D2?

O
P
O
>6?82=2>:
82I2
C63282:
36B:<ED

, 
,
P
-


P

)

O
P
O


O  
) .2?52
?682D:7
>6?E?;E<<2?
,
2B29?I2
<6
<:B:

B(+q)

D(+q) EAB = EAD =

+252
D:D:< D:D:<
CE5ED

52?

C63E29
A6BC68:

>2C:?8 >2C:?8
5:=6D2<<2?
C63E29 A2BD:<6=
36B>E2D2?
-
82B
<E2D
>652?
=:CDB:<
5:
D:D:<


?@=
D6?DE<2?
36C2B >E2D2?
I2?8
92BEC
5:=6D2<<2?
5:
D:D:<


B C &:&' *=69
<2B6?2
-

-

-
52?




.
>2<2 

 

'  . A &E2D
>652?
5:
D:D:<

@=69
>E2D2?
5:

52?
 EAD D 252=29 ' -  ' -  EBD EAB   . . 82B



 
92BEC
C2>2
36C2B
D6D2A:
36B=2G2?2?
2B29
56?82?
 

/?DE<
:DE
-
92BEC
36B>E2D2?
?682D:7

 
 
O
  
'    '   .  - 
'   '  .
.   - 
  . .           

kq =E q2

E 2 AB  E 2 AD =

EBD =

Contoh 4.7

C

EBD = E 2 

2E 2

kq 2 q2

EAC = EBD

k
qC  2


AC  qC

2


q 2 



kq 2 q2



q 2 q2

qC q  2 2q2 q2 qC = 2q 2 Jawaban: e

-  
 -
?682D:7 %25:
36C2B
>E2D2?
I2?8
92BEC
5:=6D2<<2?
5:
D:D:<

252=29
 -

Medan Listrik, Potensial Listrik, dan Kapasitor

83

3. Hukum Gauss Luas = A

E

#E<E>
"2ECC
5:52C2B<2?
A252
<@?C6A
"(1'/
!=E6?882>32B<2?
36B2A2
32?I2<
F6652?82B:C 82B:C
82I2
I2?8
>6?6>3EC CE2DE
A6B>E<22?
52=2>
2B29
D682<
=EBEC
+6B92D:<2?
&2'&6


%:<2
D6B52A2D
82B:C 82B:C
82I2
52B:
CE2DE
>652?
=:CDB:<
9@>@86?
I2?8 >6?6>3EC
D682<
=EBEC
CE2DE
3:52?8
C6=E2C

;E>=29
82B:C
>652?
I2?8 >6?6>3EC
D682<
=EBEC
3:52?8
D6BC63ED
C2>2
56?82?
A6B<2=:2?

52?


+6B<2=:2?
2?D2B2

52?

:?:
5:?2>2<2?
"(1'/
(%/0.%'


-642B2
>2D6>2D:C 5:DE=:C<2?
C63282:
36B:<ED

Garis-garis gaya yang menembus bidang permukaan.

n

O

 

Gambar 4.10

&6D6B2?82? 
7=E
2D2E
G636B  
<E2D
>652?
=:CDB:<
) 
=E2C
3:52?8
I2?8
5:D6>3EC
>652?
=:CDB:<
> %:<2
82B:C 82B:C
82I2
D6BC63ED
>6?6>3EC
3:52?8
D:52<
C642B2
D682< =EBEC
7=E
  4@C

E

Gambar 4.11 Garis gaya yang menembus suatu permukaan membentuk sudut .

56?82?

252=29
CE5ED
2?D2B2
F6652?
52?
=E2C
A6B>E<22?
I2?8 5:D6>3EC
>652?
=:CDB:<

2B:
<@?C6A
7=E6?86>E<2<2?
9E<E>?I2 I2?8
5:?I2D2<2?
C63282:
36B:<ED

71)($
#.%/
#5
5*#
'!(1.
.%
/101
,!.)1'*
0!.0101,
/!* %*# !*#*
&1)($
)10*
(%/0.%'
5*#
%(%*#'1,%
+(!$
,!.)1'*
0!.0101,
0!./!108

-642B2
>2D6>2D:C
5:DE=:C

A6B>E<22? D6BDEDEA 



O 

Contoh 4.8 #:DE?8=29
7=E
;:<2 <E2D
>652?
=:CDB:<
9@>@86?
C636C2B

)
52?
2B29?I2 2 C6;2;2B
3:52?8 3 >6>36?DE<
CE5ED
L
D6B9252A
3:52?8 4 D682<
=EBEC
D6B9252A
3:52?8

&:&' :<6D29E: 'E2C
3:52?8
A6BC68:



4>
P

4>


4>


P
O
> &E2D
>652?
=:CDB:<



)

!=E32B
36B:<ED

2

3

4

n

n

n

E

E

2

84

37°



E

/?DE<
CE5ED


 L   4@C
 L


)

P
O
>
 



Mudah dan Aktif Belajar Fisika untuk Kelas XII

3

4

/?DE<
CE5ED

L   4@C L



)

P
O
>
 


13

/?DE<
CE5ED

L   4@C L


)

P
O
>
 


13

Kata Kunci

4. Perhitungan Medan Listrik dengan Menggunakan Hukum Gauss a. Medan Listrik pada Keping Sejajar (652?
=:CDB:<
5:
2?D2B2
A6=2D
C6;2;2B
52A2D
5:9:DE?8
56?82?
>E529 >6?88E?2<2?
#E<E>
"2ECC
E2
3E29
A6=2D
<6A:?8
I2?8
>6>:=:<:
=E2C
>2C:?8 >2C:?8
5:36B:
>E2D2?
C2>2
D6BC632B
>6B2D2
D6D2A:
36B=2G2?2? ;6?:C
I2:DE
-
52?
O-
C6A6BD:
A252
&2'&6
 
,2A2D
>E2D2?

D:2A <6A:?8
5:567:?:C:<2?
C63282:
>E2D2?
-
A6B
C2DE2?
=E2C

-642B2 >2D6>2D:C
5:DE=:C<2?
C63282:
36B:<ED





O



&E2D
>652?
=:CDB:<

A252
A6=2D
<@?5E
"2ECC
2B2?I2
56?82?
>6>3E2D
CE2DE
A6B>E<22? D6BDEDEA
C6A6BD:
C:=:?56B
E?DE<
>6>E529<2?
A6B9:DE?82?
+6B92D:<2? &2'&6


6B52C2B<2?
!*67&2&&3
 <
7=E2<2 C:=:?56B
D6BDEDEA 






 6B52C2B<2?
!*67&2&&3
 < 
5:52A2D<2?
A6BC2>22?

C:=:?56B
D6BDEDEA 
C69:?882

• • • • • • •

medan listrik garis-garis gaya fluks fluks listrik keping sejajar bola konduktor permukaan Gauss





E

Gambar 4.12 Medan listrik antara dua keping sejajar dengan rapat muatan  dan  .

E1 A1

E2 A2

E3 A3

 Gambar 4.13








    B2A2D
>E2D2?
>2<2
<E2D
>652?
=:CDB:<

I2?8 5:D:>3E=<2?
@=69
C2DE
A6=2D
<@?5E22? *=69
<2B6?2



 

Perhitungan kuat medan listrik E pada pelat konduktor menggunakan permukaan tertutup (silinder) berdasarkan Hukum Gauss.

O 

6?82?
56>:<:2?
36C2B?I2
<E2D
>652?
=:CDB:<
I2?8
5:D:>3E=<2? @=69
5E2
A6=2D
<@?5E22?



 

O 

&6D6B2?82?  
B2A2D
>E2D2?
>  
A6B>:D:F:D2C
BE2?8
92>A2

P
O 
)>

Medan Listrik, Potensial Listrik, dan Kapasitor

85

b. Kuat Medan Listrik pada Bola Konduktor Berongga +6B92D:<2?
&2'&6



%:<2
<6
52=2>
<@?5E=29
>E2D2?
A@C:D:7
2D2E
>E2D2?
?682D:7 >E2D2?
D6BC63ED
2<2?
D6BC632B
>6B2D2
92?I2
5:
A6B>E<22?
3@=2
52AE? 5:
52=2>
3@=2
D:52<
D6B52A2D
>E2D2?
=:CDB:<
6B52C2B<2?
#E<E>
"2ECC 52A2D
5:D6?DE<2?
36C2B
>652?
=:CDB:<
5:
52=2>
>2EAE?
5:
=E2B
3@=2
I2?8 36C2B?I2

R r>R r
 

permukaan Gauss I permukaan Gauss II

Gambar 4.14 Bola konduktor berongga yang memiliki jari-jari R. r = jarak titik ke pusat bola.

 E

E =k

q R2

E=k E=0 0

r=R

q r2 r

Gambar 4.15 Grafik E terhadap r untuk bola konduktor berongga.

Tes Kompetensi Subbab

B 

d

E2
3E29
A2BD:<6=
36B>E2D2?
D6B=6D2<
A252
C2DE
82B:C =EBEC
C6A6BD:
A252
82>32B
36B:<ED

14 cm

–4q



F

 '  .  .   . 

&E2D
>652?
=:CDB:<
5:
=E2B
3@=2
52A2D
5:A6B@=69
56?82?
>6?82?882A
3@=2 C63282:
>E2D2?
=:CDB:<
I2?8
D6B=6D2<
5:
AEC2D
3@=2
%25:
C642B2
<6C6=EBE92? >652?
=:CDB:<
5:
C6<:D2B
3@=2
36B@?882
252=29 M 5:
52=2>
3@=2



<2B6?2
-

M 5:
A6B>E<22?
3@=2
  '   M 5:
=E2B
A6B>E<22?
3@=2
  '  .

*6/&0&31&-
)&1&2
'909
1&8.-&3

-63E29
6=632<<2?
56?82?
<646A2D2?
2G2=


P

>C
C62B29
56?82?
<E2D
>652?
=:CDB:<
I2?8 36C2B?I2

P

)
.6?DE<2?=29 2 <2A2?
6=6AE9?I2

 +6B92D:<2?
82>32B
36B:<ED

E

–24q

%:<2
;2B2<
2?D2B2
A2BD:<6=

4>
5:
>2?2<29
=6D2<
D:D:< I2?8
<E2D
>652?
>28?6D:A2D
CE5ED
C63E29
A6BC68:
5:=6D2<<2?
6>A2D 3E29
>E2D2?
I2?8
C2>2
    
C6A6BD:
82>32B
36B:<ED

4 C

-63E29
6=6E2D2?
?682D:7
52?
5:A6B46A2D
>6?E;E
A6=2D
A@C:D:7

%:<2
;2B2<
2?D2B2
A6=2D

4>
52?
>652?
=:CDB:<

 
)
D6?DE<2?=29 2 36C2B
A6B46A2D2?
6=66?42A2: A6=2D
A@C:D:7 4 =2;E
6=6
86

O 

:
328:2?
52=2>
3@=2
56?82?
.


36C2B?I2
>652?
=:CDB:<



#2= D6BC63ED
5:C6323<2?
36C2B?I2
>E2D2?
I2?8
5:=:?8<EA:
A6B>E<22?
"2ECC
$ -


52AE?
E?DE<
A6B>E<22?
"2ECC
$$
56?82?
.


36C2B?I2
>E2D2? =:CDB:<
I2?8
5:=:?8<EA:
A6B>E<22?
"2ECC
$$
C2>2
56?82?
;E>=29
>E2D2? =:CDB:<
A252
3@=2
D6BC63ED
6?82?
56>:<:2?
>652?
=:CDB:<

5:
A6B>E<22? "2ECC
$$
252=29

R

a


2D2E
   

Mudah dan Aktif Belajar Fisika untuk Kelas XII

4 C 20 cm

4 C

4 C

.6?DE<2?
<E2D
>652?
=:CDB:<
5:
AEC2D
A6BC68:
;:<2 2 <66>A2D
>E2D2?
:DE
A@C:D:7 3 D2?52
A@C:D:7
52?
?682D:7
<66>A2D
>E2D2?
:DE 36BC6=2?8 C6=:?8



#:DE?8
7=E6?6>3EC
3:52?8
A6BC68: C:C:


4>
;:<2
<E2D
>652?
=:CDB:<
C636C2B

) I2?8
2B29?I2 2 C6;2;2B
56?82?
3:52?8 3 D682<
=EBEC
3:52?8 4 >6>36?DE<
CE5ED
 L
D6B9252A
3:52?8



-63E29
6=6
36B>E2D2?
?682D:7
=66?E;E
<6A:?8
=@82>
A@C:D:7
5:
56<2D?I2 C636C2B

P

>C
%:<2
<65E2
<6A:?8
>6>:=:<: B2A2D
>E2D2?
I2?8
C2>2
D6?DE<2?
?:=2:
B2A2D >E2D2??I2
)6


P
O
<8
-6


P
O 
  0 

P
O 
)>

C. Energi Potensial Listrik dan Potensial Listrik 1. Energi Potensial Listrik +6B92D:<2?
&2'&6
 
-63E29
>E2D2?
E;:
-
5:56<2D<2?
56?82? >E2D2?
-
<:32D?I2
D6B;25:
:?D6B2E2D2?
-
52?
-
36BEA2 82I2
4@E=@>3
I2?8
2B29?I2
D@=2< >6?@=2<
/?DE<
36BA:?529
52B:
A@C:C: .
<6
A@C:C:
.
>E2D2?
-
92BEC
>6=2<E<2?
EC292
6C2B?I2
EC292
I2?8 92BEC
5:=2<E<2?
252=29
C632?5:?8
56?82?
36C2B?I2
82I2
@E=@>3
52? A6BA:?5292??I2


O      .2?52
?682D:7
>6?E?;E<<2?
EC292
>6=2G2?
82I2
@E=@>3
%:<2
EC292 I2?8
5:=2<E<2?
-
>6=2=E:
A6BA:?5292?
I2?8
C2?82D
<64:=
>2<2
EC292
- 52A2D
5:A6B@=69
56?82?
42B2


O

.

q2

q1

+

–

Fc

q1

(2) r1 r2

F

– (1)

r

Gambar 4.16 Interaksi antara muatan q1 dan q2.

    

. .

     .

  

'
-
- 
. .

'
-
-  . . .

     '
-
-      . . 

O 

-6CE2:
56?82?
567:?:C:?I2
EC292
252=29
AB@C6C
DB2?C76B
6?6B8:
2D2E 36C2B?I2
A6BE3292?
6?6B8:
2D2E
52=2>
36?DE<
>2D6>2D:C?I2
C6A6BD: 36B:<ED

   ,   ,   ,

O 

2A2D
<:D2
C:>AE=<2?
329G2
A6BC2>22?
6?6B8:
A@D6?C:2=
5:
CE2DE D:D:<
252=29

, 

'
-
-  .

O 

&6D6B2?82?  
EC292
=:CDB:<
;@E=6  
82I2
@E=@>3
) < 
<@?CD2?D2
@E=@>3


P

)> - 
>E2D2?
E;:
 - 
>E2D2?
CE>36B
 . 
;2B2<
2?D2B2
-
52?
-
>

Medan Listrik, Potensial Listrik, dan Kapasitor

87

2. Potensial Listrik +@D6?C:2=
=:CDB:<
252=29
36C2B?I2
6?6B8:
A@D6?C:2=
=:CDB:<
A6B
C2DE2? >E2D2?
+252
&2'&6
 
A@D6?C:2=
=:CDB:<
I2?8
5:>:=:<:
@=69
-
252=29 C63282:
36B:<ED



 , -

'

-
-  '
-   -
. .

O 

-642B2
E>E>
A6BC2>22?
A@D6?C:2=
=:CDB:<
5:
CE2DE
D:D:<
I2?8
36B;2B2<
. 52B:
>E2D2?
CE>36B

252=29



+ q2

–

,   *  '

r2 P

q3

r3

O 

%:<2
>E2D2?
=:CDB:<
I2?8
>6?82<:32D<2?
>E?4E=?I2
A@D6?C:2=
=:CDB:< ;E>=29?I2
=63:9
52B:
C2DE
A@D6?C:2=
=:CDB:<
5:
C63E29
D:D:<
>6BEA2<2? ;E>=29
2=;232B
A@D6?C:2=
D6B9252A
C6D:2A
>E2D2?
=:CDB:<
6C2B?I2
>E2D2? A@D6?C:2=
5:
D:D:<
,
I2?8
5:C6323<2?
@=69
>E2D2?
D:D:<
- 
-



-*
252=29

q1

r1

'
 .

+

-* .*

O 

+6B92D:<2?
&2'&6
 
%:<2
92?I2
252

>E2D2?
A@D6?C:2=
=:CDB:< 5:
D:D:<
+
252=29

- -  ,  '        . . . 

Gambar 4.17 Potensial listrik oleh 3 muatan, q1, q2, q3 di titik P.

O

3. Hubungan Usaha dan Beda Potensial Listrik +6B92D:<2?
&2'&6
 
/C292
I2?8
5:=2<E<2?
E?DE<
>6>:?529<2? >E2D2?
-
52B:
A@C:C:
.
<6
A@C:C:
.
52B:
2B29
>E2D2?
252=29 2

r2

    ,   ,   ,

Q

   r1

1

 '
 '
      -  .   .

q

   -
   

Gambar 4.18 Muatan q, bergerak dari titik 1 ke titik 2 dipengaruhi muatan Q.

'
-
 '
-
  . .

O

56?82?

O
 
252=29
3652
A@D6?C:2=
=:CDB:<
2?D2B2
D:D:<

52?
D:D:<


Contoh 4.9 -63E29
>E2D2?
A@C:D:7
-


P
O

5:86B2<<2?
>6?E;E
C63E29
:?D:
2D@>
I2?8 36B>E2D2?

%2B2<
A:C29
2G2=
<65E2
A2BD:<6=
D6BC63ED
252=29

P
O


>
52?
;2B2< A:C29
2<9:B?I2
252=29

P
O


>
%:<2
EC292
I2?8
5:A6B=E<2?
E?DE<
>6>:?529<2?


P
O 
%
D6?DE<2?
>E2D2?
:?D:
2D@>
D6BC63ED

&:&' :<6D29E: - 

P
O 
 . 

P
O


> O

. 

P

>   

P
O 
%

88

Mudah dan Aktif Belajar Fisika untuk Kelas XII

56?82?
!*67&2&&3
 < 
5:A6B@=69
 

'-

   .   . 

Tantangan




P
%


P

 
P

   



  >     O  



P

 %25:
>E2D2?
:?D:
2D@>
252=29

P
O 


O 




untuk Anda

O 


Hitunglah usaha yang diperlukan untuk memindahkan muatan positif yang besarnya 10 C dari suatu titik yang potensialnya 10 V ke suatu titik yang potensialnya 60 V.

Contoh 4.10 -68:D:82

C:<E C:<E
5:

56?82?



4>
52?




4>
-63E29
>E2D2? =:CDB:<
-

O O

2<2?
5:A:?529<2?
52B:
D:D:<

<6
D:D:<

I2?8
D6B=6D2<
A252 A6BD6?8292?

%:<2
>E2D2?
-

O

52?
>E2D2?
-

O O

2?882A
<6925:B2? -
D:52<
36BA6?82BE9
D6B9252A
A@D6?C:2=
5:

D6?DE<2?=29 2 A@D6?C:2=
5:


@=29
-
52?
-  3 A@D6?C:2=
5:


@=29
-
52?
-  4 EC292
I2?8
5:A6B=E<2?
E?DE<
>6>:?529<2?
>E2D2?
-
52B:

<6


C &:&' O
O
:<6D29E:
-

 
-

O  -

O O




4>


>

10 cm 6 cm 2 +@D6?C:2=
=:CDB:<
5:

2<:32D
-
52?
-
252=29 - '   




' B A 

 4 cm D 4 cm         

P

)>
  

F@=D

    >   >  3 +@D6?C:2=
=:CDB:<
5:

2<:32D
>E2D2?
-
52?
-

*=69
<2B6?2
-

6-
52?



>2<2


O
C69:?882   





O



4 /C292
E?DE<
>6>:?529<2?
>E2D2?
252=29 

-

O


O O


O

0


P
O
;@E=6

4. Hukum Kekekalan Energi Mekanik dalam Medan Listrik #E<E>
<6<6<2=2?
6?6B8:
>6<2?:<
AE?
36B=2<E
A252
86B2<
A2BD:<6= C6A6BD:
86B2<
AB@D@?
52?
6=6
>652?
=:CDB:<
#2=
D6BC63ED 36B=2<E
<2B6?2
>652?
=:CDB:<
>6BEA2<2?
>652?
<@?C6BF2D:7

?6B8:
D@D2=
C63E29
A2BD:<6=
56?82?
>2CC2
)
52?
>E2D2?
-
I2?8 36B86B2<
52=2>
>652?
=:CDB:<
252=29 ,

'

,

' 2D2E



-  )2  -  )2 O  

Ingatlah Hukum kekekalan energi mekanik partikel dalam medan listrik berlaku jika pada partikel tersebut tidak ada gaya lain yang bekerja selain gaya Coulomb.

(6?8:?82D
6?6B8:
A@D6?C:2=
=:CDB:<
,

-
52?
6?6B8:
<:?6D:<
'


)2
;:<2
<646A2D2?
2G2=
A2BD:<6=
2


!*67&2&&3
 < 
>6?;25:



-   )2  O 

!*67&2&&3
 <
>6?E?;E<<2?
A6BE3292?
6?6B8:
A@D6?C:2=
>6?;25: 6?6B8:
<:?6D:<

Medan Listrik, Potensial Listrik, dan Kapasitor

89

Kata Kunci • energi potensial listrik • potensial listrik • volt

Contoh 4.11 +6B92D:<2?
82>32B
36B:<ED

652
A@D6?C:2=
5:
2?D2B2
5E2
A6=2D
C6;2;2B
A252
82>32B
D6BC63ED V 252=29

0
-63E29
AB@D@?
2G2=?I2
5:
<6A:?8

%:<2
5: 2?D2B2
<65E2
A6=2D
92>A2
E52B2
9:DE?8
<646A2D2?
AB@D@? C636=E>
>6?I6?DE9
<6A:?8

(2CC2
AB@D@?
)


P
O
<8 >E2D2?
AB@D@?
-


P
O 


&:&' (6=2=E:
#E<E>
&6<6<2=2?
?6B8:
(6<2?:<
52=2>
>652? =:CDB:<
5:A6B@=69 ?6B8:
>6<2?:<
5:


6?6B8:
>6<2?:<
5:
 A

B

-  )2  -  )2  
2  2    -
         2  2 
    ) 6B52C2B<2?
36C2B2?
I2?8
5:<6D29E:
A252
C@2=
52A2D
5:A6B@=69  2  


     
  0 

   <8

2      >C     >C &646A2D2?
AB@D@?
C636=E>
>6?I6?DE9
A6=2D

252=29
   >C

5. Potensial di antara Dua Keping Sejajar %:<2
5E2
3E29
<6A:?8
C6;2;2B
5:9E3E?8<2?
56?82?
CE>36B
D682?82? 32D6B2:
>2<2
<65E2
<6A:?8
2<2?
>6>:=:<:
>E2D2?
I2?8
C2>2
D6D2A: 36B=2G2?2?
;6?:C
/C292
I2?8
5:=2<E<2?
82I2
=:CDB:<


-
E?DE< >6>:?529<2?
>E2D2? >E2D2?
C6;2E9

252=29
C636C2B 


52AE?
52=2>
=:CDB:<
CD2D:C
36C2B?I2
EC292
252=29 

-  6?82?
>6?8823E?8<2?
<65E2
A6BC2>22?
EC292
D6BC63ED
5:A6B@=69

d

 
-  E

- 
- 

 
 A

B


O


 V

Gambar 4.19 Dua buah keping sejajar dan terpisah sejauh d diberi muatan yang sama.

90

O

&6D6B2?82? 
;2B2<
2?D2B2
<65E2
<6A:?8
>  
<E2D
>652?
=:CDB:<
0> 652
A@D6?C:2=
 
O

52A2D
5:?I2D2<2?
C63282:
3652
A@D6?C:2= 
2?D2B2
<65E2
<6A:?8
C6;2;2B

Mudah dan Aktif Belajar Fisika untuk Kelas XII

Contoh 4.12 -63E29
6=632<<2?
>2CE<
5:
2?D2B2
5E2
<6A:?8
C6;2;2B
56?82?
<646A2D2? 2G2=
2
C6A6BD:
A252
82>32B
36B:<ED

x

1d 2

V

%:<2
3652
A@D6?C:2=
<65E2
<6A:?8
252=29

52?
;2B2< 2?D2B2
<6A:?8

D6?DE<2? 2 G26?I6?DE9
<6A:?8
C636=29
2D2C 3 ;2B2<
>6?52D2B
4
I2?8
5:D6>AE9
6=6
d

Tantangan untuk Anda

&:&' 2B:
#E<E>
$$
)6GD@?
52?
!*67&2&&3

<

5:A6B@=69   )  -  ) -

56?82?
2B29
<6
2D2C

-   )    ) 2 -6CE2:
A6BC2>22?
"'
329G2 0


Sebuah elektron dengan massa 9,11 × 10–31 kg dan muatan listrik –1,6 × 10–19 C, lepas dari katode menuju anode yang jaraknya 2 cm. Jika kecepatan awal elektron = 0 dan beda potensial antara anode dan katode = 200 V, hitunglah kecepatan elektron ketika sampai di anode.

-


/ 
C652?8<2?




>2<2
E?DE<
/



2<2?
5:A6B@=69   )

)  -
 -
 ) 2B:
A6BC2>22?
4

2
0
56?82?
4


5:A6B@=69 0

3

 4

2  

) 

2 -
 

) -


Tes Kompetensi Subbab

C

*6/&0&31&-
)&1&2
'909
1&8.-&3

+6B92D:<2?
82>32B
36B:<ED

q

q



r = 50 × 10–10 m





%:<2
5:<6D29E:
.


P
O

>
52?
-


P
O 
 52?
C63E29
AB@D@?
D6B=6D2<
5:
D:D:<

D6?DE<2?=29
6?6B8: A@D6?C:2=
AB@D@?

E2
>E2D2?
D:D:<
-

O
*
52?
-


*
D6BA:C29 A252
;2B2<

4>
.6?DE<2?
A@D6?C:2=
5:
D:D:<
D6?829 82B:C
9E3E?8
2?D2B2
<65E2
>E2D2?
D6BC63ED

>A2D
3E29
>E2D2?
=:CDB:<
- 
-
-
52?
-
D6B=6D2< A252
D:D:<
CE5ED
A6BC68:
I2?8
A2?;2?8
C:C:?I2

4> C6A6BD:
A252
82>32B
36B:<ED

q1



.6?DE<2?
A@D6?C:2=
D:D:<
5:
AEC2D
A6BC68:
:DE
;:<2 2 -

-

-

-

   3 -

  
-

O   -

  
-

O  

E2
3E29
>E2D2?
D:D:<
A252
<@@B5:?2D
2BD6C:EC 56?82?
<@@B5:?2D
C63282:
36B:<ED
-

  
5: D:D:<


52?
-

O   
5:
D:D:<
O

.6?DE<2? A@D6?C:2=
=:CDB:<
5:
D:D:<
 


-63E29
6=632<<2?
56?82?
<646A2D2?


>C
>2CE<
<6
52=2>
5E2
<6A:?8
C6A6BD:
82>32B

%:<2
<E2D
>652?
2?D2B2
5E2
<6A:?8


P

) D6?DE<2?
;2B2<
>6?52D2B
4
C2>A2:
6=66?E>3E<
<6A:?8
I2?8
2D2C
)6


P
O
<8 -6


P
O 

52?
#


>C

q2 x=? E

q4

V

y = 1 cm

q3

Medan Listrik, Potensial Listrik, dan Kapasitor

91

D. Kapasitor 1. Muatan pada Kapasitor e– E

–

+ Gambar 4.20

Proses pengisian muatan pada kapasitor.

+6B?29<29
?52
>6>6B92D:<2?
<@>A@?6? <@>A@?6?
I2?8
252
5: 52=2>
2=2D
6=6A@?6?
6=6?I2
+6B92D:<2? &2'&6
 
<2A2C:D@B
2D2E
<@?56?C2D@B
252=29
5E2
3E29
A6?892?D2B A6=2D
<@?5E6>A6B@=69
>E2D2?
I2?8
C2>2
D6D2A:
36B=2G2?2?
;6?:C
&2A2C:D@B 36B7E?8C:
E?DE<
>6?I:>A2?
6?6B8:
A@D6?C:2=
=:CDB:<

?52
52A2D
>6>32I2?8<2?
C63E29
EC292
E?DE<
>6>36B:<2?
>E2D2? A252
<2A2C:D@B
56?82?
42B2
<EDE3
A@C:D:7
32D6B2:
>6?2B:<
6=66?;25:
36B>E2D2?
A@C:D:7 52?
>6?5@B@?8
6=66?;25:
36B>E2D2?
?682D:7
9:?882
<65E2
<6A:?8
>6>:=:<:
>E2D2?
C2>2 92?I2
;6?:C?I2
I2?8
36B3652
+B@C6C
A6BA:?5292?
6=66=2=E:
6?6B8:
&:>:2
52B:
32D6B2:
I2?8
36B=2?8CE?8 D6BEC >6?6BEC
C2>A2:
3652
A@D6?C:2=

2?D2B2
<6A:?8
C2>2
56?82?
3652 A@D6?C:2=
I2?8
5:>:=:<:
32D6B2:
-2>A2:
2<9:B?I2
<2A2C:D@B
D6=29
A6?E9 >E2D2?
52?
D:52<
52A2D
5::C:
=28:

&2A2C:D@B
5:C:>3@=<2?
56?82?



&2A2C:D@B
5:8E?2<2?
E?DE< 2 >6?I:>A2?
>E2D2?
2D2E
6?6B8:
=:CDB:< 3 C63282:
C2=29
C2DE
<@>A@?6?
52=2>
B2?8<2:2?
A6?2=2
I2?8
36B8E?2 E?DE<
>6>:=:9
7B6<E6?C:
A252
A6C2G2D
B25:@
52? 4 >6?46829
=@?42D2?
=:CDB:<
A252
B2?8<2:2? B2?8<2:2?
I2?8
>6?82?5E?8 <E>A2B2?
;:<2
D:32 D:32
2BEC
=:CDB:<
5:AEDEC<2?

2

3 keping konduktor

isolator

keping konduktor

Gambar 4.21 (a) Jenis kapasitor; (b) Skema kapasitor.

2. Kapasitas Kapasitor ?52
>E?8<:?
A6B?29
>6=:92D
<2A2C:D@B
D6BED2>2
52=2>
A6B2=2D2? 6=636B
3652 A@D6?C:2=
I2?8
36BE329 E329
>2<2
>E2D2?
I2?8
D6BC:>A2?
5:
52=2>?I2 AE?
36BE329 E329

- -  -  - *    
<@?CD2?    * +6B32?5:?82?
-
52?

D6BC63ED
5:?2>2<2?
<2A2C:D2C
<2A2C:D@B
I2?8 92B82?I2
<@?CD2?
E?DE<
D:2A
<2A2C:D@B
C69:?882
52=2>
36?DE<
>2D6>2D:C 5:DE=:C
C63282:
36B:<ED

92

Mudah dan Aktif Belajar Fisika untuk Kelas XII







O

&6D6B2?82? - 
>E2D2?
  
3652
A@D6?C:2=
0  
<2A2C:D2?C:
0
2D2E
! &2A2C:D2C
<2A2C:D@B
D:52<
5:A6?82BE9:
@=69
A6BE3292?
>E2D2?
52? A@D6?C:2=
=:CDB:<

52
D:82
;6?:C
<2A2C:D@B
I2?8
32?I2<
5:8E?2<2?
I2:DE 2 ,/%0+.
'!.0/
&6BD2C
A252
<2A2C:D@B
:?:
36B7E?8C:
C63282:
A6?I6<2D
5: 2?D2B2
<65E2
A6=2D
=@82> 3 ,/%0+.
2.%!(
&2A2C:D@B
:?:
5:8E?2<2?
52=2>
B2?8<2:2?
A6?2=2 A252
A6C2G2D
B25:@ 4 ,/%0+.
!(!'0.+(%0
!(+
&2A2C:D@B
;6?:C
:?:
>6>:=:<:
<2A2C:D2?C:
A2=:?8 D:?88:
I2:DE
C2>A2:
56?82?

A!

kertas

Tugas Anda 4.2 Berdasarkan kepolarannya, dikenal dua jenis kapasitor, yaitu kapasitor polar dan kapasitor non-polar. Coba Anda cari informasi mengenai perbedaan di antara keduanya.

keping bergerak

Gambar 4.22

lempeng logam

2

keping tetap

3

kapasitor elektrolit

(a) Kapasitor kertas; (b) Kapasitor variabel; (c) Kapasitor elektrolit.

4

Contoh 4.13 -63E29
<2A2C:D@B
5::C:
@=69
32D6B2:

F@=D
C69:?882
36B>E2D2?

  
.6?DE<2?=29 2 <2A2C:D2C
<2A2C:D@B
D6BC63ED 3 >E2D2?
I2?8
D6BC:>A2?
52=2>
<2A2C:D@B
;:<2
5:>E2D:
32D6B2:

F@=D

&:&' :<6D29E:
 

F@=D

-


P
O


2 &2A2C:D2C
<2A2C:D@B
5:9:DE?8
56?82?
!*67&2&&3
 < 

3

-
        
! 
0 

(E2D2?
I2?8
D6BC:>A2?
5:A6B@=69
56?82?
A6BC2>22? -




P
O
!

0

  

3. Kapasitor Keping Sejajar +252

.(-&*1
&6&)&;
>6=2<E<2?
A6?6=:D:2?
D6?D2?8
A6?82BE9 A6?8:C:2?
BE2?8
5:
2?D2B2
A6=2D A6=2D
<2A2C:D@B
56?82?
>6?88E?2<2?
3292? 5:6=66?88E?2<2?
5E2
<2A2C:D@B
I2?8
:56?D:<
:
C2=29 C2DE
<2A2C:D@B
5:D6>A2D<2?
CE2DE
3292?
5:6=62=
&6>E5:2?
<65E2
<2A2C:D@B D6BC63ED
5:36B:
A@D6?C:2=
=:CDB:<
I2?8
C2>2
36C2B?I2
C6A6BD:
5:DE?;E<<2? @=69
&2'&6

(6=2=E:
66?
D6BC63ED
&6&)&;
>6>A6B@=69
92C:= CA66?82?5E?8
5:6=6E2D2?
5:
<2A2C:D@B
I2?8
>6?82?5E?8
E52B2

Medan Listrik, Potensial Listrik, dan Kapasitor

93

udara

2

bahan dielektrik

*=69
<2B6?2
>E2D2?
-
=63:9
36C2B
E?DE<

I2?8
C2>2
A252
<2A2C:D@B I2?8
>6?82?5E?8
5:6=622?



5:A6B@=69  92C:=
329G2
<2A2C:D2C
C63E29
<2A2C:D@B
2<2?
36BD2>329
36C2B
2<:32D A6?6>A2D2?
3292?
5:6=6
#&'*1


-:72D C:72D
52B:
636B2A2
:6=6 /52B2 :B &6BD2C (:<2
>6B29
56=:>2 ,E33I
>:42 +@BC6=6? &G2BC2
I2?8
5:=63EB "6=2C
A:B6H 2<6=:D +@=:6D=6? >36B .67=@? )6@AB6? (:?I2<
DB2?C7@B>2D@B .:D2?:E>
5:@
3 Gambar 4.23 (a) Baterai B menyebabkan perbedaan potensial yang sama pada setiap kapasitor; kapasitor dengan bahan dielektrik memiliki muatan yang lebih banyak. (b) Kedua kapasitor mengangkut muatan yang sama. Kapasitor dengan bahan dielektrik memiliki perbedaan potensial lebih rendah, seperti ditunjukkan oleh pembacaan alat ukur.

4378&38&
.*1*086.0

*09&8&3
.*1*086.0






0%22



    

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Sumber: Fundamental of Physics, 2001



-:72D C:72D
:?:
252=29
<:B2 <:B2
A252
D6>A6B2DEB
<2>2B
52?
E?DE<
<@?5:C: <@?5:C: C656>:<:2?
BEA2
C69:?882
>652?
=:CDB:<

5:
52=2>
5:6=6
 $?:=29
8B25:6?
A@D6?C:2=
>2E>
I2?8
>E?8<:?
D6B52A2D
5:
52=2>
5:6=6A2D<2?
5:
2?D2B2
A=2D A=2D
A6?892?D2B
E?DE<
>6?8:B:><2?
CE2DE
A6B36522? A@D6?C:2=
I2?8
=63:9
D:?88:
E?DE<
5:A2<2:<2?
5:
2?D2B2
A=2D A=2D
D6BC63ED
52B:A252 A6B36522?
A@D6?C:2=
I2?8
>E?8<:?
5:52A2D
56?82?
>6?88E?2<2?
E52B2
C63282: 5:6=6
E +q

A

–q

+6B92D:<2?
&2'&6
 
<2A2C:D@B
<6A:?8
C6;2;2B
D6B5:B:
2D2C
5E2 <6A:?8
=@82>
I2?8
D6BA2C2?8
C6;2;2B
A252
;2B2<
A:C29

>6D6B
I2?8
;2E9 =63:9
<64:=
52B:
=E2C
<6A:?8

>
&65E2
<6A:?8
D6BC63ED
5:36B:
>E2D2?
I2?8
C2>2
36C2B
D6D2A:
D:52<
C6;6?:C

%:<2
5:
2?D2B2
<6A:?8
C6;2;2B
252=29
E52B2F2<E>
<2A2C:D2?C:
<2A2C:D@B >6>6?E9:
A6BC2>22?
36B:<ED

d



 V

Gambar 4.24 Kapasitor keping sejajar dihubungkan dengan sumber tegangan V.

94



&6D6B2?82? 
=E2C
<6A:?8
> 
;2B2<
2?D2B<6A:?8
>  
A6B>:D:F:D2C
F2<E>
E52B2


P
O 
)>

Mudah dan Aktif Belajar Fisika untuk Kelas XII

O

4. Dielektrik pada Kapasitor

Tugas Anda 4.3

:
2?D2B2
5E2
A6=2D
<6A:?8
A252
<2A2C:D@B
3:2C2?I2
5:C:C:A:5:A2C2?8 CE2DE
3292?
:C@=2D@B
I2?8
5:C63ED
%!(!'0.%'
292?
:?:
5:8E?2<2?
<2B6?2 52A2D
>6>A6B36C2B
?:=2:
<2A2C:D2C
<2A2C:D@B
@?D@9
3292?
5:6=6:<2
<6BD2C
52?
<2B6D

292?
5:6=66>:=:<:
92B82
A6B>:D:F:D2C
I2?8
36B3652
56?82? 92B82
A6B>:D:F:D2C
F2<E>
6B52C2B<2?
!*67&2&&3
 <
52A2D 5:<6D29E:
329G2
A6B>:D:F:D2C
3292?
   . 
C69:?882
<2A2C:D2C
<2A2C:D@B I2?8
>6?88E?2<2?
3292?
5:6=6
 




2D2E


 . 



Mengapa dielektrik tidak menggunakan bahan konduktor? Coba tanya pada guru Anda atau cari informasi dari internet.

O

6?82?
>6>6B92D:<2?
!*67&2&&3
 <
52?
!*67&2&&3
 < 5:52A2D
9E3E?82?
<2A2C:D@B
D2?A2
5:6=6
O 

   . 

Contoh 4.14

Ingatlah

-63E29
<2A2C:D@B
<6A:?8
C6;2;2B
=E2C
D:2A
<6A:?8


4>
52?
D6BA:C29

>>
C2DE C2>2
=2:?
652
A@D6?C:2=
5:
2?D2B2
<6A:?8


F@=D
&6>E5:2?
<2A2C:D@B
5:=6A2C 52B:
CE>36B
D682?82?
52?
BE2?8
5:
2?D2B2
<6A:?8
5::C:
56?82?
5:6=66>:=:<: A6B>:D:F:D2C
B6=2D:7
3292?



#:DE?8=29 2 <2A2C:D2C
>E=2 >E=2
  3 >E2D2?
-
A252
D:2A
<6A:?8 4 <E2D
>652?
=:CDB:<

C636=E>
5:36B:
3292?
5:6=6:D:F:D2C
5:6=6652?
=:CDB:<

C6D6=29
5:36B:
>E2D2?
5:6=6
&:&' :<6D29E:  

P
O 
)>
 

>>


P
O

> 

4>


P
O

>  

0  




2

3

       
          >      

!    >  -




 P
O


!

0


P
O


4

 
0>     
0   
>

5

.   

6

Jika permitivitas dielektrik disisipkan di antara keping kapasitor, kapasitas dan beda potensialnya berubah, tetapi muatannya (q) tetap.



 .



 P
O


!



P
O
!    . 




P
O 

)>



P
O
)>

Medan Listrik, Potensial Listrik, dan Kapasitor

95

7

8

-        
0     ! 5:A6B@=69
-

-
<2B6?2
9E3E?82?
56?82?
32D6B2:
5:AEDEC
C69:?882
D:52<
>6>:=:<: ;2=2?
=2:?

      
0  
0>   
>  

5. Kapasitor Bentuk Bola bola 2 bola 1 R1 R2

+6B92D:<2?
&2'&6

E2
<E=:D
<@?5E
&E=:D
328:2?
=E2B
36B>E2D2?
?682D:7
56?82? ;2B: ;2B:
=E2B

C652?8<2?
<E=:D
I2?8
C636=29
52=2>
36B>E2D2?
A@C:D:7 56?82?
;2B: ;2B:

+@D6?C:2=
 
A252
3@=2
328:2?
52=2>
5:D:>3E=<2? @=69
>E2D2?
A@C:D:7
-
52?
>E2D2?
3@=2
328:2?
=E2B
O-
52AE?
A@D6?C:2= 
52B:
3@=2
<@?5E
&:D2
<6D29E:
329G2
<E2D
>652?
52?
A@D6?C:2=
=:CDB:<
@=69
3@=2 <@?5E6>:=:<:
A6BC2>22? 

'

Gambar 4.25 Kapasitor bola konsentris dengan V2 dibumikan

Informasi untuk Anda


52?
0  '  

O 

6?82?
56>:<:2?
3652
A@D6?C:2=
2?D2B2
<65E2
3@=2
<@?5EE2D2?
D6BC63ED
252=29 


O




  
'-        '



O


>6BEA2<2?
D6D2A2?
<6
C632?5:?82?
E?DE<
BE2?8
92>A2
'
 


P

)>O
6?82?
56>:<:2?
<2A2C:D2C

52B:
<2A2C:D@B
3@=2
252=29 



 

Sumber: Physics for You, 2001

Generator Van de Graaff adalah alat untuk mengumpulkan dan menyimpan muatan listrik statik dalam jumlah yang sangat besar. Rambut wanita ini akan saling tolakmenolak setelah memegang generator Van de Graaff dan ujungujungnya berdiri karena termuati oleh muatan yang sama.

96



 '-       



  '
   

C69:?882 



 
   

O

E?DE<
<E=:D
3@=2
<@?5E2?2



5:A6B@=69 



Information for You Van de Graff is a tool to collect and store statics electric charge enormously in quantity. The woman’s hair could be repulsive each other after held the Van de Graff generator and a hair would be repulsives caused by loading the common charges.

-

 '

O

6. Susunan Kapasitor a. Susunan Seri 2=2>
AB26?8823E?8<2?
<2A2C:D@B
E?DE< >6?52A2D<2?
?:=2:
<2A2C:D2C
I2?8
5::?8:?<2?
2B2
A6?8823E?82?
<2A2C:D@B I2?8
C2?82D
52C2B
252=29
42B2
A6?8823E?82?
56?82?
CECE?2?
C6B:
52? A2B2=6=

Mudah dan Aktif Belajar Fisika untuk Kelas XII

+6B92D:<2?
&2'&6

2=2>
CECE?2?
C6B:
<2A2C:D@B
36B=2<E 2DEB2?
36B:<ED

 (E2D2?
A252
C6D:2A
<2A2C:D@B
C2>2
I2:DE
C2>2
56?82?
>E2D2? <2A2C:D@B
A6?882?D:

O -D@D

-

-

-
-

C1 +

C2 –

V1

O

C3 –

+

V2

–

V3

V

 652
A@D6?C:2=
A252
E;E?8 E;E?8
<2A2C:D@B
A6?882?D:
C2>2
56?82? ;E>=29
D@D2=
3652
A@D6?C:2=
E;E?8 E;E?8
C6D:2A
<2A2C:D@B

D@D

 

+

Gambar 4.26 Susunan seri kapasitor

2B:
!*67&2&&3
 <
5:<6D29E:
329G2 




2D2E


 


  
  
C652?8<2?
D@D



D@D    %:<2
92B82 92B82
 


52?

5:>2CE<<2?
<6
52=2>
!*67&2&&3  < 
2<2?
5:A6B@=69
A6BC2>22?  

  


D@D   

O

6C2B
<2A2C:D2C
A6?882?D:
E?DE<
*
3E29
<2A2C:D2D@B
I2?8
5:CECE? C6B:
252=29

  



 D@D    *

O

Tugas Anda 4.4 Ketika Anda sedang merangkai peralatan elektronik, kadang-kadang sulit menemukan nilai kapasitas kapasitor yang tepat dengan kebutuhan, maka teknik penyusunan kapasitor bermanfaat sekali dalam hal ini. Jika Anda hanya memiliki kapasitor bernilai besar, susunan kapasitor apakah yang akan Anda gunakan untuk mendapatkan kapasitor bernilai kecil?

Contoh 4.15 .:82
3E29
<2A2C:D@B
I2?8
<2A2C:D2C?I2
>2C:?8 >2C:?8

!

!
52?

!
5:CECE?
C6B: 52?
5:9E3E?8<2?
<6
CE>36B
D682?82?

0
#:DE?8=29 2 <2A2C:D2C
A6?882?D: 3 >E2D2?
-
52?
3652
A@D6?C:2=
C6D:2A
<2A2C:D@B

&:&' :<6D29E:



!



!



!
D@D


0
 
  2 &2A2C:D2C
<2A2C:D@B
A6?882?D:

           ! D@D D@D    
! 
! 
! 
! 3 (E2D2?
<2A2C:D@B
5:A6B@=69
56?82?
>6?88E?2<2?
!*67&2&&3
 < -D@D

D@D
D@D


!

0


 -D@D

-

-

-


 C69:?882
D682?82?
C6D:2A
<2A2C:D@B
252=29    
  0     
  0     
  0 
! 
!


!   

Ingatlah • Kapasitas total dua buah kapasitor seri adalah

CS 

C1  C2 C1  C2

• Perbandingan potensial pada kapasitas seri adalah V1 : V2 : V3 : ... : Vn  1 : 1 : 1 : ... : 1 C1 C2 C3 Cn

b. Susunan Paralel 2=2>
CECE?2?
A2B2=6=
<2A2C:D@B
36B=2<E
92= 92=
36B:<ED

 652
A@D6?C:2=
C6D:2A
<2A2C:D@B
252=29
C2>2 D@D








O

Medan Listrik, Potensial Listrik, dan Kapasitor

97

+

V

– V1

C1

C2

+ –

V2

C3

+ –

 (E2D2?
D@D2=
<2A2C:D@B
A6?882?D:
C2>2
56?82?
;E>=29
D@D2=
>E2D2? O  -D@D

-

-

- +

V3

–

Gambar 4.27 Susunan paralel kapasitor

+6B92D:<2?
&2'&6

(E2D2?
A252
>2C:?8 >2C:?8
<2A2C:D@B 252=29
- 
 
-


52?
-


C652?8<2?
-

D@D2=

%:<2 92B82 92B82
- 
-
-
52?
-
5:>2CE<<2?
<6
52=2>
!*67&2&&3
 <  2<2?
5:A6B@=69 D@D





 O  6C2B
<2A2C:D2C
A6?882?D:
E?DE<
*
3E29
<2A2C:D@B
I2?8
5:CECE?
A2B2=6= 252=29 D@D










* O

Contoh 4.16

Tantangan untuk Anda Perhatikan gambar rangkaian kapasitor berikut. y C1 3 F 3 F 4 F 5 F 3 F 4 F 4 F C2 x Tentukan: a. kapasitas pengganti antara titik x dan titik y, b. beda potensial antara titik x dan y jika muatan pada kapasitor 5  F adalah 60  C.

.:82
3E29
<2A2C:D@B
I2?8
<2A2C:D2C?I2
>2C:?8 >2C:?8

!

!
52?

!
5:CECE? A2B2=6=
52?
5:9E3E?8<2?
<6
CE>36B
D682?82?

F@=D
#:DE?8=29 2 <2A2C:D2C
D@D2=
D@D 3 3652
A@D6?C:2=
C6D:2A
<2A2C:D@B 4 >E2D2?
C6D:2A
<2A2C:D@B 5 >E2D2?
D@D2=
<2A2C:D@B

&:&' :<6D29E:



!



!



!
D@D


0

2 &2A2C:D2C
D@D2=?I2
252=29
D@D








!


!


!


!

3 +252
CECE?2?
A2B2=6=
36?52
A@D6?C:2=
5:
C6D:2A
<2A2C:D@B
C2>2
;E82
56?82? A@D6?C:2=
D@D2=?I2
C69:?882
D@D









0 4 >E2D2?
5:
C6D:2A
<2A2C:D@B
252=29 -









!
P

0


 -







!
P

0

 
 -







!
P

0


 5 (E2D2?
D@D2=
<2A2C:D@B
252=29
-D@D

-

-

-




 



 



7. Energi yang Tersimpan dalam Kapasitor V (V)

V

q C

dq

Q

q(C)

Gambar 4.28 Grafik perubahan potensial (V) sebagai fungsi perubahan muatan selama proses pengisian berlangsung.

&2A2C:D@B
;E82
36B7E?8C:
E?DE<
>6?I:>A2?
6?6B8:
A@D6?C:2=
I2?8 5:36B:<2?
52B:
CE>36B
A@D6?C:2=
=:CDB:<
?6B8:
A@D6?C:2=
D6BC63ED
5:8E?2<2? <6D:<2
B2?8<2:2?
=:CDB:<
>E=2:
36<6B;2
-6;E>=29
>E2D2?
<6
52=2>
<2A2C:D@B D6BEC
36BD2>329
C6=2>2
AB@C6C
A6?8:C:2?
D6BEC
36B=2?8CE?8
36BC2>22? 56?82?
A6BD2>3292?
A@D6?C:2=
A252
<65E2
<6A:?8
C6;2;2B
9:?882
A6?E9 C636C2B
 

 &6?2:<2?
D@D2=
6?6B8:
A@D6?C:2=
A252
<2A2C:D@B
,
252=29
A6?;E>=292? 2D2E
:?D68B2=
>E2D2?
-!O
52B:
?@=
C2>A2:

A6B92D:<2?
&2'&6


6?82?
>6>6B92D:<2?
!*67&2&&3
< 
6?6B8:
A@D6?C:2=
I2?8
D6BC:>A2? 52=2>
<2A2C:D@B
52A2D
5:A6B@=69
C63282:
36B:<ED


,



-


   ,  



 ,


 

98

Mudah dan Aktif Belajar Fisika untuk Kelas XII

O

Contoh 4.17

Kata Kunci

E2
3E29
<2A2C:D@B
>2C:?8 >2C:?8
<2A2C:D2C?I2
   52?
   5:CECE?
C642B2
C6B: <6>E5:2?
5:9E3E?8<2?
56?82?
D682?82?



0
.6?DE<2? 2 <2A2C:D2C
D@D2= 3 6?6B8:
I2?8
D6BC:>A2?
52=2>
C:CD6>

&:&' :<6D29E:



   


   



0

        

      D@D2=

  D@D2=   







2

• • • • • • • • • •

kapasitas kapasitor kapasitor kertas kapasitor variabel kapasitor elektrolit farad kondensator isolator dielektrik katode anode

D@D2=          D@D2=       !



0      
%   






3

Mari Mencari Tahu ?52
D6=29
>6?86D29E:
86?6B2D@B
F2?
56B
"B227
52B:
3+462&7.
93890
3)&
=2D D6BC63ED
>6?892C:=<2?
>E2D2?
C6;6?:C
52=2>
;E>=29
36C2B
.E82C
?52
252=29 >6?42B:
:?7@B>2C:
>6?86?2:
42B2
<6B;2
2=2D
D6BC63ED
5:C6BD2:
56?82?
82>32B
+292>: 7E?8C: 7E?8C:
C6D:2A
328:2?
52=2>
2=2D
D6BC63ED
%:<2
D6=29
C6=6C2:
AB6C6?D2C:<2?
5: 56A2?
<6=2C

Tes Kompetensi Subbab

D

*6/&0&31&-
)&1&2
'909
1&8.-&3

.:82
3E29
<2A2C:D@B
I2?8
<2A2C:D2C?I2
>2C:?8 >2C:?8   !
  !
52?
  !
5:CECE?
C6B:
=2=E
5:9E3E?8<2?
<6 CE>36B
D682?82?

F@=D
#:DE?8=29 2


<2A2C:D2C
<2A2C:D@B
A6?882?D: 3


>E2D2?
-
52?
3652
A@D6?C:2=
C6D:2A
<2A2C:D@B

 .:82
3E29
<2A2C:D@B
56?82?
<2A2C:D2C
>2C:?8 >2C:?8 
!

!
52?

!
5:9E3E?8<2?
C642B2
A2B2=6=
<65E2 E;E?8?I2
5:9E3E?8<2?
56?82?
CE>36B
D682?82?
F@=D
#:DE?8=29

2 <2A2C:D2C
D@D2= 3 3652
A@D6?C:2=
C6D:2A
<2A2C:D@B 4 >E2D2?
C6D:2A
<2A2C:D@B 5 >E2D2?
D@D2=
<2A2C:D@B

 .6?DE<2?
<2A2C:D2C
A6?882?D:
2?D2B2

52?

52B: >E2D2?
D@D2=?I2
52B:
<2A2C:D@B
I2?8
5:A2C2?8
C6A6BD: A252
82>32B
36B:<ED
;:<2
5:9E3E?8<2?
<6
CE>36B D682?82?

F@=D

a

a

2 F

b

2 F b

2

5 mm 10 cm

5 F

6F



-63E29
<2A2C:D@B
  !
5:9E3E?8<2?
56?82?
C63E29 CE>36B
D682?82?

0
9:?882
D6B:C:
A6?E9
&6>E5:2? <EDE3 <EDE3?I2
5:9E3E?8<2?
<6
<2A2C:D@B
  !
I2?8 D:52<
36B>E2D2?
#:DE?8=29 2 >E2D2?
D@D2= 3 3652
A@D6?C:2=
A252
E;E?8 E;E?8
C6D:2A
<2A2C:D@B 4 6?6B8:
2<9:B
D@D2= 5 A6?8EB2?82?
6?6B8:
<6D:<2
<2A2C:D@B
C2=:?8 36B9E3E?82?

-63E29
<2A2C:D@B
<6A:?8
C6;2;2B
36B36?DE<
=:?8<2B2? 56?82?
B25:EC

4>
52?
;2B2<
2?D2B2
<6A:?8

>>

%:<2
5:9E3E?8<2?
<6
32D6B2:

0
9:DE?8=29 2 36C2B
<2A2C:D2C?I2 3 36C2B
>E2D2?
A252
<6A:?8
<2A2C:D@B 4 6?6B8:
A@D6?C:2=
I2?8
D6BC:>A2?
52=2>
<2A2C:D@B

3 F

24  F 4 F



2 F

3

4 F a 12  F 4  F

24  F

b 24V

4

Medan Listrik, Potensial Listrik, dan Kapasitor

99

Rangkuman















100

(E2D2?
=:CDB:<
D6B5:B:
2D2C
AB@D@?
I2?8
36B>E2D2? A@C:D:7
6=6E2D2?
?682D:7
52?
?6EDB@? I2?8
D:52<
36B>E2D2?

%:<2
5E2
>E2D2?
I2?8
36B=2:?2?
;6?:C
36B:?D6B26?2B:<
2?D2B2
<65E2
>E2D2? D6BC63ED
-632=:E2D2?
=:CDB:<
C6;6?:C 36B:?D6B26?@=2<

#E<E>
@E=@>3
>6?I2D2<2?
329G2
36C2B
82I2
D2B:< >6?2B:<
2D2E
D@=2< >6?@=2<
2?D2B2
5E2
>E2D2?
=:CDB:< 36B32?5:?8
=EBEC
56?82?
92C:=
<2=:
36C2B
<65E2 >E2D2?
52?
36B32?5:?8
D6B32=:<
56?82?
<E25B2D
;2B2< 2?D2B2
<65E2
>E2D2?
D6BC63ED

-  '   . 56?82?
'

<@?CD2?D2
@E=@>3

&E2D
>652?
=:CDB:<
5:
CE2DE
D:D:<
5:567:?:C:<2?
C63282: 82I2
@E=@>3
A6B
C2DE2?
>E2D2?
I2?8
5:2=2>:
@=69 C63E29
>E2D2?
5:
D:D:<
D6BC63ED

  - #E<E>
"2ECC
>6?I2D2<2?
329G2
;E>=29
82B:C
82I2 I2?8
<6=E2B
52B:
CE2DE
A6B>E<22?
D6BDEDEA
C632?5:?8 56?82?
;E>=29
>E2D2?
=:CDB:<
I2?8
5:=:?8<EA:
@=69 A6B>E<22?
D6BDEDEA
D6BC63ED

A6B>E<22?
D6BDEDEA   &E2D
>652?
=:CDB:<

I2?8
5:D:>3E=<2?
@=69
C63E29 A6=2D
<@?5E22?    56?82?
.

B2A2D
>E2D2? &E2D
>652?
=:CDB:<
5:
C6<:D2B
3@=2
36B@?882
252=29 2 5:
52=2>
3@=2



<2B6?2
-

3 5:
A6B>E<22?
3@=2
  '   4 5:
=E2B
A6B>E<22?
3@=2
  '  . 652
A@D6?C:2=
=:CDB:<
5:
D:D:<
%
<2B6?2
A6?82BE9
>E2D2? 
5:?I2D2<2?
56?82?
A6BC2>22?  %  ' .

Mudah dan Aktif Belajar Fisika untuk Kelas XII



/C292
E?DE<
>6>:?529<2?
C63E29
>E2D2?
=:CDB:<
- 52B:
C2DE
D:D:<
<6
D:D:<
=2:??I2
5:?I2D2<2?
56?82?   

- O

&2A2C:D@B
252=29
<@>A@?6?
=:CDB:<
I2?8
36B7E?8C: E?DE<
>6?I:>A2?
>E2D2?
52?
6?6B8:
A@D6?C:2=
=:CDB:<

&2A2C:D@B
>6BEA2<2?
5E2
A6=2D
<@?5E
&2A2C:D2C
<2A2C:D@B
252=29
<6>2>AE2?
CE2DE
<2A2C:D@B E?DE<
>6>A6B@=69
2D2E
>6?I:>A2?
>E2D2?
=:CDB:<

  56?82?  
<2A2C:D2?C:
!2B25 - 
>E2D2?
  
3652
A@D6?C:2=
0@=D

 6C2B?I2
<2A2C:D2C
<2A2C:D@B
56?82?
3292?
5:6=622? 

 .  56?82?
 .

A6B32?5:?82?
2?D2B2
A6B>:D:F:D2C
3292?   
52?
A6B>:D:F:D2C
F2<E>
  

 %:<2
*
3E29
<2A2C:D@B
5:CECE?
C6B:
36C2B
>E2D2?
A252 C6D:2A
<2A2C:D@B
252=29
C2>2

-D@D

-

-

-





652
A@D6?C:2=
A252
E;E?8 E;E?8
<2A2C:D@B
A6?882?D: C2>2
56?82?
;E>=29
D@D2=
3652
A@D6?C:2=
<2A2C:D@B

D@D











* 6C2B
<2A2C:D2C
<2A2C:D@B
A6?882?D:
E?DE<
*
3E29 <2A2C:D@B
I2?8
5:CECE?
C6B:
252=29

  



 D@D    *

 %:<2
*
3E29
<2A2C:D@B
5:CECE?
A2B2=6=
36C2B
3652 A@D6?C:2=
A252
C6D:2A
<2A2C:D@B
252=29
C2>2

D@D







 6C2B
>E2D2?
D@D2=
<2A2C:D@B
A6?882?D:
C2>2
56?82? 36C2B
>E2D2?
D@D2=
<2A2C:D@B

-D@D

-

-

- 6C2B
<2A2C:D2C
<2A2C:D@B
A6?882?D:
E?DE<
*
3E29 <2A2C:D@B
I2?8
5:CECE?
A2B2=6=
252=29 D@D











*

 6C2B
6?6B8:
=:CDB:<

I2?8
D6BC:>A2?
52=2> <2A2C:D@B
252=29

-  

-      

Peta Konsep 9&8&3
.786.0

>6?:>3E=<2?

36BEA2

(652?
':CDB:< >6?I6323<2?

652
+@D6?C:2=
':CDB:< 

.

>6?I:>A2?

?6B8:
+@D6?C:2=
':CDB:< ,

-
.

(E2D2?
.:D:<

(E2D2?
.6B5:CDB:3EC:

A6BC2>22?
<E2D
>652??I2

4@?D@9

M @=2
&@?5E
Refleksi Setelah mempelajari bab ini, Anda tentu menjadi paham tentang listrik statis, medan listrik, potensial listrik, energi potensial listrik, kapasitor dan jenis-jenisnya, serta bagaimana cara kapasitor bekerja. Dari semua materi pada bab ini, bagian mana yang menurut Anda sulit untuk dipahami? Coba Anda diskusikan dengan teman atau guru Fisika Anda.

Pemasangan kapasitor pada sakelar-sakelar listrik yang memiliki daya tinggi dapat menghilangkan terjadinya percikan bunga api. Pemanfaatan ini berguna sekali untuk keamanan. Coba Anda tuliskan manfaat lain penggunaan kapasitor.

Medan Listrik, Potensial Listrik, dan Kapasitor

101

Tes Kompetensi Bab 4 *6/&0&31&-
5&)&
'909
1&8.-&3  !.1.-1&-
7&1&-
7&89
/&:&'&3
;&3,
5&1.3,
8*5&8











@=2

52?

>2C:?8 >2C:?8
>E2D2?


52?



%2B2<
2?D2B3@=2
252=29

>
%:<2
<65E2
3@=2
D6BC63ED 36B252
5:
E52B2
>2<2




2 6?6B8:
A@D6?C:2=
3@=2



P



;@E=6 3 6?6B8:
A@D6?C:2=
3@=2



P



;@E=6

4 ,
3@=2



,
3@=2
  5 ,
3@=2


,
3@=2




 6 6?6B8:
A@D6?C:2=
3@=2



P

;@E=6 :<6D29E:

3E29
36?52
D:D:<
I2?8
36B>E2D2?
I2:DE 


52?

%:<2

>6?2B:<


>6?@=2<

52?
 >6?2B:<

C652?8<2?

36B>E2D2?
?682D:7
A6B?I2 D22?
I2?8
36?2B
252=29




2 >E2D2?

A@C:D:7
>E2D2?

?682D:7 3 >E2D2?

A@C:D:7
>E2D2?

A@C:D:7 4 >E2D2?

?682D:7
>E2D2?

A@C:D:7 5 >E2D2?

?682D:7
>E2D2?

?682D:7 6 >E2D2?

A@C:D:7
>E2D2?

?682D:7 -2DE2?
>652?
=:CDB:<
52A2D
5:?I2D2<2?
52=2>
N

2 ) 5 %) 3 0> 6 ) 4 )> E2
3E29
<6A:?8
C6;2;2B
5:36B:
>E2D2?
I2?8
C2>2
36C2B 52?
36B=2G2?2?
;6?:C
&E2D
>652?
=:CDB:<
5:
2?D2B2
5E2 <6A:?8
D6BC63ED
N

2 36B32?5:?8
=EBEC
56?82?
B2A2D
>E2D2??I2 3 36B32?5:?8
D6B32=:<
56?82?
B2A2D
>E2D2??I2 4 36B32?5:?8
D6B32=:<
56?82?
<E25B2D
;2B2< 2?D2B<6A:?8 5 36B32?5:?8
=EBEC
56?82?
;2B2<
<65E2
<6A:?8 6 2B29?I2
>6?E;E
<6
<6A:?8
I2?8
36B>E2D2?
A@C:D:7 -63E29
6=6
CE2DE
>652?
=:CDB:<
D:52<
>6?8 2=2>:
82I2
;:<2
6=6


2 36B86B2<
52=2>
2B29
D682<
=EBEC
D6B9252A
>652? =:CDB:< 3 36B86B2<
C6;2;2B
56?82?
2B29
>652?
=:CDB:< 4 36B86B2<
56?82?
2B29
C6>32B2?8 5 36B86B2<
>6=:?8<2B 6 D:52<
36B86B2< E2
>E2D2?
D:D:<
C6;6?:C
52?
C2>2
36C2B
-

-
 
 
36B252
A252
;2B2<

4>
C2DE
C2>2
=2:?
%:<2 '


P

)>
36C2B
82I2
D@=2<
I2?8
5:2=2>: <65E2
>E2D2?
D6BC63ED
252=29




2 
) 3 
P

) 4 
P
O
) 5 
P
O
) 6 
P

)



+6B92D:<2?
82>32B
36B:<ED

d

A



B

+B@D@?
I2?8
36B86B2<
52B:
<6A:?8

<6
<6A:?8

C6A6BD: D2>A2<
A252
82>32B
D6BC63ED
>6>:=:<:
<646A2D2?

P


>C
%:<2
;2B2<
2?D2B2
<6A:?8
52=2>
F2<E>


4>
>2CC2
AB@D@?

O
<8
>E2D2?
AB@D@?

P O

 
3652
A@D6?C:2=
<6A:?8
C6;2;2B
D6BC63ED
252=29




2 
0 5 
0 3
0 6 
0 4 
0 $!#
 E2
3E29
36?52
36B>E2D2?
O-
52?
-
36B;2B2<
.
C2DE C2>2
=2:?
%:<2
;2B2<
.
5:E329 E329
>2<2
8B27:< 9E3E?82?
82I2
@E=@>3
56?82?
.
252=29




2

F

r

3

F

r

4

F

5

F

r

r

6

F

r

102

Mudah dan Aktif Belajar Fisika untuk Kelas XII

 -63E29
<2A2C:D@B
 F 5:9E3E?8<2?
56?82?
CE>36B D682?82?
9:?882
52A2D
>6?I:>A2?
6?6B8:
C636C2B
 ;@E=6
6C2B
>E2D2?
I2?8
D6BC:>A2?
52=2>
<2A2C:D@B D6BC63ED
252=29



4@E=@>3

2 
P
O 5 
P
O O 3 
P
6 
P
O O 4 
P


+6B92D:<2?
82>32B
36B:<ED

A

B O

+2q

–q

%:<2
,
252=29
=6D2<
D:D:<
I2?8
B6CE=D2?
<E2D
>652? =:CDB:2<2
,
252=29




2 D6A2D
5:
D:D:<
* 3 2?D2B2
D:D:<

52?
* 4 2?D2B2
D:D:<
*
52?
 5 A252
A6BA2?;2?82?

56?82?
,
  , 6 A252
A6BA2?;2?82?

56?82?
,
  ,

+252
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5:36B:<2?
>E2D2? C636C2B
-
C69:?882
3652
A@D6?C:2=?I2

6C2B
6?6B8: 52=2>
<2A2C:D@B






 - 5

-  2

 

 3

 6

-   

4

 

 +6B92D:<2?
82>32B
36B:<ED

C1

F

a

b C3

C2

F

F

%:<2
<2A2C:D@B
<6D:82
<2A2C:D2?C:?I2
C2>2
<2A2C:D@B A6?882?D:
2?D2B2
E;E?8
2
52?
3
252=29



!
72B25

 2 
 5

   3 
 6

 

4


 

 E2
3E29
<2A2C:D@B
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C6B:
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A6?882?D:?I2




  2

5

   3

 6


     4

 

 -63E29
<2A2C:D@B
A6=2D
C6;2;2B
D2?A2
5:6=6:<2
I2?8
>6>:=:<:
<@?CD2?D2 5:6=66?;25:




2    3    4    5    6  

 +6B32?5:?82?
82I2
@E=@>3
52?
82I2
8B2F:D2C:
2?D2B2 
3E29
6=6
BE2?8
F2<E>
252=29
N 2 
P

) 3 
P

) 4 
P

) 5 
P

) 6 C2>2
36C2B

 /C292
E?DE<
>6>32G2
C63E29
6=6


2 
P
O
;@E=6 3 
P
O 
;@E=6 4 
P
O 
;@E=6 5 
P O 
;@E=6 6 
P
O 
;@E=6

 &2A2C:D@B
   5:36B:
>E2D2?
9:?882

<:=@F@=D
6C2B 6?6B8:
I2?8
D6BC:>A2?
52=2>
<2A2C:D@B
252=29




2 
;@E=6 3 
;@E=6 4 
;@E=6 5 
;@E=6 6 
;@E=6

 (6?EBED
>@56=
@9B
D6?D2?8
2D@>
9:5B@86?
6=66?86=:=:?8:
AB@D@?
-

6
56?82?
;2B: ;2B: 
P
O


%:<2
!


P
O 

>2<2
82I2
@E=@>3 2?D2B2
6=6
9:5B@86?
C636C2B


?6GD@?

2 
P
O 3 
P
O

4 
P
O 5 
P
O 6 
P
O

 -63E29
6=66?86=:=:?8:
AB@D@?
A252
;2B2<

K

%:<2
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6=62<2
<646A2D2? 6=66?86=:=:?8:
AB@D@?

252=29



>C

2 
P
 3 
P
 4 
P
 5 
P
 6 
P
  2=2>

4@E=@>3
D6B52A2D



3E29
6=6
2 
P
 3 
P
 4 
P
 5 
P
 6 
P


Medan Listrik, Potensial Listrik, dan Kapasitor

103



&:&'1&-
5*68&3;&&3
'*6.098
.3.
)*3,&3
8*5&8



+6B92D:<2?
82>32B
36B:<ED

6 cm

O

4  C



8 cm

3 C

.6?DE<2? 2 <E2D
>652?
5:
D:D:<
 3 82I2
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>E2D2?  

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a



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P
O
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>>
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P
O
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5:=2<E<2?
A252
>E2D2?
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282B
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>>
52B:
>E2D2?
D6BC63ED  E2
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=@82>
I2?8
C6;2;2B
52?
;2B2 C2DE
52B:
I2?8
=2:?
5:36B:
>E2D2?
=:CDB:<
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36B=2G2?2? 9:?882
3652
A@D6?C:2=

0

r C1

C2

C4

C3 C5 15 cm

b

s C6





C7

.6?DE<2? 2 <2A2C:D2?C:
A6?882?D:
D@D2=
2?D2B2
D:D:<
2
52?
3 ;:<2
C6D:2A
<2A2C:D@B
>6>:=:<:
<2A2C:D2C
3 F  3 >E2D2?
A252
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<2A2C:D@B
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2
52?3 5:36B:
3652
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A252
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<2A2C:D@B
;:<2

52?
/
5:36B: 3652
A@D6?C:2=

052?

52?
.
5:36B:
3652 A@D6?C:2=

0

-63E29
<2A2C:D@B
<2A2C:D2?C:?I2
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5:6=6
52?
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5:6=6
A@BC6=6?

+6B92D:<2?
82>32B
36B:<ED

B

d=10 cm

A

6B2A2
82I2
I2?8
5:2=2>:
C63E29
6=6
2

C C A

B C

3

C

C

C

C

C

C

C

C

A

B

4

40  F

12  F

A

v

B

13 F

5

VB – VA = 200 volt

%:<2
>E2D2?
6=62CC2 6=6A2:
5:
D6A:
 3 <646A2D2?
C22D
D:32
5:
D6A:


 -63E29
6=632<<2?
56?82?
<646A2D2?

P

>C
52=2> 2B29
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<E2D
>652?
=:CDB:<

)
6B2A2
;2B2< I2?8
5:D6>AE9
6=6
36B96?D:  @=2
<64:=
36B>E2D2?

?
O
?

?
52? O
?
5:=6D2<<2?
5:
D:D:< D:D:<
CE5ED
C63E29
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A2?;2?8
5:28@?2=

>

2 #:DE?8
A@D6?C:2=
=:CDB:<
5:
AEC2D

3 #:DE?8
36C2B
>652?
=:CDB:<
5:
AEC2D
3@=2
<6>E5:2? 82>32B<2?
2B29
52B:
A@D6?C:2=
5:
D:D:<
AEC2D

104

Mudah dan Aktif Belajar Fisika untuk Kelas XII

6F 6F A

16  F

B

28  F 12  F

&2A2C:D@B    5:9E3E?8<2?
56?82?
CE>36B D682?82?

0
C6=2?;ED?I2
CE>36B
D682?82?
5:AEDEC 52?
<2A2C:D@B
<6>E5:2?
5:9E3E?8<2?
A2B2=6=
<6 <2A2C:D@B  

2 6B2A2
>E2D2?
D@D2=
<2A2C:D@B
D6BC63ED 3 6B2A2
3652
A@D6?C:2=
C6D:2A
<2A2C:D@B 4 6B2A2
6?6B8:
I2?8
9:=2?8
A252
<2A2C:D@B    <6D:<2
5:9E3E?8<2?
56?82?
<2A2C:D@B
 

Bab

5 Sumber: Physics For Scientists and Engineers, 2000

Jarum pada kompas menyimpang mengikuti arah medan magnetik yang ditimbulkan oleh kawat berarus.

Medan Magnetik Hasil yang harus Anda capai: menerapkan konsep kelistrikan dan kemagnetan dalam berbagai penyelesaian masalah dan produk teknologi. Setelah mempelajari bab ini, Anda harus mampu: menerapkan induksi magnetik dan gaya magnetik pada beberapa produk teknologi.

)490@07
:4<06=4B0=
B4;07
38@0=6
<4=4@0=6
30?0B
<4<1C0B
30=
<4<0=500B:0=
14=30
A4?4@B8
8BC
30= <4=F41CB=F0
A410608
<06=4B 8:0
=30
?4@70B8:0=
<06=4B
10=F0:
38<0=500B:0=
?030
#)'$& :>B>@
;8AB@8:
<4<>@8
:>@
;8AB@8:
0?0B:07 =30
<4=F41CB:0=
14=3014=30
;08=
F0=6
<4=66C=0:0=
<06=4B 8:0
=30
<4<8;8:8
<06=4B
10B0=6
=30
30?0B
<4=60<0B8 10608<0=0
14=3014=30
F0=6
B4@6>;>=6
54@@><06=4B8:
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14A8
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90@C<
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8BC
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30;0<
70;
0?0
A090:07
6490;0
:4<06=4B0=
8BC
&030
101
8=8
=30
0:0= <4;>68

A. Medan Magnetik B. Gaya Magnetik C. Aplikasi Gaya Magnetik D. Sifat Magnet Bahan

Medan Magnetik

105

Tes Kompetensi Awal %"%+3,
,%,/%+!)!0(
*.-1%/
%$!-
!&-%2(*
*%0)!*!-+!'
1.!+1.!+
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#4=C@CB
=30
0?0:07
:0E0B
14@0@CA
30?0B
<4=6 70A8;:0=
<430=
<06=4B8: ?0
F0=6
0:0=
B4@9038
98:0
:0E0B
14@0@CA
38A810
94;0A:0=
54=><4=0 B4@A41CB
A4AC08
?4=64B07C0=
F0=6
B4;07
=30
<8;8:8





>10
=30
A41CB:0=
:>=4=:>=4=
4;4:B@>=8: F0=6
:4@90=F0
<4=670A8;:0=
<430=
<06=4B8: )41CB:0=
30=
94;0A:0=

94=8A
1070=
<06=4B8: 4@8:0=
2>=B>7=F0
<0A8=6<0A8=6
A0BC
1C07

A. Medan Magnetik 1. Medan Magnetik di Sekitar Magnet Permanen

tarik-menarik

tolak-menolak

Gambar 5.1 Pola garis gaya magnetik

0

1

Sumber: Physics, 1995

Gambar 5.2 Pola garis gaya magnetik pada (a) magnet U dan (b) magnet batang.

#06=4B8A<4
38:4B07C8
>@0=6
A490:
14@01030103
F0=6
;0;C
/0<0= 3C;C
A49C<;07
>@0=6
<4=64B07C8
107E0
18987
14A8
F0=6
14@E0@=0
<4B0;8: 0B0C
F0=6
38A41CB
<06=4B8:
<4<8;8:8
A850B
<4=0@8:
?0@B8:4;?0@B8:4;
14A8 :4
0@07=F0
!0B0
<06=4B
14@0A0;
30@8
J "'K
F08BC
AC0BC
304@07
38 A80
!428;
B4;0:<4=>;0:
30?C=
98:0
:CBC1:CBC1 B830:
A494=8A
3834:0B:0=
:43C0
:CBC1
B4@A41CB
0:0=
B0@8:<4=0@8: &4@70B8:0=
!,"!0
 #430=
<06=4B8:
030;07
304@07
38
A4:8B0@
<06=4B
38
<0=0
<06=4B <06=4B
;08=
0B0C
14=3014=30
<06=4B8:
F0=6
38;4B0::0=
30;0<
304@07
8=8 <0A87
38?4=60@C78
<06=4B
B4@A41CB
)C<14@
<430=
<06=4B8:
381430:0= <4=9038
3C0
94=8A
F08BC
<06=4B
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30=
<06=4B
8=3C:A8 8
A4:8B0@
<06=4B
?4@<0=4=
B4@30?0B
<430=
<06=4B8:
F0=6
30?0B 3860<10@:0=
34=60=
60@8A60@8A
60F0
<06=4B8:
0@8A60@8A
60F0
<06=4B8:
A4;0;C :4;C0@
30@8
:CBC1
CB0@0
<06=4B
30=
<0AC:
:4
:CBC1
A4;0B0=
<06=4B
8
30;0< <06=4B
60@8A60@8A
60F0
<06=4B8:
<4<8;8:8
0@07
30@8
A4;0B0=
:4
CB0@0
04@07 F0=6
<4<8;8:8
<430=
<06=4B8:
:C0B
3860<10@:0=
34=60=
60@8A60@8A
60F0 F0=6
@0?0B
A430=6:0=
304@07
F0=6
<430=
<06=4B8:=F0
;4<07
3860<10@:0= 34=60=
60@8A60@8A
60F0
<06=4B8:
F0=6
@4=660=6
&>;0?>;0
60@8A
60F0 <06=4B8:
<06=4B
+
30=
<06=4B
10B0=6
38BC=9C::0=
?030
!,"!0
  04@07
0B0C
B4
1

B

B

n

Gambar 5.3 (a) Garis gaya magnetik tegak lurus menembus bidang A (fluks magnetik); (b) Medan magnetik B menembus bidang A dengan membentuk sudut terhadap garis normal.

A

 0

garis normal

106

Mudah dan Aktif Belajar Fisika untuk Kelas XII

A

garis normal

;C:A
<06=4B8:
  
030;07
10=F0:=F0
60@8A
<430=
<06=4B8:
F0=6 38;8=6:C?8
>;47
AC0BC
;C0A
304@07
B4@B4=BC

30;0<
0@07
B460:
;C@CA )420@0
<0B4<0B8A
30?0B
38BC;8A
107E0
C=BC:

:>=AB0=
30=

:>=AB0=  
:>=AB0=
?4@A0<00==F0
030;07   - 






2>A


Contoh 5.1 4@0?0:07
:C0B
<430=
<06=4B8:
F0=6
3870A8;:0=
>;47
5;C:A
<06=4B8:

E414@
F0=6 <4=4<1CA
1830=6
A4;C0A


<
A420@0
B460:
;C@CA !5!" 8:4B07C8
 

-1



< 4=60=
<4=66C=0:0=
%01!,!!-
 9 
38?4@>;47     
-1<   038
:C0B
<430=
<06=4B8:
F0=6
3870A8;:0=
030;07

-1<

2. Medan Magnetik di Sekitar Arus Listrik +=BC:
<4;0
60@8A60@8A
60F0
<06=4B8: !4@0?0B0=
60@8A60@8A
60F0
F0=6
180A0
38A41CB
 )'
!"(
B4@A41CB <4=C=9C::0=
14A0@=F0
:C0B
<430=
<06=4B8:
38
B4B>=60=
30=
A4;0;C <4=90C78
:CBC1
CB0@0
30=
<4=34:0B8
:CBC1
A4;0B0= &030
A4:8B0@
 
!-1
'0(12(!-
%012%$
  I  
58A8:0E0= 0A0;
4=<0@:
<4=4;47
A4B4;07
%012%$ <4;0:C:0=
AC0BC
?4@2>100=
A434@70=0
60@
;4187
94;0A
<4<070<8 ?4@2>100=
F0=6
38;0:C:0=
%012%$
;0:C:0=;07
*2(4(2!1
(1(*!
14@8:CB

Medan Magnetik di Sekitar Kawat Berarus Tujuan Percobaan Mengamati medan magnetik yang disebabkan oleh kawat berarus listrik. Alat-alat Percobaan 1. Sebuah baterai 2. Kabel 3. Sebuah saklar

4. 5.

untuk Anda

I 

8:0
AC3CB


0B0C

B460:
;C@CA
B4@7030?
1830=6

%01!,!!-
 9 <4=9038 


 I !4B4@0=60= - 
D4:B>@
A0BC0=
A40@07
60@8A
=>@<0;  
5;C:A
<06=4B8:
-1  
;C0A
1830=6
F0=6
<4;8=6:C?8
>;47

<  
<430=
<06=4B8:
-1<


AC3CB
0=B0@0

30=
-

Aktivitas Fisika 5.1

Informasi

Sumber: Conceptual Physics, 1998

Baru-baru ini di dalam tengkorak burung merpati ditemukan sejenis sensor magnet yang tersusun dari magnet-magnet elementer. Tiap magnet tersebut terhubung dengan otak melalui saraf. Sensor tersebut dapat menentukan posisi arah melintang sepanjang magnet Bumi, selain itu juga dapat menentukan ketinggian posisinya dari permukaan Bumi.

Information for You Pigeons have recently been found to have multiple-domain magnetite magnets within their skulls. That are connected with a large number of nerves to the pigeon’s brain. Pigeons have a magnetic sense, and not only can they discern longitudinal directions along the earth’s magnetic field, but they can also detect latitude by the dip of the earth’s field.

I

I

Gambar 5.4 Garis-garis gaya magnetik yang terbentuk oleh serbuk besi ketika ditempatkan di tempat yang dipengaruhi oleh kawat berarus listrik.

Kawat penghantar Kompas kecil

Medan Magnetik

107

U

S

U

U

U

Langkah-langkah Percobaan 1. Susunlah peralatan sehingga terlihat seperti Gambar (a). Dalam keadaan sakelar S terbuka, letakkan kawat penghantar di atas kompas dengan arah memanjang sejajar dengan jarum kompas. 2. Perhatikan arah penunjukan jarum kompas. 3. Kemudian, tutuplah sakelar S seperti pada Gambar (b). Amati jarum kompas, apa yang terjadi? 4. Baliklah polaritas baterai seperti pada Gambar (c). Lakukanlah pengamatan seperti langkah 2 dan 4. Amati yang terjadi dengan jarum kompas.

U

U

S

S

S

S I

S

S

0

1

S I

S

I

2 Gambar 5.5

(a) Penghantar tidak dilalui arus listrik; (b) Penghantar dialiri arus listrik dari utara; (c) Penghantar dialiri arus listrik dari arah selatan. Perhatikan arah simpangan jarumnya.

Sumber: Conceptual Physics, 1998

Gambar 5.6 Pola garis medan magnetik yang dibentuk oleh serbuk besi di sekitar kawat lurus berarus listrik.

I

0 5.

1

2

Apakah kesimpulan Anda dari kegiatan ini?

%012%$
?4@B0<0
:0;8
<4A8A8
A48<10=6
F08BC
CB0@0A4;0B0=
A4?4@B8
!,"!0
 ! !4B8:0
0@CA
<4=60;8@
30@8
0@07
CB0@0
90@C<
:>100=
F0=6
38;0:C:0==F0
%012%$
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107E0
38 A4:8B0@
?4=670=B0@
14@0@CA
B4@30?0B
<430=
<06=4B8:
F0=6
0@07=F0 14@60=BC=6
:4?030
0@07
0@CA
;8AB@8:
F0=6
<4=60;8@
<4;0;C8
?4=670=B0@ B4@A41CB
+=BC:
<4=4=BC:0=
0@07
0@CA
;8AB@8:
30=
60@8A60@8A
<430= <06=4B8:
=30
30?0B
<4=66C=0:0=
0BC@0=
B0=60=
:0=0= 8:0
0@CA
;8AB@8:
380;8@:0=
?030
:0E0B
;C@CA
F0=6
<4=4<1CA
B460: ;C@CA
?030
1830=6
F0=6
B4;07
38B01C@8
A4@1C:
14A8
A4@1C:A4@1C:
14A8 0:0=
<4=C=9C::0=
?>;0
60@8A60@8A
<430=
<06=4B8:
F0=6
14@14=BC: ;8=6:0@0=
A4?CA0B
A4?4@B8
38BC=9C::0=
?030
!,"!0
  @07
60@8A60@8A
60F0
<06=4B8:
38
A4:8B0@
?4=670=B0@
14@0@CA
30?0B 38B4=BC:0=
34=60=
:08307
B0=60=
:0=0=
A4?4@B8
!,"!0

14@8:CB

Gambar 5.7 (a) Kaidah tangan kanan; (b) Jika arus menuju pengamat, arah medan magnetik berputar ke kiri; (c) Jika arah arus menjauhi pengamat, arah medan magnetik berputar ke kanan.

B

B

B

B

0

1

2

!4B4@0=60= 
0@07
0@CA
<4=4<1CA
1830=6
60<10@
<4=34:0B8
?4=60<0B 
0@07
0@CA
<4=4<1CA
1830=6
60<10@
<4=90C78
?4=60<0B

108

Mudah dan Aktif Belajar Fisika untuk Kelas XII

8:0
B0=60=
:0=0=
<4=664=660<
?4=670=B0@
;C@CA
34=60=
81C
90@8 <4=C=9C::0=
0@07
0@CA
;8AB@8:
0@07
;8?0B0=
90@890@8
F0=6
;08==F0
<4=C= 9C::0=
0@07
?CB0@0=
60@8A60@8A
60F0
<06=4B8:

3. Hukum Biot-Savart )4B4;07
?4=4@0=6
58A8:0E0=
&@0=28A
14@=0<0 %!-
!/2(12%
(.2
30=
%+(6
!4!02
14@70A8;
<4=4=BC:0=
14A0@=F0
<430= <06=4B8:
F0=6
38B8<1C;:0=
>;47
0@CA
;8AB@8:
AB018;
(.2
30=
!4!02 <4;0:C:0=
?4=60<0B0=
B4@7030?
:>=B@81CA8
8=3C:A8
<06=4B8:
?030
AC0BC B8B8:
F0=6
38B8<1C;:0=
>;47
A41C07
?4=670=B0@
14@0@CA
;8AB@8: !C0B
<430=
<06=4B8:
38=F0B0:0=
30;0<
8=3C:A8
<06=4B8: &4@70B8:0=
!,"!0

C:C<
8>B)0D0@B
<4=F0B0:0=
14A0@=F0 8=3C:A8
<06=4B8:
F0=6
38A4101:0=
>;47
4;4<4=
0@CA
;8AB@8:

 14@10=38=6
;C@CA
34=60=
:C0B
0@CA
;8AB@8:
  14@10=38=6
;C@CA
34=60=
?0=90=6
4;4<4=
?4=670=B0@ 
  14@10=38=6
B4@10;8:
34=60=
:C03@0B
90@0:
&
0=B0@0
A41C07
B8B8:
B4<06=4B8:
34=60=
4;4<4=
?4=670=B0@ 


 A410=38=6
34=60=
A8=CA
AC3CB
0?8B
0=B0@0
0@07
0@CA
30=
60@8A
?4=6 7C1C=6
B8B8:
&
34=60= 
 )420@0
<0B4<0B8A
?4@A0<00=
8>B)0D0@B
30?0B
38BC;8A
A410608
14@8:CB

 

 

A8=

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i d




r

dB

d


P

Gambar 5.8 Kawat d
yang dialiri arus listrik.

I 

8<0=0

030;07
AC3CB
0=B0@0
0@07
30@8


30=
&

030;07
:>=AB0=B0 ?4@<4018;8B0A
F0=6
<4<8;8:8
=8;08
A410608
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*
<   
*
<

I 

4. Induksi Magnetik di Sekitar Kawat Lurus &4@70B8:0=
!,"!0

)4CB0A
:0E0B
;C@CA
34=60=
?0=90=6


380;8@8 0@CA
;8AB@8:
A478=660
B8<1C;
8=3C:A8
<06=4B8:
38A4:8B0@
:0E0B
B4@A41CB 8:0
380<18;
4;4<4=


?030
:0E0B
;C@CA
B4@A41CB
30=
A41C07
B8B8:
& F0=6
14@90@0:
&
30@8


AC3CB
F0=6
3814=BC:
>;47
4;4<4=


34=60=
& 030;07
 4@30A0@:0=
?4@A0<00=
8>B)0D0@B
14A0@=F0
8=3C:A8
<06=4B8:
38 B8B8:
&
030;07  
A8=

 

 & 8:0
?0=90=6
:0E0B


030;07
30@8
=>;
A0
   








 
A8=


&









 
A8=

&

  






r




0

dB

a P

I

I Gambar 5.9

&4@70B8:0=
:4<10;8
!,"!0

%01!,!!-
 9 
<4@C?0:0=
8=3C:A8 <06=4B8:
:0@4=0
?4=60@C7
:0E0B


30@8
=>;
:4
10E07
38?4@78BC=6:0= <0:0
8=3C:A8
<06=4B8:
B>B0;
38
&
030;07     

d


Induksi magnetik akibat kawat lurus berarus di sebuah titik yang berjarak a dari kawat lurus tersebut.

I

Medan Magnetik

109

Tugas Anda 5.1 Coba Anda cari informasi apakah di sekitar kawat lurus berarus timbul medan magnetik juga jika arus yang mengalir arus bolak-balik.

,0@8014;



30=
&
30;0<
?4@A0<00=
8=8
A0;8=6
B4@8:0B
A4AC08
14=BC: A468B860
F0=6
3814=BC:=F0

& 
    30=

A8=  A8=
   

 


 

4=60=
<4=AC1AB8BCA8:0=
70@60
D0@8014;D0@8014;
8=8
?030
%01!,!! 9 
30=
<4=68=B46@0;:0==F0
3830?0B
?4@A0<00=
10@C
A410608
14@8:CB 




 

     



 


  

      






I   +=BC:
<4=4=BC:0=
0@07
8=3C:A8
<06=4B8:=F0
30?0B
386C=0:0= 0BC@0=
B0=60=
:0=0=
&4@70B8:0=
!,"!0
 

i

Gambar 5.10 Arah induksi magnetik di sekitar kawat lurus berarus. (a) Arah arus ke bawah; (b) Arah arus ke atas.

B

B

i

0

1

Contoh 5.2

Tantangan untuk Anda Jika pada kumparan yang dialiri arus listrik ditambahkan inti besi lunak, medan magnetik yang ditimbulkannya semakin besar. Mengapa demikian?

110

)41C07
:0E0B
;C@CA
F0=6
A0=60B
?0=90=6
380;8@8
0@CA
;8AB@8:


*4=BC:0=
14A0@ 8=3C:A8
<06=4B8:
38
0
B8B8:
&
F0=6
14@90@0:

2<
30@8
:0E0B
1
B8B8:
'
F0=6 14@90@0:

2<
30@8
:0E0B !5!" 4=60=
<4=6C=0:0=
%01!,!!-
 9
38?4@>;47  

-1< 

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L
I
*  
 
 
<  

d


P r

0



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L
I
*   
 

<  038
8=3C:A8
<06=4B8:
38
B8B8:
&
030;07

L
I
*
30=
38
B8B8:
'
030;07

L
I
*

Mudah dan Aktif Belajar Fisika untuk Kelas XII

5. Induksi Magnetik di Sekitar Kawat Melingkar 8:0
A4CB0A
:0E0B
F0=6
<4;8=6:0@
380;8@8
0@CA
;8AB@8:
38
A4:8B0@
:0E0B B4@A41CB
0:0=
B8<1C;
8=3C:A8
<06=4B8:
F0=6
14A0@=F0
30?0B
38B4=BC:0= 34=60=
<4=66C=0:0=
?4@A0<00=
8>B)0D0@B
&4@70B8:0=
!,"!0
 0@0
F0=6
B)0D0@B
0:0=
3830?0B
<430= <06=4B8:
38
?CA0B
;8=6:0@0=
A410608
14@8:CB



A8=  

  





   &4@;C
=30
?4@70B8:0=
107E0
?0=90=6
:0E0B
A0<0
34=60=
:4;8;8=6 :0E0B
A478=660
8=B46@0;
<4<8;8:8
10B0A
30@8
=>;
A0
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Id
z

r r I

0

a

x

y

Gambar 5.11 Induksi magnetik di pusat kawat melingkar berarus.




   

  



I  %01!,!!-
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14@;0:C
C=BC:
A0BC
:0E0B
<4;8=6:0@
8:0
B4@30?0B 
;8;8B0=
:0E0B
<4;8=6:0@
<0:0
8=3C:A8
<430=
<06=4B8:
38
?CA0B ;8=6:0@0=
:0E0B
<4=9038 $ 

  I  @07
<430=
<06=4B8:
30?0B
38B4=BC:0=
34=60=
0BC@0=
B0=60=
:0=0= A4?4@B8
?030
!,"!0
 $ 

Contoh 5.3 &4@70B8:0=
60<10@
14@8:CB
8:0
8=3C:A8
<06=4B8:
38
?CA0B
;8=6:0@0=
14@0@CA
;8AB@8:


F0=6
14@90@890@8

030;07

B4=BC:0=
8=3C:A8
<06=4B8:
38
B8B8:
& 0 I 1 I a

Tantangan untuk Anda Perhatikan gambar berikut. I

2I

2I

a2

a P

I

P

!5!"

0 %;47
:0@4=0
14=BC:
:0E0B
70=F0
14@C?0

;8=6:0@0=
14A0@
8=3C:A8
<06=4B8:
38


030;07
$


34=60=
0@07 <0AC:
1830=6



1 (4AC;B0=
:C0B
<430=
<06=4B8:
38
B8B8:

030;07
$
 
I


 

I

a1 A

Penghantar berbentuk dua setengah lingkaran konsentris, memiliki jari-jari a1 = 2 cm dan a2 = 4 cm. Penghantar dialiri oleh arus listrik I = 2 A.Tentukanlah besar induksi magnetik total yang timbul di titik pusat A.

Medan Magnetik

111

6. Induksi Magnetik di Pusat dan Ujung Solenoida # "#
030;07
AC0BC
;8;8B0=
:0E0B
?4=670=B0@
F0=6
?0=90=6
0B0C :C;0
A0<0
34=60=
F0=6
3870A8;:0=
>;47
<06=4B
10B0=6 8:0
?0=90=6
A>;4=>830


30=
B4@38@8
0B0A

1C07
;8;8B0=
9C<;07 ;8;8B0=
C=BC:
A4B80?
A0BC0=
?0=90=6
<4=9038
"  
30=
90@890@8
:C;4=>830 30=
38A4101:0=
>;47
4;4<4=
A>;4=>830
A4?0=90=6
,
030;07

P

0 a C

P A

r

x

F

  D

d

1 Gambar 5.12 (a) Solenoida yang dialiri arus listrik; (b) Penampang irisan membujur solenoida.

Hans Oersted (1777–1851)

I 


030;07
AC3CB
0=B0@0
&
30=
,
A478=660   A8= &     2>A42 & A8= ,  2>B0= ,   2>B0= ,    2>A42   

4A0@=F0
8=3C:A8
<06=4B8:
38
&
>;47
A4;C@C7
?0=90=6
:0E0B
A>;4=>830 34=60=
<4=AC1AB8BCA8:0=
70@60
&
30=
,
:4
30;0<
%01!,!!-
 9
030;07   "

       A8=  



Tokoh

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2>A   2>A   

I



8:0
A>;4=>830
B4@A41CB
A0=60B
?0=90=6
A478=660


H
30=


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8=3C:A8
<06=4B8:
38
?CA0B
A>;4=>830
030;07      "

0B0C

   I 
4A0@
8=3C:A8
<06=4B8:
38
C9C=6
A>;4=>830
34=60=


H
30=


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0B0C

   I   


Contoh 5.4 Hans Oersted adalah seorang ahli Fisika berkebangsaan Denmark. Pada tahun 1819 ia menemukan hubungan mendasar antara kemagnetan dan kelistrikan. Percobaannya menunjukkan terjadinya pembelokan pada jarum kompas ketika didekatkan dengan kawat berarus. Fenomena ini dikenal dengan elektromagnetisme. Sumber: www.allautobiographies.com

112

)41C07
A>;4=>830
F0=6
?0=90=6=F0



<
30=
90@890@8
&


<
<4<8;8:8

;8;8B0= A>;4=>830
B4@A41CB
380;8@8
0@CA
A414A0@

 0 *4=BC:0=
8=3C:A8
<06=4B8:
38
C9C=6
A>;4=>830 1 8:0
A>;4=>830
38@4=660=6:0=
A478=660
?0=90=6=F0
3C0
:0;8
A4;4=>830
B4@A41CB !5!" 0 =3C:A8
<06=4B8:
38
C9C=6
A>;4=>830
38B4=BC:0=
34=60=
<4=66C=0:0= %01!,!!-
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F08BC 
 
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* $   

Mudah dan Aktif Belajar Fisika untuk Kelas XII

1

8:0
?0=90=6=F0
389038:0=
3C0
:0;8
A4;4=>830
<4=9038   

 
-1< 

  
  " 
<    
* $   

6. Induksi Magnetik di Sumbu Toroida #&#
030;07
:C@>830
380;8@8
0@CA
;8AB@8:
0:0=
B8<1C; 60@8A60@8A
<430=
<06=4B8:
14@14=BC:
;8=6:0@0=
38
30;0<
B>@>830
B4@A41CB 4A0@=F0
8=3C:A8
<06=4B8:
38
B8B8:
&
F0=6
14@030
38
?CA0B
AC<1C B>@>830
030;07

Kata Kunci • • • • • • • •

medan magnetik garis gaya magnetik fluks magnet kaidah tangan kanan hukum Biot-Savart solenoida toroida induksi magnetik

lingkaran arus

I



   "

$8;08
"
030;07
9C<;07
;8;8B0=

?4@A0BC0=
:4;8;8=6
;8=6:0@0=
 2 a  A478=660
%01!,!!-
 9
30?0B
38BC;8A
A410608
14@8:CB

 !4B4@0=60=  
90@890@8
B>@>830
<  
9C<;07
;8;8B0=

Tes Kompetensi Subbab

 

 

a

I  B

Gambar 5.13 Toroida yang dialiri arus listrik.

A

%0)!*!-+!'
$!+!,
"3*3
+!2('!-



*4=BC:0=
0@07
<430=
<06=4B8:
38
B8B8:
&
?030
60<10@ 14@8:CB 0 1 2 P



P

I

I

I

8:0
8=3C:A8
<06=4B8:
38
?CA0B
:0E0B
;8=6:0@0=
F0=6 14@0@CA

30=
14@90@890@8

030;07

B4=BC:0=
8=3C:A8 <06=4B8:
38
B8B8:
&
?030
60<10@
14@8:CB 0 9I 6I

P





a

2I P

)41C07
:0E0B
;C@CA
14@0@CA
;8AB@8:
38;4B0::0=
<4 <0=90=6
?030
0@07
CB0@0A4;0B0=
8:0
0@CA
<4=60;8@ ?030
:0E0B
30@8
CB0@0
:4
A4;0B0=
B4=BC:0=
0@07
<430= <06=4B8:
38
B8B8:
&
F0=6
14@90@0:
&
38
0B0A
:0E0B
B4@A41CB &030
60<10@
14@8:CB

2a 3a

1 I 2a

y+ P

B


x–

I

x+



A z y–

&4=670=B0@

F0=6
14@0@CA
;8AB@8:
14@78A8B85
*4=BC:0=
0@07
<430= <06=4B8:
38
B8B8:

30=


)41C07
:C
Medan Magnetik

113

B. Gaya Magnetik










F












I









Gambar 5.14 Seutas kawat berarus I berada di ruang yang dipengaruhi medan magnetik B sehingga timbul gaya magnetik (F) pada kawat tersebut.

8
!4;0A
.
=30
B4;07
<4=64B07C8
107E0
98:0
?4=670=B0@
F0=6 380;8@8
0@CA
;8AB@8:
30=
38B4;47
:0E0B 14@0@CA
;8AB@8:
38
30;0<
<430=
<06=4B8:
14@10=38=6
;C@CA
34=60=
:C0B 0@CA
?0=90=6
:0E0B
?4=670=B0@
:C0B
<430=
<06=4B8:
30=
A8=CA
AC3CB 0=B0@0
0@07
0@CA
30=
0@07
8=3C:A8
<06=4B8:
)420@0
<0B4<0B8A
14A0@ 60F0
<06=4B8:
30?0B
38BC;8A:0=
A410608
14@8:CB I   
A8=

!4B4@0=60= 
60F0
<06=4B8:
$


:C0B
0@CA
;8AB@8:
  
8=3C:A8
<06=4B8:
*


AC3CB
F0=6
3814=BC:
>;47

34=60=


?0=90=6
:0E0B
?4=670=B0@ 0;0<
14=BC:
D4:B>@
%01!,!!-
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30?0B
38BC;8A
A410608
14@8:CB 


L



I 

1. Menentukan Arah Gaya Magnetik I


B

=30
B4;07
<4=64B07C8
107E0
98:0
A4CB0A
:0E0B
F0=6
<4<8;8:8 ?0=90=6


380;8@8
0@CA
;8AB@8:

30=
38B4
Aktivitas Fisika 5.2 Gambar 5.15 Penghantar panjang
berarus listrik I di dalam medan magnetik homogen B.

Gaya Magnetik Tujuan Percobaan Mengamati gaya magnetik Alat-Alat Percobaan 1. Aluminium foil 2. Sebuah magnet U 3. Sebuah baterai

4. 5.

Sebuah sakelar Dua buah penjepit

Langkah-Langkah Percobaan 1.

Susunlah peralatan tersebut, seperti pada gambar berikut. pita aluminium

penjepit

– + sakelar

114

Mudah dan Aktif Belajar Fisika untuk Kelas XII

sebelum sakelar ditutup, pita aluminium dalam keadaan mendatar

F P

Q – + sakelar

2. 3. 4. 5. 6.

Tutuplah sakelar agar arus listrik mengalir melalui pita aluminium. Apakah yang terjadi dengan pita aluminium tersebut? Ubahlah polaritas dari baterai, kemudian amati yang terjadi pada pita aluminium. Balikkan posisi kutub magnet U, kemudian amati yang terjadi pada pita aluminium. Apakah kesimpulan Anda dari kegiatan ini?

&4@70B8:0=
!,"!0
 
8:0
:0E0B
?4=670=B0@
F0=6
380;8@8
0@CA ;8AB@8:

38;4B0::0=
38
30;0<
304@07
<430=
<06=4B8:

:0E0B
B4@A41CB 0:0=
<4=60;0<8
60F0
<06=4B8:

F0=6
0@07=F0
A4?4@B8
38BC=9C::0=
?030 60<10@
B4@A41CB
A478=660
:0E0B
0:0=
<4;4=6:C=6
:4
0B0A :0=
B4B0?8
98:0
?>;0@8B0A
10B4@08
3810;8:
:0E0B
0:0=
<4;4=6:C=6 :4
10E07
0:810B
30@8
60F0
<06=4B8:
F0=6
0@07=F0
:4
10E07
468BC
?C;0 70;=F0
98:0
0@07
8=3C:A8
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Gambar 5.16 Gaya magnetik yang dialami kawat penghantar berarus listrik. I F

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Gambar 5.17 Arah gaya magnetik sesuai dengan kaidah tangan kanan.

Informasi untuk Anda

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Sumber: Physics for Scientist and Engineers

Ion bermuatan dari Matahari mendekati Bumi dan masuk ke atmosfer menuju daerah kutub pada umumnya dikatakan adanya pengaruh medan magnetik Bumi. Kadang-kadang kejadian ini menimbulkan fenomena yang disebut aurora borealis atau cahaya utara di daerah lintang utara.

Information for You Charged ions approach the Earth from the Sun and enter the atmosphere mainly near the poles, sometimes causing a phenomenon called the aurora borealis or northern lights in northern latitudes.

Medan Magnetik

115

B

v

F

R

A

Gambar 5.18 Gaya yang dialami muatan bergerak dalam medan magnetik.

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Gambar 5.19 Pengukuran massa ion menggunakan spektroskop massa.

116

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untuk Anda Suatu partikel bermuatan 0,04 C bergerak sejajar dengan kawat berarus listrik 10 A. Jika jarak antara partikel dan kawat 5 cm, hitunglah berapa gaya yang dialami partikel tersebut?

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117

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garis medan karena I1

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Gambar 5.20 Dua buah penghantar lurus sejajar dialiri oleh arus listrik. (a) I1 searah dengan I2. (b) I1 berlawanan arah dengan I2.

Q

garis medan karena I2

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118

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Tugas Anda 5.2 Anda mengenal kipas angin, mesin cuci, dan motor penggerak mesin jahit. Alat-alat tersebut dihubungkan dengan listrik arus AC. Carilah cara kerjanya. Apakah cara kerjanya sama dengan motor listrik yang dibahas berdasarkan Gambar 5.21?

sik k o m at u ka t a t rb o r on

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Sumber: Science Discovery 1, 2004

Gambar 5.23 Kereta api terbang mampu melaju dengan kecepatan lebih dari 400 km/jam.

Medan Magnetik

121

D. Sifat Magnet Bahan 070=
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Gambar 5.24 (a) Gerak melingkar elektron mengelilingi inti; (b) Spin elektron tidak berpasangan; (c) Spin elektron berpasangan.

U

U

0

Gambar 5.25 Bahan ferromagnetik dalam solenoida

S

U

Gambar 5.26 Bahan paramagnetik dalam solenoida

1

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2. Bahan Paramagnetik

S

U

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3. Bahan Diamagnetik Gambar 5.27 Bahan diamagnetik dalam solenoida

122

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Mudah dan Aktif Belajar Fisika untuk Kelas XII

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Sumber: Physics for Scientists and Engineers, 3rd Edition, 2000

5. Penerapan Bahan Ferromagnetik a. Relai Magnetik (Penghubung)  
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;4
inti besi solenoida armatur (lempengan)

Gambar 5.28 sakelar

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Skema rangkaian relai magnetik

bel

pegas baja

besi U

interuptor (pemutar arus)

kumparan sakelar

Gambar 5.29 Skema rangkaian bel listrik

Medan Magnetik

123

Rangkuman







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Mudah dan Aktif Belajar Fisika untuk Kelas XII



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Refleksi Setelah Anda mempelajari materi bab ini, Anda tentu dapat memahami bahwa medan magnetik dihasilkan dari magnet permanen dan kawat berarus. Anda tentu memahami juga pengaruh medan magnetik pada kawat berarus serta pemanfaatan fenomena tersebut dalam

teknologi. Dari materi yang ada pada bab ini, bagian manakah yang menurut Anda sulit dipahami? Coba Anda diskusikan dengan teman atau guru Fisika Anda. Selain itu, coba Anda cari manfaat dan aplikasi lain dari materimateri bab ini.

Medan Magnetik

125

Tes Kompetensi Bab 5  (+('+!'
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6 Sumber: Physics Today, 1989

Generator listrik mampu mengubah energi kinetik menjadi energi listrik melalui proses induksi elektromagnetik.

Induksi Elektromagnetik Hasil yang harus Anda capai: menerapkan konsep kelistrikan dan kemagnetan dalam berbagai penyelesaian masalah dan produk teknologi. Setelah mempelajari bab ini, Anda harus mampu: memformulasikan konsep induksi Faraday dan arus bolak-balik, keterkaitannya, serta aplikasinya.

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A. Gejala Induksi Elektromagnetik B. Aplikasi Induksi Elektromagnetik Faraday C. Arus dan Tegangan Listrik Bolak-balik

Induksi Elektromagnetik

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A. Gejala Induksi Elektromagnetik 1. Pengertian Induksi Elektromagnetik

1

2

3 Gambar 6.1 Induksi elektromagnetik (a) ketika magnet diam, jarum galvanometer juga diam, (b) ketika magnet dimasukkan ke dalam kumparan, jarum galvanometer bergerak ke kanan, dan (c) ketika magnet dikeluarkan dari dalam kumparan, jarum galvanometer bergerak ke kiri.

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Aktivitas Fisika 6.1 Induksi Elektromagnetik Tujuan Percobaan Mengamati arus induksi Alat-Alat Percobaan 1. Kumparan 2. Galvanometer 3. Magnet batang Langkah-Langkah Percobaan 1. Susunlah peralatan seperti gambar berikut.

magnet

kumparan

galvanometer

132

Mudah dan Aktif Belajar Fisika untuk Kelas XII

2. Siapkan magnet batang, kemudian dekatkan pada kumparan, dan diamkan. Catat hasil pengamatan Anda. 3. Gerakkan magnet mendekati kumparan atau masukkan ke dalam kumparan. Amati jarum galvanometer dan catat hasil pengamatannya. 4. Diamkan magnet batang di dalam kumparan. Amati jarum galvanometer dan catat hasil pengamatannya. 5. Gerakkan magnet batang dengan arah menjauhi kumparan atau keluar kumparan, amati jarum galvanometer, dan catat hasil pengamatannya. 6. Balikkan polaritas magnet batang, lakukan lagi percobaan langkah 2 dan langkah 5. Catat hasil pengamatannya. 7. Tulis kesimpulan Anda tentang kegiatan ini.

Tokoh Michael Faraday (1791–1867)

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Michael Faraday adalah seorang ahli Fisika, ahli Kimia, dan ahli Filsafat berkebangsaan Inggris, tepatnya lahir di Newington Butts, Surey, pada tahun 1971. Ia bersama-sama dengan Joseph Henry menemukan fenomena induksi elektromagnetik. Penemuan ini merupakan kebalikan dari penemuan Oersted, yaitu dari medan listrik dapat menghasilkan arus listrik.

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Sumber: www.allbiographies.com

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Induksi Elektromagnetik

133

Tantangan untuk Anda Bagaimana pengaruh perubahan lilitan terhadap ggl induksi?

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Informasi untuk Anda

Para teknisi ini sedang memasang generator pada instalasi pembangkit listrik di Meksiko. Bagian yang terlihat di sini adalah stator, yang tidak bergerak. Selanjutnya, ke dalam stator itu dimasukkan rotor. Rotor merupakan sebuah elektromagnetik kuat yang dapat berputar dengan kecepatan tinggi. Akibat perputaran rotor, terjadi perubahan medan magnetik dan membangkitkan arus listrik pada kumparan di dalam stator.

Information for You The pisture have shown the technician placing generator on electrical instalation in Mexico. You can see a part, called stator. The rotor is put inside it. Rotor is a strong electromagnetic device that can spin with cause the velocity. That spin is what cause the change of magnetic field and generate electrical element on coil inside stator. Sumber: The Oxford Children’s Book of Science, 1995

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Induksi Elektromagnetik

135

Tantangan untuk Anda Sebuah penghantar berbentuk U terletak di dalam daerah berinduksi magnetik homogen B = 4 × 10–3 T berarah menembus bidang kertas. Penghantar PQ sepanjang 40 cm menempel pada penghantar U dan digerakkan ke kanan dengan kecepatan tetap v = 10 m/s. Hitung ggl induksi antara ujung-ujung PQ.

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4.

Sumber: www.linearmagnetics.com

Induktor adalah sebuah elektromagnet, yaitu sebuah kumparan kawat dengan atau tanpa inti magnetik yang menghasilkan magnetisme jika arus mengalir melewatinya.

Information for You Inductor is an electromagnet, made of wire coil with or without magnetic core which can produce magnetism when electric current through.

Induktansi Diri

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25B1A>G1 :D71
25AD218
C5A8141@
F1;CD
)5AD2181>
65C9;
B521>49>7
45>71> ;D1C
1ADB
,531A1
=1C5=1C9B
49CD<9B
B521719
25A9;DC  ' 
1C1D
 '  


45>71>

=5AD@1;1>
BD1CD
;?>BC1>C1
G1>7
49B52DC
45>71>
9>4D;C1>B9 49A9
;D=@1A1>
C5AB52DC
">4D;C1>B9
49A9
C571;
CD; 75?=5CA9;
;D=@1A1>>G1
,1CD1>
9>4D;C1>B9
91<18
85>AG
!
5>CD; D=D=
9>49;1C?A
25AD@1
B?<5>?941
B531A1
<1>7BD>7
9>4D;C1>B9
49A9 B?<5>?941
9>9
41@1C
4989CD>7
B5@5AC9
25A9;DC
9>9

Sumber: Jendela Iptek, 1997

136

J1

Mudah dan Aktif Belajar Fisika untuk Kelas XII



'

J2

(<58
;1A5>1
65C9;
D>CD;

<9<9C1>
@141
B?<5>?941
141<18  '





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=1;1
9>4D;C1>B9
49A9
B?<5>?941
141<18 B521719
25A9;DC 
  (
J "-.)*
 3
25A<1;D
:D71
D>CD;
9>4D;C1>B9
49A9
@141
@DB1C C?A?941
;1A5>1
25B1A
9>4D;B9
=17>5C
49
1>C1A1
;54D1>G1
B1=1
G19CD B525B1A
 (
 #9;1
@5>1=@1>7
B?<5>?941
1C1D
C?A?941
499B9
?<58
2181>
G1>7
=5=9<9;9 @5A=5129<9C1B
A5<1C96
+
49=1>1
@5A=5129<9C1B
2181>
141<18

   + 

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">4D;C1>B9
49A9
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41>
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141<18

  
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$5C5A1>71>  
9>4D;C1>B9
49A9
B?<5>?941
1C1D
C?A?941
45>71>
9>C9
2181>
!  
9>4D;C1>B9
49A9
B?<5>?941
1C1D
C?A?941
C1>@1
9>C9
!

Pembahasan Soal Kuat arus listrik dalam suatu rangkaian tiba-tiba turun dari 10 A menjadi 2 A dalam waktu 0,1 s. Selama peristiwa ini terjadi, timbul ggl induksi sebesar 32 V dalam rangkaian. Induktansi rangkaian adalah .... a. 0,32 H b. 0,40 H c. 2,5 H d. 32 H e. 40 H UMPTN 1993 Pembahasan: Diketahui: I = 10 A – 2 A = 8 A

 ind = 32 V dI d
3,2 Vs 32 V = 8A 3,2 Vs L= = 0,4 H 8A

 ind  L

Contoh 6.3 -?A?941
45>71>
1=@1>7

3=
41>
@1>:1>7>G1

3=
=5=9<9;9



<9<9C1> -5>CD;1><18
9>4D;C1>B9
49A9
C?A?941
C5AB52DC 1 9;5C18D9 

3=


K
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=   





3=


=  

K

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02 = B589>771
49@5A?<58

Jawaban: e







   

 
02 = 



  
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9>4D;C1>B9
49A9
C?A?941
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G1>7
C5A49A9
1C1B
B52D18
;D=@1A1>
9>4D;C?A
 B52D18
<1=@D
)
41>
BD=25A
C571>71>

G1>7
49<5>7;1@9
45>71>
B52D18 B1;<1A

C1=@1;
@141
)-
 
$5C9;1
B1;<1A
498D2D>7;1>
<1=@D C941;
B575A1
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=1;B9=D=
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141
:541
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B5B11C
)141
B11C 1ADB
49@DCDB;1>
@141
B1;<1A
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:541
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)5A81C9;1>
)-
 
)5A9BC9F1
9>9 =5>D>:D;;1>
C9=2D<>G1
1ADB
9>4D;B9
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49B5212;1>
?<58
141>G1 @5AD2181>
65C9;
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9>4D;C?A
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1ADB
@141
B52D18
;D=@1A1>
41@1C
=5>9=2D<;1>
77< 9>4D;B9
G1>7
25B1A>G1
25A21>49>7
71>
35@1C>G1
@5AD2181>
;D1C 1ADB>G1
,531A1
=1C5=1C9B
@5A>G1C11>
G1>7
49;5=D;1;1>
@5AC1=1
;1<9 ?<58
+.",$
"*-2
9>9
49CD<9B;1>
B521719
25A9;DC

    -

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1

P L

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2 Gambar 6.5 Sebuah rangkaian tertutup (a) saklar s ditutup, lampu menyala dan (b) saklar s dibuka, lampu masih menyala beberapa saat kemudian padam.

Induksi Elektromagnetik

137

-1>41
>571C96
=5AD@1;1>
@5>G5BD191>
45>71>
!D;D=
%5>H
#9;1    <1:D
@5AD2181>
;D1C
1ADB
 
;?>BC1>
=1;1
"-.)*
 3
41@1C  -  49CD<9B;1>
B521719
25A9;DC

  

         -  - 

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$5C5A1>71>

 
;D1C
1ADB
<9BCA9;
@141
;51411>
1F1<



;D1C
1ADB
<9BCA9;
@141
;51411>
1;89A
 - 
-
J
-

B5<1>7
F1;CD
@5AD2181>
;D1C
1ADB
<9BCA9;
B

Contoh 6.4 ,52D18
;D=@1A1>
45>71>
9>4D;C1>B9
49A9

!
491<9A9
1ADB
B521719
6D>7B9
F1;CD




J
- 

41<1=
1=@5A5
41>
-
41<1=
B5;?>
-5>CD;1>
77<
9>4D;B9
49A9
G1>7
C5A:149 B11C
-


B 1 9;5C18D9  

!



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- 5>71>
=5>77D>1;1>
"-.)*
 3 
49@5A?<58    
  -  
J

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- 
J
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/ #149
77<
9>4D;B9
49A9
@141
B11C
-


141<18

/  
 

5. Energi yang Tersimpan dalam Induktor

a

b

I

">4D;C?A
=1=@D
=5>G9=@1>
5>5A79
>5A79
C5AB52DC
C5AB9=@1>
41<1= 25>CD;
=541>
=17>5C
)5A81C9;1>
)-
 
#9;1
B52D18
9>4D;C?A
 491<9A9
1ADB
<9BCA9;

G1>7
B5<1G1
C571>71> 1>C1A1
C9C9;

41>

141<18

  


Gambar 6.6 Sebuah kawat induktor yang dialiri arus listrik.

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J

B589>771
25B1A>G1
41G1
G1>7
4925A9;1>
@141
9>4D;C?A
141<18

              - 

J

>5A79
G1>7
4925A9;1>
@141
9>4D;C?A
49@5A?<58
45>71>
=5>7 9>C57A1B9;1>
41G1
@141
9>4D;C?A
B589>771
49@5A?<58
@5AB1=11>

  

-

  



138

Mudah dan Aktif Belajar Fisika untuk Kelas XII

 -



 


   


J

6. Rapat Energi dalam Bentuk Medan Magnetik -5<18
49;5C18D9
25B1A>G1
9>4D;B9
=17>5C9;
49
@DB1C
B?<5>?941 =5=5>D89
@5AB1=11>
25A9;DC



 (



1C1D







(

41@D>
9>4D;C1>B9
49A9
@141
B?<5>?941
141<18

  (

#9;1
>9<19
41>
>9<19
9>4D;C1>B9

>41
BD2BC9CDB9;1>
;5
41<1=
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49@5A?<58
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5>71>
E??941 C?A?941
141<18




#9;1
A1@1C
5>5A79 1C1D
5>5A79
@5A
B1CD1>
E??C1B9

25A<1;D
8D2D>71>





  

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Tantangan

Contoh 6.5 ,52D18
B?<5>?941
=5=9<9;9
:D=<18
<9<9C1>
B521>G1;



2D18
@1>:1>7
41>
1=@1>7
B?<5>?941
141<18

3=
41>

3= 
!9CD>7<18 1 9>4D;C1>B9
49A9
B?<5>?941 2 5>5A79
G1>7
4981B9<;1>
@141
B?<5>?941
:9;1
1ADB
<9BCA9;
G1>7
=5>71<9A
B525B1A

 3 25B1A
77<
9>4D;B9
49A9
G1>7
C5A:149
:9;1
41<1=
F1;CD

B
;D1C
1ADB
25AD218
=5>:149 
 1 9;5C18D9  



<9<9C1>
 

3=

K J

= 


3=

K
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=



   1 ">4D;C1>B9
49A9
B?<5>?941
141<18

 


Sebuah kumparan (solenoida) memiliki induktansi 500 mH. Hitunglah besar induksi ggl yang dibangkitkan dalam kumparan itu jika ada perubahan arus listrik dari 100 mA menjadi 40 mA dalam waktu 0,01 s secara beraturan.



  
02 = 



    = 

2

 
  
!

  
= >5A79
G1>7
4981B9<;1>
@141
B11C




141<18

3

     
  
! 
   
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5B1A
%
9>4D;B9
49A9
:9;1
1ADB
25AD218
=5>:149


141<18

 


untuk Anda





  

              
   !     
 
/   
B - 

Induksi Elektromagnetik

139

8. Induktansi Silang  S

P

G

R

Gambar 6.7 Induktor silang antara dua buah kumparan.

)5A81C9;1>
)-

-1=@1;
4D1
2D18
;D=@1A1>
G19CD ;D=@1A1>
@A9=5A
)
41>
;D=@1A1>
B5;D>45A
,
G1>7
25A45;1C1>
$D=@1A1> @A9=5A
)
498D2D>7;1>
45>71>
BD=25A
C571>71>

41>
81=21C1>
E1A9125< 
$D=@1A1>
B5;D>45A
,
498D2D>7;1>
45>71>
B52D18
71<E1>?=5C5A
  #9;1
@141
;D=@1A1>
@A9=5A
C5A:149
@5AD2181>
;D1C
1ADB
45>71> =5>7D218D218
81=21C1>

1;1>
C5A:149
@5AD2181>
65C9;
@141 ;D=@1A1>
@A9=5A
5>71>
45=9;91>
65C9;
G1>7
49<9>7;D@9 ;D=@1A1>
B5;D>45A
@D>
1;1>
25AD218
@D<1
)5AD2181>
65C9; C5AB52DC
1;1>
=5>9=2D<;1>
77<
@141
;D=@1A1>
B5;D>45A 7<
G1>7
C9=2D<
@141
;D=@1A1>
@A9=5A
=1D@D>
;D=@1A1>
B5;D>45A 49B52DC
#(.%-(,#
,#&(!
1C1D
#(.%-(,#
-#'&
&#%
5B1A>G1
77<
9>4D;B9 G1>7
C9=2D<
@141
;D=@1A1>
B5;D>45A
;1A5>1
@5>71AD8
@5AD2181>
65C9;
41<1=
F1;CD
-
=5=5>D89
@5AB1=11>

I

Gambar 6.8 Perubahan arus pada kumparan 1 akan memengaruhi arus pada kumparan 2.

    

     -

J

   

    -

J 

#9;1
"-.)*
 3
49C9>:1D
<5298
<1>:DC
49@5A?<58      

1C1D

     5>71>
=5C?45
9>C57A1B9
1;1>
49@5A?<58

           



 



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5>71>
=5>77D>1;1>
@5>1<1A1>
G1>7
B1=1
=1;1
41A9
"-.)*  3 
1;1>
49@5A?<58



Kata Kunci • • • • • • • • • •

gaya gelak listrik (ggl) induksi elektromagnetik Hukum Faraday fluks medan magnetik Hukum Lenz gaya Lorentz ggl induksi diri kumparan primer kumparan sekunder induktansi silang





J 

5B1A
65C9;

G1>7
49C9=2D<;1>
?<58
1ADB
<9BCA9;

45>71>

   
#9;1
>9<19

49BD2BC9CDB9
;1>
;5
41<1=
"-.)*
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49@5A?<58
@5AB1=11>
25A9;DC

:D=<18
<9<9C1>

141<18
   









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$5C5A1>71>

  ;D1C
1ADB
<9BCA9;
;D=@1A1>
@A9=5A
   9>4D;C1>B9
B9<1>7
!   :D=<18
<9<9C1>
;D=@1A1>
B5;D>45A    :D=<18
<9<9C1>
;D=@1A1>
@A9=5A   65C9;
G1>7
49<9>7;D@9
?<58
;D=@1A1>
B5;D>45A
G1>7
49 C9=2D<;1>
?<58
;D=@1A1>
@A9=5A
02

140

Mudah dan Aktif Belajar Fisika untuk Kelas XII

Contoh 6.6 ,52D18
B?<5>?941

;D=@1A1>

=5=9<9;9
@1>:1>7




3=
45>71>
1=@1>7


K

J
=
41>
21>G1;
<9<9C1>





<9<9C1>
9
B5;9C1A
@DB1C
B?<5>?941 49<9<9C;1>
;D=@1A1>

45>71>
21>G1;
<9<9C1>





<9<9C1>
-5>CD;1> 1 9>4D;C1>B9
B9<1>7
;54D1
;D=@1A1> 2 77<
G1>7
C9=2D<
@141
;D=@1A1>

:9;1
@141
;D=@1A1>

=5>71<9A
1ADB
B525B1A

 G1>7
25A21<9;
1A18
41<1=
F1;CD

B 1 1 5>71>
=5>77D>1;1>
"-.)*
 3 
25B1A
9>4D;C1>B9
B9<1>7
41@1C 49C5>CD;1>
B521719
25A9;DC 





  

 
02 = 






  
=  



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K
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! )141
;D=@1A1>

D>CD;
B5<1>7
F1;CD

-


B
;D1C
1ADB

25A21<9;
1A18 B589>771
#

J 

J 

J

5B1A
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9>4D;B9
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Tes Kompetensi Subbab


  
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49
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= B
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R = 0,5  v = 20 m/s Q



B

1<1=
B52D18
;D=@1A1>
C5A41@1C
@5AD2181>
;D1C
1ADB B525B1A


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@5AD2181>
65C9;

02
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;D=@1A1>
;54D1
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9>4D;C1>B9
B9<1>7 ;54D1
;D=@1A1>
:9;1
;D=@1A1>
;54D1
=5=9<9;9


<9<9C1>

Induksi Elektromagnetik

141

B. Aplikasi Induksi Elektromagnetik Faraday @<9;1B9
9>4D;B9
5<5;CA?=17>5C9;
1A141G
41<1=
;5894D@1>
B581A9 81A9
1>C1A1
<19>
B521719
25A9;DC

1. Generator )5A>18;18
>41
25A;D>:D>7
;5
C5=@1C
)5=21>7;9C
%9BCA9;
-5>171 9A
G1>7
141
49
#1C91
C5A41@1C
1<1C
<9BCA9;
G1>7 49>1=1;1>
!(+-)+
5>5A1C?A
141<18
1<1C
G1>7
25A6D>7B9
=5>7D218 5>5A79
;9>5C9;
=5>:149
5>5A79
<9BCA9;
5>5A1C?A
25;5A:1
25A41B1A;1> @A9>B9@
9>4D;B9
5<5;CA?=17>5C9;
-5A41@1C
4D1
:5>9B
75>5A1C?A
G19CD 75>5A1C?A
1ADB
2?<1;21<9;
41>
75>5A1C?A
1ADB
B51A18 a. Generator Arus Bolak-Balik )-

=5>D>:D;;1>
B;5=1
75>5A1C?A
1ADB
2?<1;21<9;
1A1 ;5A:1
75>5A1C?A
C5AB52DC
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B521719
25A9;DC  $D=@1A1>
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49<5C1;;1>
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=541>
=17>5C9;
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@5AD2181>
:D=<18
71A9B
71G1
=17>5C9;
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9>4D;B9
G1>7
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=5>781B9<;1>
1ADB
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1ADB
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.++(- ;1A5>1
B5C91@
;1A2?>
1;1>
=5>41@1C;1>
@?<1A9C1B
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9>4D;B9
G1>7 25AD218D218
;DCD2
@?B9C96
41>
;DCD2
>571C96
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25A71>C91>  .>CD;
=5>G1
1ADB
<9BCA9;
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;54D1
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;D=@1A1> 49@1B1>79
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C5A@9B18
41>
49C5=@5<;1>
@141
B9;1C
;1A2?> G1>7
498D2D>7;1>
45>71>
;125<
@5>G1
Gambar 6.9 Skema generator arus bolak-balik

arah putaran

cincin luncur

b. Generator Arus Searah )141
41B1A>G1
@A9>B9@
;5A:1
75>5A1C?A
1ADB
B51A18
B1=1
45>71>
@A9>B9@ ;5A:1
75>5A1C?A
1ADB
2?<1;21<9;
)5A25411>>G1
@141
75>5A1C?A
1ADB B51A18
49@1B1>7
%)'.--)+
25AD@1
B52D18
39>39>
25<18
D>7B9
;?=DC1C?A 9>9
=5>71CDA
171A
B5C91@
B9;1C
;1A2?>
B5<141@1C;1>
@?<1A9C1B
77< 9>4D;B9
G1>7
C5C1@
B9;1C
@5AC1=1
@?B9C96
41>
B9;1C
<19>>G1
>571C96 kumparan

Gambar 6.10 Skema generator arus searah sikat karbon komutator arah putaran

2. Transformator (Trafo) +(, )+'-)+
141<18
1<1C
G1>7
25A6D>7B9
=5>7D218
=5>19;;1>
1C1D =5>DAD>;1>
C571>71>
<9BCA9;

(<58
;1A5>1
9CD
CA1>B6?A=1C?A
492179 =5>:149
4D1
:5>9B
G19CD
CA1>B6?A=1C?A
,-*.*
41>
CA1>B6?A=1C?A
,-*)0(

142

Mudah dan Aktif Belajar Fisika untuk Kelas XII

-A1>B6?A=1C?A
C5A49A9
1C1B
;D=@1A1>
@A9=5A
;D=@1A1>
B5;D>45A
41>
9>C9 25B9
?>C?8
CA1>B6?A=1C?A
49CD>:D;;1>
@141
)-


Gambar 6.11 Transformator

Sumber: PHYWE, 1989

a. Transformator Step-up -A1>B6?A=1C?A
,-*.*
25A6D>7B9
=5>19;;1>
C571>71>
<9BCA9;

171> B545A81>1
B52D18
CA1>B6?A=1C?A
,-*.*
49CD>:D;;1>
@141
)-
  )141
CA1>B6?A=1C?A
BC5@D@
:D=<18
<9<9C1>
B5;D>45A
<5298
21>G1;
41A9@141 :D=<18
<9<9C1>
@A9=5A
,
* b. Transformator Step-down -A1>B6?A=1C?A
,-*)0(
25A6D>7B9
=5>DAD>;1>
C571>71>
<9BCA9;
 171>
B545A81>1
B52D18
CA1>B6?A=1C?A
,-*)0(
49CD>:D;;1>
@141 )-
 
)141
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,-*)0(
:D=<18
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@A9=5A
<5298 21>G1;
41A9@141
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*
, )A9>B9@
;5A:1
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141<18
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@A9=5A
498D2D>7;1>
45>71>
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=1BD;1>
25AD@1 1ADB
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25B1A
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25AD218D218
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25B9
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71G1
=17>5C9;
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C5A:149
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*
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*
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,
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@A9=5A
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C571>71>
B5;D>45A
/  , 
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<9<9C1>
B5;D>45A #9;1
@141
CA1>B6?A=1C?A
491>771@
C941;
141
5>5A79
<9BCA9;
G1>7
25A D218
=5>:149
5>5A79
25>CD;
<19>
CA16?
9451<
25A<1;D
8D2D>71>
B521719 25A9;DC 41G1
@A9=5A

41G1
B5;D>45A @

B
1C1D
@
@

B
B

primer

sekunder Ns

Np

Gambar 6.12 Bagan transformator step-up

primer

sekunder

Np

Ns

Gambar 6.13 Bagan transformator step-down

J 

$5C5A1>71> * 
41G1
@A9=5A
0 , 
41G1
B5;D>45A
0

* 
1ADB
@A9=5A


, 
1ADB
B5;D>45A


Induksi Elektromagnetik

143

Aktivitas Fisika 6.2 Membuat Transformator Tujuan Percobaan Membuat transformator step-up. Alat-Alat Percobaan

kumparan

1. Kawat tembaga kira-kira 1 m 2. Sumber tegangan 12 V (AC)

inti besi

voltmeter Langkah-Langkah Percobaan 1. Susunlah peralatan seperti gambar berikut. 2. Lilitkan 10 lilitan kawat pada bagian kiri inti besi, sebagai kumparan primer. Kemudian, sebanyak 20 lilitan pada bagian karena inti besi yang lain, sebagai kumparan sekunder. Hubungkan dengan sumber tegangan. 3. Catat besar tegangan yang terbaca pada voltmeter. 4. Teruskan menambah 10 lilitan pada kumparan sekunder hingga menjadi 50 lilitan. Amati dan catat besar tegangan yang terbaca pada voltmeter. 5. Masukkan data hasil pengamatan Anda pada tabel. Kemudian, buatlah grafik tegangan sekunder terhadap jumlah lilitan sekunder.

Data Percobaan

Tantangan untuk Anda Anda memiliki walkman yang menggunakan baterai 3 V. Bagaimana caranya agar walkman tersebut dapat menggunakan listrik dari PLN yang besarnya 220 V?

No

Jumlah Lilitan Sekunder

Besar Tegangan Sekunder

1. 2. 3. 4.

20 30 40 50

.... .... .... ....

6. Berikan kesimpulan Anda dari kegiatan ini.

Contoh 6.7 ,52D18
CA16?
BC5@4?F>
1;1>
497D>1;1>
D>CD;
=5>G1<1;1>
A149?

/
41A9
BD=25A )%'

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<9<9C1>
@A9=5A
41>
B5;D>45A
CA16?
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<9<9C1>
@A9=5A
:9;1
<9<9C1>
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G1>7
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:9;1
49
;D=@1A1>

@A9=5A
C5A41@1C
1ADB
B525B1A

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 *


/

,




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,


/

*



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144

Mudah dan Aktif Belajar Fisika untuk Kelas XII

)141
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CA1>B6?A=1C?A
G1>7
9451<
,521791>
5>5A79 <9BCA9;
25AD218
=5>:149
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41<1=
CA1>B6?A=1C?A
=9B1<>G1 5>5A79
@1>1B
#149
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CA1>B6?A=1C?A
C941;
9451<
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41A9@141
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@A9=5A
,*
B589>771
569B95>B9>G1 <5298
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41A9


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494569>9B9;1>
B521719
81B9<
2179
41G1 <9BCA9;
B5;D>45A
45>71>
41G1
<9BCA9;
@A9=5A
69B95>B9
CA1>B6?A=1C?A 49>G1C1;1>
41<1=
@5AB5>
B531A1
=1C5=1C9B
49CD<9B



,  

 *

Ingatlah Tidak ada transformator yang efisiensinya 100% karena sebagian energi listrik berubah menjadi bentuk energi lain.

J 

$5C5A1>71> 
569B95>B9
CA1>B6?A=1C?A , 
41G1
<9BCA9;
B5;D>45A * 
41G1
<9BCA9;
@A9=5A

Contoh 6.8 ,52D18
CA1>B6?A=1C?A
41<1=
B52D18
A149?
CA1>B9BC?A
41@1C
=5>7D218
C571>71>
41A9
/
=5>:149

/
#D=<18
<9<9C1>
B5;D>45A>G1

<9<9C1>
C5A>G1C1
;D1C
1ADB
G1>7 =5>71<9A
@141
A149?


#9;1
569B95>B9
CA1>B6?A=1C?A
C5AB52DC
 
C5>CD;1> 1 ;D1C
1ADB
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@141
;D=@1A1>
@A9=5A 2 41G1
@A9=5A
41>
41G1
B5;D>45A 1 9;5C18D9
 *


/

,


<9<9C1>

,


/

,



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C941;
25AB961C
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=1;1
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0 , 
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141<18

0

c.

Arus Pusar pada Transformator )5A81C9;1>
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;?>4D;C?A
1>C1A1
4D1
=D;1
;DCD2
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G1>7
49CD>:D;;1>
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)-

#9;1
=541>
=17>5C9;


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67
=5<17
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C5ACDCD@
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1;1>
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1ADB
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49>1=1;1>
45>71>
+.,
2
1C1D
1ADB
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@DB1A
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CA16?
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5>5A79
1C1D
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9>9
3D;D@ C9>779
CA16?
41@1C
C5A21;1A
B589>771
ADB1; .>CD;
=5>35718
;5AD791>
41G1
21C1>7
;?>4D;C?A
CA1>B6?A=1C?A 492D1C
25A<1@9B<1@9B
B5@5AC9
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)-

9
1>C1A1
<1@9B1> @5<1C
771
<9>7;1A1>
1ADB
@DB1A
=5>:149

kumparan

arus pusar

inti besi

1

2 Gambar 6.14 (a) Trafo dengan inti besi. (b) Trafo dengan inti besi berupa lempengan yang terpisahkan oleh laker.

Induksi Elektromagnetik

145

Kata Kunci • • • • • • • •

generator Alternating Current komutator transformator/trafo transformator step-up transformator step-down efisiensi transformator arus pusar (arus eddy)

;539<
5>71>
31A1
B5@5AC9
9>9
;5AD791>
41G1
1;921C
1ADB
@DB1A
41@1C 49;DA1>79 -941;
B5=D1
1ADB
@DB1A
25AB961C
=5AD79;1>
@141
2525A1@1
@5A1<1C1> G1>7
=5>77D>1;1>
@A9>B9@
9>4D;B9
5<5;CA?=17>5C9;
1ADB
@DB1A
G1>7 C9=2D<
41@1C
=5>7D>CD>7;1>
B5@5AC9
@141
A5=
=17>5C9;
CD>7;D
<9BCA9; 41>
B5CA9;1
<9BCA9;

3. Transmisi Daya Listrik Jarak Jauh -A1>B=9B9
41G1
<9BCA9;
:1A1;
:1D8
=5AD@1;1>
@A?B5B
@5>G1
41G1 <9BCA9;
;5
C5=@1C
CD:D1>
C5AC5>CD
G1>7
:1A1;>G1
:1D8
41A9
@DB1C
@5=21>7;9C <9BCA9;
%9BCA9;
G1>7
B1=@19
49
AD=18AD=18
@5>4D4D;
141<18
81B9< CA1>B=9B9
41G1
<9BCA9;
:1A1;
:1D8 41
4D1
31A1
=5>CA1>B=9B9;1>
41G1
<9BCA9;
:1A1;
:1D8
G19CD  45>71>
;D1C
1ADB
<9BCA9;
25B1A
G1>7
4981B9<;1>
?<58
75>5A1C?A  45>71>
C571>71>
G1>7
<5298
C9>779
41A9
C571>71>
G1>7
4981B9<;1> ?<58
75>5A1C?A )141
D=D=>G1
CA1>B=9B9
41G1
<9BCA9;
:1A1;
:1D8
=5>77D>1;1>
C571>7 1>
C9>779
$5D>CD>71>
G1>7
49@5A?<58
45>71>
=5>77D>1;1>
CA1>B=9B9 C571>71>
C9>779
141<18
1ADB
<9BCA9;
G1>7
=5>71<9A
=5<1771
D;DA1>
;1F1C
CA1>B=9B9
G1>7
;539<
:D71
41@1C
=5=25A9;1> 1>771A1>
291G1
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41>
41G1
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CA1>B=9B9
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B589>771
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41G1
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25B1A )141
@A?B5B
CA1>B=9B9
41G1
<9BCA9;
:1A1;
:1D8
497D>1;1>
CA1>B6?A=1C?A BC5@D@
G1>7
C5A41@1C
49
@DB1C
@5=21>7;9C
G1>7
=5>19;;1>
C571>71> 89>771
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B5;9C1A
 


/
$5=D491>
41G1
<9BCA9;
49CA1>B=9B9;1> ;5
;?C1;?C1
CD:D1>
9
;?C1;?C1
CD:D1>
497D>1;1>
CA1>B6?A=1C?A
BC5@ 4?F>
G1>7
49@1B1>7
49
71A4D71A4D
<9BCA9;
B589>771
C571>71>
<9BCA9;
B91@ 497D>1;1>
@141
C571>71>
B5;9C1A

/ pabrik

transmisi tegangan tinggi pusat pembangkit listrik tiang listrik generator

rumah

Gambar 6.15 Bagan transmisi daya listrik jarak jauh transformator

transformator

gardu listrik Sumber: Jendela IPTEK, 1995

Contoh 6.9 ,52D18
75>5A1C?A
G1>7
41@1C
=5>781B9<;1>
41G1
<9BCA9;

&0
49CA1>B=9B9;1>
=5<1781>C1A
45>71>
81=21C1>


 
!9CD>7<18
@5AB5>C1B5
41G1
G1>7
89<1>7 =5>:149
;171>
G1>7
497D>1;1>
CA1>B=9B9

;/ 2 C571>71>
49>19;;1>
?<58
CA16?
BC5@
D@
=5>:149



;/ 1 9;5C18D9




&0

+

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41G1
<9BCA9;
G1>7
89<1>7
:9;1
497D>1;1>
C571>71>
CA1>B=9B9

;/ 41@1C
4989CD>7
B521719
25A9;DC    
0  




CA1>B=9B9      
/

146

Mudah dan Aktif Belajar Fisika untuk Kelas XII

1G1
G1>7
89<1>7
=5>:149
;17
141<18 89<1>7









 


&0 
&0  

  
&0 1G1
G1>7
89<1>7
:9;1
C571>71>
CA1>B=9B9
49>19;;1>
=5<1:149 


;/
141<18    
0  


CA1>B=9B9    

 
/ 1G1
G1>7
89<1>7
=5>:149
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141<18 89<1>7






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0  
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@5AB5>C1B5
89<1>7    

     
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89<1>7 

2

Tes Kompetensi Subbab

B

"-&'*($
!()
0'0
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CA1>B6?A=1C?A
BC5@D@
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@A9=5A>G1
41>



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41>
;D1C
1ADB>G1


-5A>G1C1 C571>71>
;5>G1
B525B1A

/
41>
;D1C
1ADB>G1






!9CD>7 1 569B95>B9
CA1>B6?A=1C?A 2 @5A21>49>71>
:D=<18
<9<9C1>
@A9=5A
41> B5;D>45A ,52D18
CA1>B6?A=1C?A
G1>7
49@1B1>7
@141
A149? =5=9<9;9
B@5B969;1B9

/
C5A=9>1<
@A9=5A>G1 498D2D>7;1>
45>71>
BD=25A
)%'

/
#9;1
:D=<18 <9<9C1>
B5;D>45A
CA1>B6?A=1C?A
C5AB52DC

<9<9C1>
41> 1ADB
G1>7
=5>71<9A
41<1=
A149?


C5>CD;1> 1 :D=<18
<9<9C1>
@A9=5A 2 ;D1C
1ADB
@141
;D=@1A1>
@A9=5A 3 41G1
@A9=5A
41>
41G1
B5;D>45A
:9;1
569B95>B9 CA1>B6?A=1C?A
 

C. Arus dan Tegangan Listrik Bolak-balik ADB
41>
C571>71>
<9BCA9;
2?<1;21<9;
=5=9<9;9
>9<19
G1>7
B5<1 1A18>G1
5B1A1>
1ADB
41>
C571>71>
2?<1;21<9;
49<1=21>7;1>
45>71> 
5F1B1
9>9
81=@9A
B5=D1
@5A1<1C1>
AD=18
C1>771
49?@5A1B9;1> 45>71>
5>5A79
<9BCA9;
1ADB
2?<1;21<9;
,5@5AC9
>41
;5C18D9
@5A25411> =5>41B1A
1>C1A1
1ADB
2?<1;21<9;
41>
1ADB
B51A18
141<18
@?<1A9C1B>G1 .>CD;
=5>75C18D9
@?<1A9C1B
1ADB
B51A18
G1>7
B5<1
1ADB 2?<1;21<9;
G1>7
B5<11;1>
?B99
41@1C
:D71
491=1C9
>9<19
6A5;D5>B9
41> @5AD2181>
C571>71>
@141
B5C91@
B11C
41<1=
25>CD;
7A169;
41@D>
D>CD; =5>7D;DA
>9<19
C571>71>
565;C96
41>
;D1C
1ADB
<9BCA9;
565;C96
41@1C 497D>1;1>
E?
1=@5A5=5C5A


1. Sumber Tegangan Bolak-Balik (AC) 5>CD;
C571>71>
1ADB

G1>7
C5A2131
@141
=?>9C?A
?B9DADC
6D>7B9
B9>DB
,1<18
B1CD
BD=25A
 G19CD
75>5A1C?A
1ADB

1<1=
75>5A1C?A
141
;D=@1A1>
G1>7
41@1C 25A@DC1A
2521B
41<1=
=541>
=17>5C
8?=?75>

Induksi Elektromagnetik

147

V

Vm

V = Vm sint

T

2T

1 I

Im

 J

I = Im sint

T

)5A81C9;1>
)-
 
(B9<1B9
71G1
75A1;
<9BCA9;
41>
1ADB
2?<1; 21<9;
4981B9<;1>
41A9
B52D18
;D=@1A1>
G1>7
25A@DC1A
45>71>
<1:D
C5C1@ )141
71=21A
7A169;
C5AB52DC
@D>31;
41>
B9=@D<
757
=5>D>:D;;1> ;54D4D;1>
;D=@1A1>
C5A8141@
25B1A
BD4DC
G1>7
49C5=@D8>G1 5B1A
77<
G1>7
4981B9<;1>
41A9
B52D18
75>5A1C?A
G1>7
25A@DC1A =5=5>D89
@5AB1=11>
25A9;DC

2T

 
1C1D
D>CD;
-  
  J  -

;1A5>1




3?B
=1;1 
3?B  

3?B  - 
1C1D
 
J  #9;1
;9C1
CDAD>;1>
"-.)*
 3 
4941@1C

 J

2 Gambar 6.16 Grafik tegangan dan arus listrik bolak-balik terhadap waktu.

J  J 

    B9>  C1D
41@1C
:D71
49CD<9B;1>
45>71>

J 1

   =1;B B9>  -

$1A5>1
77<
9>4D;B9
B1=1
45>71>
2541
C571>71>
49
1>C1A1
4D1
;DCD2
77< 9>4D;B9
=1;1
41@1C
49CD<9B J 2

  =1;B B9> 
-

$5C5A1>71>  '%, 
71G1
75A1;
<9BCA9;
=1;B9=D=
E?71>
=1;B9=D=
E?5C9;
02  
:D=<18
<9<9C1>
;D=@1A1> 
7
;D=@1A1>
= 


25B1A>G1
9>4D;B9
=17>5C9;
-   
6A5;D5>B9
BD4DC
@DC1A1>
;D=@1A1>
A14 B

#9;1
75>5A1C?A
C5AB52DC
498D2D>7;1>
45>71>
A1>7;191>
C5ACDCD@ 1;1>
=5>71<9A
1ADB

bagian atas yang berlogam

bagian tengah yang transparan

2. Alat Ukur Tegangan dan Arus Bolak-Balik

lampu

logam

Gambar 6.17 Cara pengukuran dengan testpen

.>CD;
=5>75C18D9
141>G1
C571>71>
<9BCA9;
@141
BD1CD
A1>7;191> B1<18
B1CD>G1
497D>1;1>
B52D18
1<1C
G1>7
49B52DC
-,-*(
)-
 ;1>
C5C1@9
C5BC@5>
C941;
41@1C
=5>75C18D9
25B1A>G1
C571>71>
G1>7 =5>71<9A
@141
A1>7;191>
C5AB52DC
41@D>
D>CD;
=5>7D;DA
>9<19
C571>71> 41>
;D1C
1ADB

497D>1;1>
E?
1=@5A5=5C5A

'9<19 G1>7
C5A2131
@141
1<1C
D;DA
9>9
=5>D>:D;;1>
81A71
565;C96
41A9
C571>71> 41>
;D1C
1ADB
&5<141
41@1C
=5=181=9
81B9< @5>7D;DA1>
1ADB
<9BCA9;
41<1=
B52D18
A1>7;191>

Gambar 6.18 Pengukuran arus listrik dengan menggunakan volmeter.

148

1

Mudah dan Aktif Belajar Fisika untuk Kelas XII

2

)141
)-

C5A=9>1<
;DCD2
@?B9C96
498D2D>7;1>
45>71> 1ADB
=1;B9=D=


@141
1=@5A5=5C5A
C5A2131
>9<19
1ADB


B541>7 ;1>
)-

C5A=9>1<
@?B9C96
498D2D>7;1>
45>71>
1ADB
=1;B9=1<



C5A>G1C1
C5A2131
>9<19
1ADB


)5>7D;DA1>
9>9
=5>D>:D;;1>
218F1 1>7;1
G1>7
C5A2131
@141
1<1C
D;DA
141<18
>9<19
565;C96>G1
5=9;91> @D<1
41<1=
=5>7D;DA
C571>71>
B52D18
A1>7;191>
B5@5AC9
)-
 $5C9;1
1ADB
=5>71<9A
:1AD=
@5>D>:D;
E?7
C5A2131
=5>D>:D;;1>
81A71
565;C96
C571>71>
A1>7;191>
<9BCA9;
>41 41@1C
:D71
=5>7D;DA
;D1C
1ADB
C571>71>
41>
81=21C1>
<9BCA9;
219; 1ADB

=1D@D>
1ADB

45>71>
=5>77D>1;1>
=D
(8==5C5A
41>
?B9
resistor

voltmeter

Gambar 6.19 Pengukuran tegangan pada sebuah rangkaian

Contoh 6.10 1A9
@5>71=1C1>
=5<171>

/ 3= B541>7;1>
B5<1>7
F1;CD
=5>D>:D;;1>

=B 3=
,;1<1
C91@
;?C1;
=5=9<9;9
D;DA1> 
3=
K

3=
-5>CD;1> 1 C571>71>
=1;B9=D= 2 6A5;D5>B9
BD=25A 1 9;5C18D9

**


/ 3= 1 )141
71=21A
C5A2131
C571>71>
@D>31;
**
141<18

3=
;1A5>1
B;1<1
E5AC9;1<


/ 3=
=1;1 **



3=
K

/ 3=


E?71>
=1;B9=D=>G1
141<18   '%,

**


/


/ 2 )5A9?45
41<1=
7A169;

3=
,;1<1
8?A9H?>C1<

=B 3=
=1;1 


3=
K

=B 3=


=B




B     
!H #149
6A5;D5>B9
BD=25A
  





3. Nilai Efektif Tegangan dan Kuat Arus AC <1C
D;DA
G1>7
497D>1;1>
D>CD;
=5>75C18D9
:D=<18
41G1
G1>7 C5A@1;19
@141
;52DCD81>
5>5A79
<9BCA9;
49
AD=18
141<18
;08
=5C5A 5A5C1>
1>7;1
=5>D>:D;;1>
25B1A
5>5A79
<9BCA9;
G1>7
C5A@1;19 )5A81C9;1>
)-
 
-1=@1;
:5<1B
1>7;1
G1>7
=5>D>:D;;1> 25B1A
5>5A79
<9BCA9;
G1>7
C5A@1;19
41<1=
B1CD1>
;08
%#&)
--
").+ >5A79
9>9
49@5A?<58
41A9
>9<19

41>
 
!1A71
565;C96
1ADB
1C1D
C571>71> 2?<1;21<9;
141<18
;D1C
1ADB
1C1D
C571>71>
2?<1;21<9;
G1>7
491>771@ B5C1A1
45>71>
1ADB
1C1D
C571>71>
B51A18
G1>7
=5>781B9<;1>
:D=<18
;17
B1=1
;5C9;1
=5<1781>C1A
41<1=
F1;CD
G1>7
B1=1 A

R

B

Gambar 6.20 kWh meter yang ada di rumah digunakan untuk mengukur daya listrik yang terpakai.

I Im I ef t

Gambar 6.21

   m sint

1

2

1 t 2

(a) Sebuah rangkaian dengan hambatan R; (b) Grafik hubungan I dengan t.

,52D18
@5>781>C1A
B5@5AC9
)-
 
=5=9<9;9
81=21C1>

41<1=  F1;CD
-
5B1A
5>5A79
;17
4981B9<;1>
?<58
1ADB
565;C96>G1
141<18

Induksi Elektromagnetik

149









J 

>5A79
;17
4981B9<;1>
41<1=
A1>7;191>
1ADB

=5=9<9;9
>9<19  G1>7
B5<1
=5=5>D89 @5AB1=11> 

-

   



J 

"-.)*
 3 
41>
"-.)*
 3 
=5=9<9;9
>9<19
G1>7
B1=1 B589>771
41@1C
49@5A?<58
25B1A
1ADB
565;C96>G1  
  



56


- 






56




 

-

=1;B 

J 

45>71>
=5>77D>1;1>
31A1
=1C5=1C9;1
49@5A?<58
1ADB
565;C96
  
B521719 25A9;DC

=1;B




J

 5>71>
31A1
G1>7
B1=1
C571>71>
565;C96
 
1;1>
49@5A?<58
@5AB1=11> 



 



=1;B

J 

Contoh 6.11

Tantangan untuk Anda Apakah yang menyebabkan sengatan listrik, arus listrik, atau tegangan listrik?

150

,52D18
A1>7;191>
G1>7
498D2D>7;1>
45>71>
BD=25A
<9BCA9;
2?<1;21<9;
49D;DA 45>71>
E?
=5>D>:D;;1>
1>7;1

E?71>
2?<1;21<9;
BD=25A
41>
CD<9B;1>
@5AB1=11>>G1
:9;1
6A5;D5>B9>G1

!H 1 9;5C18D9  

/ 

!H 1 '9<19
C571>71>
=1;B9=D=
=1;B
4989CD>7
45>71>
=5>77D>1;1>
"-.)*  3 
B589>771
49@5A?<58  =1;B 
 2



/ )5AB1=11>
C571>71>
49@5A?<58
45>71>
=5>77D>1;1>
"-.)*
 3 
G19CD  
=1;B
B9>  
B9>

-

Mudah dan Aktif Belajar Fisika untuk Kelas XII

Contoh 6.12 +74F3:
C3@9=3;3@
3CFD
4A>3=43>;=
?7?;>;=;
:3C93
E793@93@
D74393;
8F@9D;
H3=EF J3;EF


 D;@

.
!;EF@9>3: 3 E793@93@
?3=D;?F?
%#* 4 E793@93@
BF@53=
=7
BF@53=
(( 5 E793@93@
787=E;8

  6 8C7=F7@D;
3@9F>7C
  7 B7C;A67


 8 8C7=F7@D;


 9 E793@93@
D7E7>3:
+

 
D7=A@ #7#$ 7@93@
?7@J3?3=3@
B7CD3?33@
E793@93@
J3@9
6;=7E3:F;
E7C:363B
'34#/##0  9
?3=3
6;B7CA>7: 

%
D;@ +



 D;@

+ 63C;
B7CD3?33@
E7CD74FE
6;B7CA>7: 


.
63@



C36 D 3

%


.

4

((


%

 .



.

5





6




C36 D

7



  


D

8






9



 D;@

+

 D;@
   



.

%

 
.  
.









  
!K 
I Imaks

4. Harga Kuat Arus dan Tegangan Rata-Rata !3C93
=F3E
3CFD
4A>3=43>;=
C3E3C3E3
363>3:
=F3E
3CFD
4A>3=43>;= J3@9
@;>3;@J3
D7E3C3
67@93@
=F3E
3CFD
D73C3:
F@EF=
?7?;@63:=3@ ?F3E3@
>;DEC;=
J3@9
D3?3
63>3?
H3=EF
J3@9
D3?3
)7C:3E;=3@
9C38;=  D;@FDA;63
63C;
3CFD

D7>3?3

D7=A@
#/$#3
  -@EF=
3CFD
D73C3:
D7E7>3:

D7E7@93:
B7C;A67
?F3E3@
J3@9 6;>3>F;@J3
?7?7@F:;
B7CD3?33@  

) 

Ir t 1 t 2

Gambar 6.22 Grafik hubungan I dan t Imaks = kuat arus maksimum Ir = kuat arus rata-rata.

N 

63BF@
3:
?F3E3@
J3@9
6;>3>F;
3CFD
4A>3=43>;=
63>3?
C3@9=3;3@ 363>3: 

 



%#*

D;@ ++

N 



Induksi Elektromagnetik

151

3C;
'34#/##0
 9 
63@
'34#/##0
 9 
3=3@
6;B7CA>7:
B7C D3?33@
3CFD
C3E3C3E3
D74393;
47C;=FE

Tugas Anda Arus listrik bolak-balik dapat diamati menggunakan osiloskop. Bersama kelompok belajar Anda, gunakan osiloskop yang ada di laboratorium sekolah Anda untuk mempelajari penggunaan osiloskop tersebut. Catatlah halhal yang Anda anggap penting. Mintalah bimbingan guru Fisika Anda untuk melakukan tugas ini.

)



 



%#*

D;@ ++

N 



7@93@
?7@99F@3=3@
53C3
?3E7?3E;=3
6;B7CA>7: )



 %#*


N 

7@93@
B7@<343C3@
J3@9
D3?3
3=3@
6;B7CA>7:
B7CD3?33@
E793@93@ C3E3C3E3
F@EF=
3CFD

D747D3C )



%#*


N 

$7E7C3@93@ ) 
E793@93@
C3E3C3E3
. %#* 
E793@93@
?3=D;?F?
.

Contoh 6.13 )7CD3?33@
E793@93@
4A>3=43>;=
DF3EF
C3@9=3;3@
>;DEC;=
?7?7@F:;
B7CD3?33@ 


D;@

.
,7@EF=3@
E793@93@
C3E3C3E3
J3@9
6;:3D;>=3@
DF?47C
E7CD74FE #7#$ ;=7E3:F;



D;@

.
%#*


. 7@93@
?7@99F@3=3@
'34#/##0
 9 
6;B7CA>7:  ) 


%#*

. 






.


5. Diagram Fasor Im Vm

t

Gambar 6.23 Grafik fasor I dan V R

a

b

I V = Vmsin t

Gambar 6.24 Rangkaian arus AC dengan hambatan murni R.

152

-@EF=
?7?B7C?F63:
?7@93@3>;D;D
3C3:
E793@93@
63@
=F3E
3CFD 6;9F@3=3@
?7EA67
6;39C3?
83DAC
3DAC
(!*
-#+')
363>3:
DF3EF
47D3C3@ J3@9
6;>F=;D=3@
D74393;
DF3EF
G7=EAC
J3@9
47D3C
DF6FE
BFE3C@J3
E7C:363B DF?4F
:AC;KA@E3>
D3?3
67@93@
DF6FE
83D7@J3
 + 
3DAC
DF3EF
G7=EAC 47CBFE3C
47C>3H3@3@
3C3:
<3CF?
<3?
67@93@
=757B3E3@
DF6FE E7E3B 6;93?43C=3@
A>7:
D74F3:
3@3=
B3@3:
F@EF=
?7@J3E3=3@
E793@93@
63@ =F3E
3CFD
63C;
DF3EF
C3@9=3;3@
>;DEC;=
)363
#/$#3

63B3E
@63 B7C:3E;=3@
6;39C3?
83DAC
63C;
E793@93@
63@
3CFD
4A>3=43>;= +F?4F
:AC;KA@E3>
B363
#/$#3

363>3:
F@EF=
?7@7@EF=3@ 47D3C@J3
DF6FE
83D7
J3@9
6;F=FC
63>3?
D3EF3@
67C3<3E
3E3F
C36;3@ D763@9=3@
DF?4F
G7CE;=3>
F@EF=
?7@7@EF=3@
@;>3;
E793@93@
63@
=F3E 3CFD
J3@9
?7CFB3=3@
47D3C3@
D=3>3C
4F=3@
G7=EAC
&7>3>F;
?7EA67 6;39C3?
83DAC
;@;
6;:3C3B=3@
@63
63B3E
?7?3:3?;
3C3:
83D7
G7=EAC 67@93@
?F63:

6. Resistor pada Rangkaian Arus Bolak-Balik )363
#/$#3

D74F3:
C7D;DEAC
?FC@;

6;B3D3@9
63>3?
C3@9=3;3@ E7CEFEFB
3CFD
4A>3=43>;=
C38;=
E793@93@
DF?47C
J3@9
47CD;83E
D;@FDA;63 3=3@
?7@9:3D;>=3@
3CFD
J3@9

Mudah dan Aktif Belajar Fisika untuk Kelas XII

)7CD3?33@
E793@93@
4A>3=43>;=
3@E3C3
F3: N  

%
D;@ +

rotasi Vm im



%
D;@ +

N

i = im sin t


2

3


t

2
5
2




7@93@
?7?7C:3E;=3@
6-6/
*/
J3@9
?7@J3E3=3@
43:H3
=F3E
3CFD   J3@9
?7@93>;C



?7?;>;=;
:F4F@93@
67@93@
%

% 
?3=3
47D3C  3CFD
4A>3=43>;=
J3@9
?7@93>;C
?7>3>F;
C3@9=3;3@
?7?7@F:;
B7CD3?33@

V = Vm sin t

3
2

Gambar 6.25 Grafik kuat arus I dan tegangan V pada rangkaian hambatan murni, arus I, dan tegangan V sefase.

)7C:3E;=3@
#/$#3

7C63D3C=3@
'34#/##0
 9 
63@ '34#/##0
 9 
3CFD
?3=D;?F?
63@
E793@93@
?3=D;?F?
?7?;>;=; 83D7
J3@9
D3?3
CE;@J3
=7E;=3
BAD;D;
=F?B3C3@
?7@53B3;
DF6FE
 M E793@93@
63@
3CFD
J3@9
?7>3>F;
C3@9=3;3@
?7@53B3;
@;>3;
?;@;?F?
%74;: <7>3D@J3
@63
B7C:3E;=3@
#/$#3
 
;39C3?
83DAC
63C;
=F3E
3CFD 63@
E793@93@@J3
?7?;>;=;
3C3:
J3@9
D3?3

Imaks

Vmaks

t

Gambar 6.26

Mari Mencari Tahu *7D;DEAC
?7CFB3=3@
D3>3:
D3EF
=A?BA@7@
>;DEC;=
J3@9
D7C;@9
6;9F@3=3@
+7E;3B C7D;DEAC
J3@9
6;
6;
B3D3C3@
?7?;>;=;
@;>3;
C7D;DE3@D;
E7CE7@EF
,F93D
@63
4F3E>3: C7DF?7
?7@97@3;
53C3
?7@9;67@E;8;=3D;
@;>3;
D74F3:
C7D;DEAC
+7E7>3:
?7?4F3E C7DF?7
;@;
6;:3C3B=3@
@63
63B3E
?7@97@3>;
C7D;DEAC
63@
?3?BF
?7@7@EF=3@ @;>3;
C7D;DE3@D;@J3
63C;
5;C;5;C;
J3@9
6;?;>;=;
C7D;DEAC
E7CD74FE
93C
>74;:
?7@3C;= D7CE3=3@
93?43C
?353??353?
C7D;DEAC
63>3?
EF>;D3@
@63

Diagram fasor arus I dan tegangan V memiliki arah yang sama.

7. Induktor Murni pada Rangkaian Arus Bolak-Balik )7C:3E;=3@
#/$#3
 
+74F3:
;@6F=EAC
?FC@;
?7?;>;=;
;@6F= E3@D;
6;C;

)363
D33E
3CFD
>;DEC;=
?7@93>;C
63>3?
C3@9=3;3@
3=3@
E;?4F> 93J3
97C3=
>;DEC;=
;@6F=D;
99>
6;
3@E3C3
F3:

     + 63BF@
4763
E793@93@
3@E3C3
F3: N  

%
D;@ + #;=3
@63
B7C:3E;=3@
C3@9=3;3@
3CFD
4A>3=43>;=
63>3?
#/$#3
 C3@9=3;3@
E7CD74FE
?7CFB3=3@
C3@9=3;3@
E7CEFEFB
63@
47C>3=F
!F=F? ""
$;C5:A88
!F=F?
""
$;C5:A88
?7@J3E3=3@
43:H3    

(>7:
=3C7@3
63>3?
C3@9=3;3@
E7CD74FE
C7D;DEAC
?FC@;
E;63=
363
J3;EF @;>3;



?3=3
6;B7CA>7:
B7CD3?33@
=F3E
3CFD
D74393;
47C;=FE 

*,%)

"&,#+') 






%
D;@ +
N

+ 


   % D;@ + + 

L

a

b

I = Im sin t

Gambar 6.27 Rangkaian arus bolak-balik dipasang seri dengan sebuah induktor murni.

V = Vm sin ( t ) Vm I = Im sin t im



2

t

2


Gambar 6.28 Grafik kuat arus I dan tegangan V pada rangkaian induktif murni. Tegangan V mendahului arus I sekitar 90°.

Induksi Elektromagnetik

153

% D;@ + +   %
N 5AD +   




 




 


% D;@
+ 








 



 % D;@  + N




  

N 

7@93@
?7?7C:3E;=3@
B7@FCF@3@
'34#/##0
 9 
6;=7E3:F;
43:H3   %  % 
&7@FCFE
!F=F?
(:?

?7CFB3=3@
@;>3;
:3?43E
;@6F=EAC

 J3@9
6;D74FE
C73=E3@D;
;@6F=E;8

D7:;@993
63B3E
6;EF>;D

VL

90°

I

Gambar 6.29 Diagram fasor arus I dan tegangan V. Tegangan mendahului 90°.


 
3E3F




 N $7E7C3@93@   
C73=E3@D;
;@6F=E;8
  


=757B3E3@
DF6FE
C36 D  
;@6F=E3@D;
6;C;
!

8C7=F7@D;
!K 7@93@
?7@93?3E;
:F4F@93@
'34#/##0
 9 
63@
'34#/##0  9 
6;=7E3:F;
B7C47633@
83D7
3@E3C3
E793@93@
63@
=F3E
3CFD
J3@9
?7>3>F;
C3@9=3;3@
;@6F=E;8
E7CD74FE
D747D3C

7@93@
DF6FE
83D7
3CFD
E7CE;@993>
363>3:
C36
63C;
DF6FE
83D7
E793@93@ @63
3=3@
>74;:
?7?3:3?;
<;=3
?7?7C:3E;=3@
9C38;=
E793@93@ D;@FDA;63>
63@
9C38;=
3CFD
D;@FDA;63>
E7C:363B
H3=EF
?7>3>F;
#/$#3 
#;=3
E793@93@
B363
F3;
3CFD 63>3?
C3@9=3;3@
3=3@
?;@;?F?



+743>;=@J3
<;=3
E793@93@
47C@;>3; ?;@;?F?
3CFD
J3@9
?7>3>F;
C3@9=3;3@
3=3@
?3=D;?F?


%#*

Contoh 6.14 


!K
6;B3D3@9
;@6F=EAC
D747D3C
 
!
*3@9=3;3@
;@;
?7@9:3D;>=3@
3CFD
?3=D;?F?


!;EF@9
E793@93@ ?3=D;?F?
;@6F=EAC
E7CD74FE #7#$ ;=7E3:F; 

8C7=F7@D;




3CFD
?3=D;?F?
%


 ;@6F=E3@D;
;@6F=EAC


 
! ,793@93@
?3=D;?F?
63B3E
6;:;EF@9
67@93@
B7CD3?33@  % 

%

)363
D74F3:
C3@9=3;3@
3CFD
4A>3=43>;=
67@93@
8C7=F7@D;




 %    !K 
 
! 

   


. 

 


154

Mudah dan Aktif Belajar Fisika untuk Kelas XII

8. Kapasitor pada Rangkaian Arus Bolak-Balik )C;@D;B
=79F@33@@J3
D3?3
67@93@
;@6F=EAC
$3B3D;EAC
63>3? C3@9=3;3@
3CFD
4A>3=43>;=
6;B3D3@9
F@EF=
?7?B7CA>7:
7@7C9;
>;DEC;=
J3@9 >74;:
47D3C
CFD
>;DEC;=
J3@9
?7>3>F;
=3B3D;EAC
J3@9
=3B3D;E3D@J3
D747D3C 
D7>3?3
+
3=3@
?7@J7434=3@
=3B3D;EAC
47C;D;
?F3E3@
D747D3C

  +

C

a

N

b

V = Vmsin t

63BF@
?F3E3@
J3@9
E7CD;?B3@
63>3?
=3B3D;EAC
#/$#3
 
363>3:



 

N

.



%
D;@ +

N

Gambar 6.30 Rangkaian kapasitor

6;
?3@3 3C;
=76F3
'34#/##0
 9 
63@
'34#/##0
 9 
63B3E
6;B7CA>7:   +

 



% D;@ +   

 + +

 

D;@ +

% 5AD +  
% 
5AD +


  V = Vm sin  t   2 

Vi

i = im sin t

 
%
5AD + 


 



%
D;@  
+ 

N

7@93@
?7?7C:3E;=3@
B7@FCF@3@
'34#/##0
 9 
6;=7E3:F;
43:H3  %

% 
&7@FCFE
!F=F?
(:?

?7CFB3=3@
@;>3;
:3?43E3@ =3B3D;EAC
J3@9
6;D74FE
C73=E3@D;
=3B3D;E;8

D7:;@993
63B3E
6;EF>;D

   
3E3F



Gambar 6.31 Grafik I dan V pada kapasitor dalam rangkaian AC.

N

$7E7C3@93@   
C73=E3@D;
=3B3D;E;8
    
=3B3D;E3D
=3B3D;EAC



8C7=F7@D;
DF6FE
C36 D 7@93@
?7@93?3E;
:F4F@93@
'34#/##0
 9


63@
'34#/##0  9 
6;=7E3:F;
B7C47633@
83D7
3@E3C3
E793@93@
63@
=F3E
3CFD
J3@9
?7>3>F;
C3@9=3;3@
=3B3D;E;8
E7CD74FE
D747D3C

7@93@
DF6FE
83D7
3CFD
?7@63:F>F;
363>3:
C36
63C;
DF6FE
83D7
E793@93@ @63
3=3@
>74;:
?7?3:3?;
<;=3
?7?7C:3E;=3@
9C38;=
E793@93@ D;@FD;A63>
63@
9C38;=
3CFD
D;@FD;A63>
E7C:363B
H3=EF
?7>3>F;
#/$#3   #;=3
E793@93@
B363
F3;
3CFD 63>3?
C3@9=3;3@
3=3@
?3=D;?F?


%
+743>;=@J3
<;=3
E793@93@ 47C@;>3;
?3=D;?F?
3CFD
J3@9
?7>3>F;
C3@9=3;3@
3=3@
?;@;?F?




I 90°

V

Gambar 6.32 Diagram fasor I dan V pada kapasitor.

Induksi Elektromagnetik

155

!#$'.
 *3@9=3;3@


63@

B363
*3@9=3;3@
CFD
A>3=3>;= 1

#0)-#+#0

'#-5#04+



*7D;DEAC






"@6F=EAC




$

+(#5

D783D7



%
D;@ +


E7CE;@993>



%
D;@ + 



D747D3C

C36



$3B3D;EAC










'056-
'34#/##0
 6056-



4+0




?7@63:F>F;
 D747D3C

C36













%
D;@ + 


!#$'.

?7@993?43C=3@
B7CD3?33@
3CFD
J3@9
?7@93>;C
63>3? E;3B
C3@9=3;3@
#;=3
DF?47C
E793@93@
4A>3=43>;=
6;3@993B
D3?3
J3;EF 

%
D;@ + +747EF>@J3
@63
63B3E
?7@7@EF=3@
E7C>74;:
63:F>F
B7CD3?33@
3CFD 43CF
=7?F6;3@
B7CD3?33@
E793@93@@J3
&;D3>@J3
3CFD
J3@9
?7@93>;C 63>3?
E;3B
C3@9=3;3@
D3?3


%
D;@ +
?3=3
3=3@
6;63B3E
B7CD3?33@ E793@93@
B363 3 C7D;DEAC


%
D;@ + 4 5

Tantangan untuk Anda





 

;@6F=EAC


%
D;@ + 
 =3B3D;EAC


%
D;@ + 


Contoh 6.15 +74F3:
=3B3D;EAC
67@93@





6;B3D3@9
B363
DF?47C
E793@93@
4A>3=43>;=

Sebuah kapasitor 100  F dihubungkan dengan sumber tegangan bolak-balik. Arus yang mengalir melalui rangkaian memenuhi persamaan I = 4 sin 200t A. Tuliskan persamaan tegangan kapasitor itu.

67@93@
8C7=F7@D;

 47C;=FE



!K
,793@93@
?3=D;?F?
%




.
D7B7CE;
B363
93?43C
c  200  F

V = Vmsin t

!;EF@9
3CFD
?3=D;?F?
B363
=3B3D;EAC
E7CD74FE #7#$ 

;=7E3:F; 







O
 N




!K
%




.










C36 D
  










C36 D 
 
   (>7:
=3C7@3
;EF
3CFD
?3=D;?F?@J3
363>3:
%

%  



 



156

Mudah dan Aktif Belajar Fisika untuk Kelas XII

9. R, L, dan C Dipasang Seri dalam Rangkaian Arus Bolak-Balik *3@9=3;3@
3CFD
>;DEC;=
4A>3=43>;=
J3@9
?7?;>;=;
:3?43E3@
?FC@;
 6;B3D3@9
D7C;
67@93@
;@6F=EAC

D7CE3
=3B3D;EAC

3=3@
?7@9:3D;>=3@ E3?43:3@
7@7C9;
>;DEC;=
J3@9
>74;:
47D3C
E7E3B;
C3@9=3;3@
//
D7C;
;@; :3@J3
63B3E
6;9F@3=3@
F@EF=
?7?B7C47D3C
63J3
>;DEC;=
J3@9
E7C43E3D CFD
?3=D;?F?
63>3?
C3@9=3;3@
?7?;>;=;
@;>3;
63@
83D7
J3@9
D3?3
B363 D7E;3B
=A?BA@7@
%74;:
<7>3D
63B3E
@63
B7C:3E;=3@
#/$#3
  #;=3
@63
E7E3B=3@
B7CD3?33@
3CFD
63>3?
C3@9=3;3@
D747D3C 

%
D;@
+  
B7CD3?33@
D7E;3B
E793@93@
67@93@
?7?7C:3E;=3@
6;39C3? 83DAC
B363
#/$#3
 
J3;EF 

?
D;@ +
?7?;>;=;
83D7
J3@9
D3?3
67@93@
3CFD






D;@
+   E7CE;@993>
83D7

D747D3C
C36

R

L

I = Im sin
t  v

Gambar 6.33 Resonansi rangkaian R–L–C seri





?
D;@
+ 
?7@63:F>F;
83D7

D747D3C
C36

VL



?

V

VL – VC

)7C:3E;=3@
C3@9=3;3@
//
D7C;
B363
#/$#3
 
+7E;3B
=A?BA@7@ ?7?;>;=;
E793@93@


63@

+753C3
83DAC
E793@93@E793@93@
;@; 63B3E
6;3:=3@
D74393;
47C;=FE N 

"  "  "  "

C

i

 VR

VC

Gambar 6.34 Diagram fasor V, VR, VL, dan VC.

(>7:
=3C7@3
" 
" 
63@
"
D3>;@9
E793=
>FCFD
:3C93

63B3E
6; E7@EF=3@
67@93@
?7@99F@3=3@
:;EF@9
G7=EAC
J3;EF







N


  
   

67@93@
?7?7C:3E;=3@
=7?43>;
#/$#3
 
:3?43E3@
EAE3>
C3@9=3;3@ D7C;
NN
J3@9
6;D74FE
67@93@
;?B763@D;
C3@9=3;3@
.
63B3E
6;:;EF@9 63C;
'34#/##0
 9

  
 



X

Z

L

X –X L

C






   

   
 

X

R

C



   
    

N 3

Gambar 6.35 Diagram fasor Z, R, XL, dan XC.

3E3F       










N 4

(>7:
=3C7@3
3CFD
?3=D;?F?
J3@9
?7@93>;C
6;
D7E;3B
=A?BA@7@
D3?3 47D3C@J3
:F4F@93@
3@E3C3
E793@93@
?3=D;?F?
63@
3CFD
?3=D;?F?
;@; ?7?7@F:;
B7CD3?33@



2
J3;EF

    
    

N 

$7E7C3@93@ % 
E793@93@
?3=D;?F?  
;?B763@D;
C3@9=3;3@  

N


C73=E3@D;
EAE3>

Induksi Elektromagnetik

157

763
83D7
3@E3C3
E793@93@
63@
=F3E
3CFD
63B3E
6;E7@EF=3@
D74393; 47C;=FE

E3@  

          

N

D7:;@993
B7CD3?33@
E793@93@@J3
363>3:
    % D;@

+     B34;>3



DF6FE
83D7

47C@;>3;
BAD;E;8
CE;@J3
E793@93@ ?7@63:F>F;
3CFD
63@
C3@9=3;3@
47CD;83E
;@6F=E;8
#;=3


DF6FE
83D7 
47C@;>3;
@793E;8
CE;@J3
E793@93@
E7CE;@993>
A>7:
3CFD
63@
C3@9=3;3@ 47CD;83E
=3B3D;E;8
#;=3



6;B7CA>7:



63@



CE;@J3 E793@93@
?7?;>;=;
83D7
D3?3
D763@9=3@
C3@9=3;3@
47CD;83E
C7D;DE;8
3>3? :3>
;@;
6;=3E3=3@
E7>3:
E7C<36;
C7DA@3@D; L

R

a

b

10. R dan L Dipasang Seri dalam Rangkaian Arus Bolak-Balik )7C:3E;=3@
#/$#3
 
+7E;3B
=A?BA@7@
?7?;>;=;
E793@93@
 63@

+753C3
83DAC
E793@93@E793@93@
;@;
63B3E
6;3:=3@
D74393; "

"

"
7@93@
?7@99F@3=3@
:;EF@93@
G7=EAC
47D3C@J3
E793@93@ EAE3>
J3;EF

I = Imsin t V

Gambar 6.36 Rangkaian R dan L dipasang seri VL

N

    

7D3C@J3
;?B763@D;
C3@9=3;3@
*N%
6;B7CA>7:
D7B7CE;
47C;=FE   
 

  

  

Vab

N

      3E3F

)

VR

   

 

Gambar 6.37 Diagram fasor VL dan VR XL

N



I

763
83D7
3@E3C3
E793@93@
63@
=F3E
3CFD
63B3E
6;E7@EF=3@
D74393; 47C;=FE

Z

E3@  

     

N

)7C:3E;=3@
#/$#3
 
E3?B3=
43:H3
E793@93@
?7@63:F>F;
3CFD D747D3C

C36
D7:;@993
<;=3


%
D;@ +
?3=3


%
D;@
+    

)

R

Gambar 6.38 Diagram fasor VL dan VR V/i Vab Iab t

Gambar 6.39 Grafik tegangan mendahului arus dengan beda fase 0<  <90°.

Contoh 6.16 +74F3:
C7D;DEAC




6;B3D3@9
D7C;
67@93@
;@6F=EAC



B363
E793@93@ 4A>3=43>;=



. 3 !;EF@9
;?B763@D;
C3@9=3;3@
83D7
C3@9=3;3@
=F3E
3CFD
63@
 4 ,7@EF=3@
83D7


63@

E7C:363B
=F3E
3CFD
" #7#$ X = 40  I R = 30  ;=7E3:F; L C7D;DEAC
?FC@;



  V V C73=E3@D;
;@6F=EAC



  L R E793@93@
DF?47C





. V = Vmsin t

158

Mudah dan Aktif Belajar Fisika untuk Kelas XII

3

"?B763@D;

6;:;EF@9
?7>3>F;
'34#/##0
 9  2 


 

 


 


 83D7
E793@93@
  
6;:;EF@9
67@93@
'34#/##0
 9

  E3@
 




     
M 


.  


 
  ,793@93@







 


. ,793@93@
;@6F=EAC
.





 


. 
D783D7
67@93@
=F3E
3CFD

 



?7@63:F>F;

D747D3C
C36
 

 M 
?7@63:F>F;

 

M

$F3E
3CFD




4

Contoh 6.17 3>3?
C3@9=3;3@
D7C;
6;B3D3@9
C7D;DEAC
?FC@;




63@
D74F3:
;@6F=EAC



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R

C

b

I = Imsin t V

Gambar 6.40 Rangkaian seri R-C dalam rangkaian arus AC.

  
 

  

  

N 

     3E3F

   





 

N

Induksi Elektromagnetik

159



V

763
83D7
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I R

ab

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V

V

ab

C

Gambar 6.41 Diagram fasor tegangan Vab terhadap Vc.

     

N 

)363
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Contoh 6.18 R



3>3?
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D7C;
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Z

X C

Gambar 6.42 Diagram fasor vektor hambatan R terhadap Xc.

R  30 I V = Vmsin t

V/i

#7#$ ;=7E3:F;  

 

i V ab

 
 


O
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Gambar 6.43 Grafik tegangan Vab terhadap arus

  


 




 





C36 D 
  
 

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Contoh 6.19 *3@9=3;3@
D7C;

N

67@93@
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160

Mudah dan Aktif Belajar Fisika untuk Kelas XII

3 4

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12. L dan C Dipasang Seri dalam Rangkaian Arus Bolak-Balik ;=7E3:F;
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Gambar 6.44 Rangkaian seri L dan C V L

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Induksi Elektromagnetik

161

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13. Resonansi pada Rangkaian R–L–C *7DA@3@D;
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Gambar 6.46 Grafik kurva impedansi Z dan grafik linear R.

 

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14. Daya pada Arus Listrik Bolak-Balik "@6F=EAC
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162

Mudah dan Aktif Belajar Fisika untuk Kelas XII



    
    

N

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Contoh 6.20

Kata Kunci • • • • • • • • • •

arus bolak-balik tegangan bolak-balik arus listrik efektif tegangan efektif osiloskop diagram fasor voltmeter impedansi resonansi daya

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Induksi Elektromagnetik

163



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164

Mudah dan Aktif Belajar Fisika untuk Kelas XII



 

 

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166

Mudah dan Aktif Belajar Fisika untuk Kelas XII

Penala pada radio merupakan pemanfaatan rangkaian impedansi. Coba Anda cari manfaat atau aplikasi konsep Fisika yang ada pada bab ini dan coba Anda diskusikan cara kerjanya bersama teman Anda.

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174

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Bab

7 Sumber: CD Image

Matahari merupakan sumber energi yang utama bagi Bumi. Matahari memberikan energi melalui radiasi gelombang yang dipancarkannya.

Radiasi Benda Hitam Hasil yang harus Anda capai: menganalisis berbagai besaran fisis dan gejala kuantum dan batas-batas berlakunya relativitas Einstein dalam paradigma Fisika Modern. Setelah mempelajari bab ini, Anda harus mampu: menganalisis secara kualitatif gejala kuantum yang mencakup hakikat dan sifatsifat radiasi benda hitam, serta penerapannya.

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A. Pengertian Radiasi Benda Hitam B. Dualisme Gelombang Partikel

Radiasi Benda Hitam

175

Tes Kompetensi Awal !*'*192
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A. Pengertian Radiasi Benda Hitam 1. Radiasi Benda Hitam

Gambar 7.1 Cahaya yang masuk ke dalam kotak yang dianggap sebagai benda hitam sempurna, akan dipantulkan terus-menerus oleh dinding kotak.

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176

Mudah dan Aktif Belajar Fisika untuk Kelas XII

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Contoh 7.1

Ingatlah

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0 >

Contoh 7.2

Dalam SI, suhu menggunakan satuan derajat Kelvin.

Ingatlah

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/
282B
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Dalam kehidupan sehari-hari tidak ada benda yang dapat menyerap seluruh radiasi yang mengenainya.

Radiasi Benda Hitam

177

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/

2. Spektrum Radiasi Benda Hitam

benda hitam celah

detektor

prisma spektrum gelombang

Gambar 7.2 Percobaan untuk mengamati spektrum radiasi benda hitam. T ºC

Warna

1.500 1.400

putih

1.300 kuning muda 1.200 1.100

kuning

1.000

oranye

900

merah-oranye

800 merah ceri 700 600

merah tua

500

baru tampak

$:<2
CE2DE
36?52
A252D
5:A2?2C<2?
36?52
D6BC63ED
2<2?
>6>2? 42B<2?
B25:2C:
<2=@B
52=2>
36?DE<
CA6
86=@>32?8
6=628?6D:< 56?82?
A2?;2?8
86=@>32?8
52?
7B6<E6?C:
I2?8
36B3652
*252
CE9E
I2?8 =63:9
D:?88:
2<2?
D6B52A2D
B25:2C:
:?7B2
>6B29
I2?8
D:52<
52A2D
5:=:92D D6D2A:
52A2D
5:B2C2<2?
A2?2C?I2
"2=
:?:
52A2D
5:2>2D:
A252
C6A@D@?8 =@82>
I2?8
5:A2?2C<2?
9:?882
>6?42A2:
CE9E

%
&@82>
36BA:;2B 56?82?
G2B?2
<E?:?8
2D2E
<6AED:9AED:92?
+25:2C:
86=@>32?8
I2?8 5:A2?42B<2?
<6>E5:2?
5:=6G2D<2?
>6=2=E:
46=29
282B
5:A6B@=69
36B<2C 86=@>32?8
I2?8
C6>A:D
C6A6BD:
D2>A2<
A252
&2'&6
 6?82?
56>:<:2?
86=@>32?8
2<2?
D6B5:CA6BC:
>6?EBED
A2?;2?8 86=@>32?8
C6D:2A
CA6
6D632B
52A2D 5:86C6B86C6B
>6?EBED
CE5ED
56F:2C:
86=@>32?8
D6B5:CA6BC:
*6B4@322? :?:
5:=2<E<2?
A252
CE9E
36?52
9:D2>
I2?8
36B36523652
*6?82>2D2? A252
:?D6?C:D2C
CA6
B25:2C:
>6?892C:=<2?
52D2
I2?8
5:DE?;E<<2? A252
&2'&6
 &2'&6

>6?E?;E<<2?
329G2
2A23:=2
CE9E
CE2DE
36?52
D6BEC 5:?2:<<2?
:?D6?C:D2C
B6=2D:7
52B:
CA6
4292I2
I2?8
5:A2?42B<2??I2 36BE329
#?:
>6?I6323<2?
A6B86C6B2?
52=2>
G2B?2G2B?:
CA6
I2?8 D6B2>2D:
I2?8
52A2D
5:8E?2<2?
E?DE<
>6?232?8
>2E>
  >22<:?
D:?88: CE9E
CE2DE
36?52
 >22<:?
36B86C6B
<6
2B29
A2?;2?8
86=@>32?8 I2?8
=63:9
A6?56< 82B
?52
=63:9
>6>292>:
"E<E>
*6B86C6B2?
0:6?
A6B92D:<2? &2'&6

36B:<ED I (Wb/m2)

Sumber: Physics, 1993

Gambar 7.3

T 2 > T1

Spektrum warna pijar suatu benda panas berubah terhadap suhunya.

Gambar 7.4 Grafik pergeseran panjang gelombang maksimum pada saat suhu berubah dari T1 ke T2.

178

Mudah dan Aktif Belajar Fisika untuk Kelas XII

T1

 (m) Sumber: Physics, 1993

2B:
&2'&6

D6B=:92D
329G2
A2?;2?8
86=@>32?8
:?D6?C:D2C >2E>
36?52
I2?8
CE9E?I2
D:?88:
=63:9
A6?56<
52B:
A2?;2?8 86=@>32?8
:?D6?C:D2C
36?52
I2?8
CE9E?I2
B6?529
!6;2=2
A6B86C6B2?
 >2
5:C63ED
"E<E>
*6B86C6B2?
0:6?
,642B2 >2D6>2D:C
"E<E>
*6B86C6B2?
0:6?
5:?I2D2<2?
52=2>
A6BE>EC2? 36B:<ED  >2
Tokoh Max Planck (1858 – 1947)

0:6?
>6?6>E<2?
329G2
92C:=
<2=:
2?D2B2
 >2ED=2< >6BEA2<2?
3:=2?82?
<@?CD2?
C6A6BD:
A252
*67&2&&3
?
52?
3:=2?82? <@?CD2?
A252
A6BE>EC2?
"E<E>
*6B86C6B2?
0:6?
36B?:=2:
P M
>% "E<E>
*6B86C6B2?
0:6?
;E82
5:8E?2<2?
E?DE<
>6>A6B<:B2<2?
CE9E C63E29
3:?D2?8
56?82?
>6=:92D
G2B?2
4292I2
C63E29
3:?D2?8
CE9E 3:?D2?8
D6BC63ED
52A2D
5:AB65:
Contoh 7.3 ,63E29
36?52
5:A2?2C<2?
9:?882
36BCE9E
 
L
-6?DE<2?=29
A2?;2?8
86=@>32?8 >2E>
B25:2C:
D6B>2=
I2?8
5:A2?42B<2?
36?52
D6BC63ED &;&' :<6D29E:


 
L

  %

 
% 36B52C2B<2?
*6BC2>22?
 M
2<2?
5:A6B@=69  >22%

    >%     
>  
% $25:
A2?;2?8
86=@>32?8
>2E>
B25:2C:
D6B>2=
252=29

P
M
>

>2
Contoh 7.4 6B52C2B<2?
8B27:<
:?D6?C:D2C
B25:2C:


 I D6B9252A
A2?;2?8
86=@>32?8
C6A6BD: T1 5:DE?;E<<2?
A252
82>32B
5:
C2>A:?8 T2 3282:>2?2<29
9E3E?82?
2?D2B2
CE9E
>ED=2< T3  
 
52?
 m1 m2 m 3  &;&' 2B:
8B27:<
A252
82>32B
5:
C2>A:?8
5:A6B@=69 329G2
EBED2?
A2?;2?8
86=@>32?8
E?DE<
:?D6?C:D2C
B25:2C:
>2E>
252=29 &  &  & 
6B52C2B<2?
9E<E>
A6B86C6B2?
0:6?
*67&2&&3
?
5:A6B@=69

Sumber: Fundamental Of Physics, 2001

Max Planck dilahirkan di Kiel, Jerman. Ia belajar di Munich dan Berlin. Seperti banyak ahli fisika lainnya, ia seorang pemain musik yang pandai. Pada tahun 1900 M, setelah 6 tahun belajar di Universitas Berlin, ia menyatakan bahwa kunci pemahaman radiasi benda hitam adalah anggapan bahwa pemancaran dan penyerapan radiasi terjadi dalam kuantum energi hf. Penemuan yang menghasilkan hadiah Nobel 1918 ini sekarang dianggap sebagai tonggak dari fisika modern. Walaupun selama Hitler berkuasa, ia tetap tinggal di Jerman. Ia memprotes perlakuan Nazi kepada ilmuwan Yahudi dan sebagai akibatnya ia harus melepaskan kedudukannya sebagai presiden Institut Kaiser Wilhelm. Setelah perang dunia kedua, institut tersebut diberi nama Institut Planck dan ia kembali menjabat sebagai presiden sampai akhir hayatnya.

&  
2D2E
&   $25:
&  &  & 

  
5:A6B@=69
 
 




   

3. Pendekatan Teori Rayleigh–Jeans *252
D29E?

>E?4E=
A6BD2?I22?
3282:>2?2
>6?;6=2C<2? CA6
B25:2C:
36?52
9:D2>
*6?82>2D2?
52?
66?
I2?8
D6=29 5:=2<E<2?
A252
>2C2
:DE
>6?4@32
>6?42B:
A6?;6=2C2?
,A6
B25:2C:

Radiasi Benda Hitam

179

n (v)

v

Gambar 7.5 Sebaran jumlah partikel yang memiliki kecepatan (v). Intensitas

s an - Je ei g h Rayl n ck iW Teo ri i Pl a Te or Teo r ie n

Panjang gelombang

Gambar 7.6 Perbandingan teori Wien, dan teori Rayleigh-Jeans dengan teori Planck yang memenuhi data percobaan.

36?52
9:D2>
5:D6B2?8<2?
@=69
46)
&=1*.,-
  M  
52?
!.6
&2*7 45;&6)
*&37
 M 
*6BE>EC2?
>6B6<2
>6?88E?2<2?
D6@B: 7:C:<2
<=2C:<
C6A6BD:
D6@B:
<:?6D:<
82C :
%6=2C
1#
?52
D6=29
>6>A6=2;2B:
>2D6B:
A2?2C
36?52
36B9E3E?82? 56?82?
6?6B8:
<:?6D:<
>@=6<E=
82C
-6@B:
6<:A2BD:C:
>6?I2D2<2?
329G2 6?6B8:
D@D2=
82C
52=2>
CE2DE
BE2?82?
D6BDEDEA
5:328:
C642B2
>6B2D2
<6 C6D:2A
>@=6<E=
I2?8
D6B52A2D
5:
52=2>
BE2?82?
$:<2
6?6B8:
D@D2=
C:CD6> 252=29

52?
32?I2@=6<E=
I2?8
36B252
52=2>
BE2?82?
252=29
 6?6B8:
B2D2B2D2
I2?8
5:>:=:<:
@=69
C6D:2A
>@=6<E=
252=29
 
?6B8:
:?: D6B<2?5E?8
52=2>
36?DE<
6?6B8:
<:?6D:<
C6D:2A
>@=6<E=
<2?
D6D2A:
D:52< 36B2BD:
C6>E2
>@=6<E=
>6>:=:<:
6?6B8:
I2?8
C2>2
I2:DE

(:=2:
D6BC63ED 92?I2=29
>6BEA2<2?
?:=2:
B2D2B2D2 &2'&6

>6>A6B=:92D<2?
C632B2?
<646A2D2?
I2?8
5:>:=:<: >@=6<E=
82C
,632B2?
6?6B8:
<646A2D2?
<2B6?2
6?6B8:
<:?6D:<
C6=2=E 36B9E3E?82?
56?82?
<646A2D2?
&=1*.,-*&37
>6=:92D
329G2
<EBF2 C632B2?
:DE
C6BEA2
56?82?
92C:=
I2?8
5:A6B@=69
A252
:?D6?C:D2C
CA6 B25:2C:
D6B>2=
'6B6<2
36B2?882A2?
252
<6>:B:A2?
2?D2B2
C:72D
A2?2C 36?52
52?
B25:2C:
D6B>2=
6?82?
A6?56<2D2?
I2?8
>6B6<2
=2<E<2? &=1*.,-*&37
>6>A6B@=69
A6BE>EC2?
329G2
:?D6?C:D2C
B25:2C: 36B32?5:?8
D6B32=:<
56?82?
A2?8<2D
6>A2D
A2?;2?8
86=@>32?8
   .?DE<

I2?8
36C2B
:?D6?C:D2C
2<2?
C6>2<:?
<64:=
A23:=2  >6?56<2D:
D2<
9:?882
:?D6?C:D2C
2<2?
>6?56<2D:
?@=
"2=
:?:
C6CE2: 56?82?
92C:=
6>A:B:C
<2?
D6D2A:
2A23:=2

>6?56<2D:
?@=
92=
:?:
C2?82D >6?I:>A2?8
52B:
92C:=
6>A:B:C
I2?8
>6?E?;E<<2?
329G2
:?D6?C:D2C >6?56<2D:
?@=
<6D:<2

C6>2<:?
>6?864:= *6?I:>A2?82?
A6BC2>22?
+2I=6:89 $62?C
I2?8
C2?82D
;2E9
:?: 5:?2>2<2?
N6?42?2
.=DB2
/:@=6DO
<2B6?2

I2?8
<64:=
36B252
52=2> G:=2I29
86=@>32?8
E=DB2F:@=6D &2'&6

>6?E?;E<<2?
9E3E?82?
:?D6?C:D2C
52?
A2?;2?8 86=@>32?8
2?D2B2
<EBF2
+2I=6:89$62?C
52?
<EBF2
0:6?
-6@B:
%.*3 52A2D
>6?6B2?8<2?
CA6
A252
A2?;2?8
86=@>32?8
I2?8
A6?56< D6D2A:
>6?I:>A2?8
<6D:<2
>6?6B2?8<2?
A2?;2?8
86=@>32?8
I2?8
A2?;2?8 ,632=:6?6B2?8<2?
CA6
A252 A2?;2?8
86=@>32?8
I2?8
A2?;2?8
D6D2A:
>6?I:>A2?8
<6D:<2
>6?6B2?8<2? A2?;2?8
86=@>32?8
I2?8
A6?56<

4. Teori Planck

Tugas Anda 7.1 Carilah dari buku-buku tentang formula yang digunakan RayleighJeans untuk menjelaskan intensitas radiasi benda hitam.

180

-6@B:
<=2C:<
D6?D2?8
4292I2
C63282:
86=@>32?8
6=628?6D:< 52A2D
>6?;6=2C<2?
52B:
>2?2
2C2=
B25:2C:
<2=@B
D6D2A:
D:52<
3:C2
56?82? D6A2D
>6>AB65:
4292I2
I2?8
5:A2?42B<2?
C63282:>2?2 I2?8
D6B2>2D:
@=69
%.*3
-6@B:
%.*3
>2>AE
>6?;6=2C<2?
B25:2C:
36?52 9:D2>
E?DE<
A2?;2?8
86=@>32?8
I2?8
A6?56<
D6D2A:
8282=
E?DE<
A2?;2?8 86=@>32?8
I2?8
A2?;2?8
,632=:6?;6=2C<2?
B25:2C:
36?52
9:D2>
E?DE<
A2?;2?8
86=@>32?8
I2?8
A2?;2?8 D6D2A:
8282=
E?DE<
A2?;2?8
86=@>32?8
I2?8
A6?56< <9:B?I2
A6?;6=2C2?
I2?8
>6>E2C<2?
=29:B
52B:
C6@B2?8
7:C:<2G2? $6B>2?
&<
1&3(0
 M 
#2
>6?4@32
>6=2<E<2?
A6?56<2D2? =2:?
56?82?
>6?82;E<2?
CE2DE
BE>EC
6>A:B:C
52?
D6B?I2D2
C2?82D
4@4@< 56?82?
92C:=
A6?82>2D2??I2

Mudah dan Aktif Belajar Fisika untuk Kelas XII

1&3(0
>6?42B:
52C2B
D6@B:
E?DE<
BE>EC
6>A:B:C?I2
E2
3E=2? <6>E5:2?
:2
36B92C:=
>6>3E2D
2CE>C:
I2?8
C2?82D
>6?86;ED<2?
A252 C22D
:DE
I2:DE
D6?D2?8
86D2B2?
>@=6<E=>@=6<E=
5:
A6B>E<22?
36?52 9:D2>
1&3(0
>6?86>E<2<2?
A@CDE=2D?I2
C63282:
36B:<ED  ?6B8:
B25:2C:
I2?8
5:A2?42B<2?
@=69
86D2B2?
>@=6<E=>@=6<E= @C:=2D@B
36?52
>6BEA2<2?
A2<6DA2<6D
<E2?D2
6?6B8:
6C2B?I2 6?6B8:
52=2>
C6D:2A
A2<6D
>6BEA2<2?
<6=:A2D2?
3:=2?82?
3E=2D
CE2DE 36C2B2?
NO
I2:DE ?

'
!
 M 56?82?
'
252=29
3:=2?82?
2C=:
  

I2?8
5:C63ED
3:=2?82? <E2?DE>
52?

252=29
7B6<E6?C:
86D2B2?
>@=6<E=>@=6<E=
C652?8 <2?
!
>6BEA2<2?
<@?CD2?D2
*=2?4<
I2?8
36C2B?I2
 
P
M
$C )=69
<2B6?2
6?6B8:
B25:2C:
36BC:72D
5:C
 '@=6<E=>@=6<E=
>6?I6B2A
2D2E
>6>2?42B<2?
6?6B8:
B25:2C:
4292I2 52=2>
A2<6D
5:C
2D2E
7@D@?
?6B8:

7@D@? <2B6?2
A6B36522?
5E2
D:?8<2D
6?6B8:
C6A6BD:
D2>A2<
A252
&2'&6 
252=29 

!
 M $:<2
>@=6<E=>@=6<E=
>6?I6B2A >6>2?42B<2?

7@D@?
D:?8<2D 6?6B8:?I2
36BD2>329
2D2E
36B<EB2?8
C636C2B
!
!282C2?
1&3(0
:?: 36B=2<E
E?DE<
36?52
9:D2> &2?8<29
1&3(0
56?82?
A@CDE=2D?I2
329G2
6?6B8:
@C:=2D@B
D6B<E2?D:C2C: >6BEA2<2?
C63E29
D6B@3@C2?
36C2B
1&3(0
C6?5:B:
36BC2>2
:=>EG2? C62?8<2D2??I2
D:52<
>6?I252B:
92=
:DE
!282C2?
*=2?4<
5:2?882A
C63282: D@?882<
2G2=
7:C:<2
<E2?DE>

ke n =


E

n

4hf

4

3hf

3

2hf

2

1hf

1

0

0

Gambar 7.7 Tingkat-tingkat energi yang diperkenankan untuk satu molekul yang frekuensinya f.

Contoh 7.5 $E>=29
>:?:>E>
7@D@?
4292I2
I2?8
A2?;2?8
86=@>32?8?I2
 
?>
E?DE< >6?:>3E=<2?
B2?8C2?82?
F:CE2=
A252
>2D2
?@B>2=
252=29
'
 
7@D@?
A6B
C6<@? $:<2
46A2D
B2>32D
4292I2


P 
> C
52?
!

P M
$C
D6?DE<2?=29
52I2 4292I2
D6BC63ED &;&' ?6B8:
4292I2
36B52C2B<2?
*67&2&&3
? 
252=29

  '!   '!   2I2
4292I2
D6BC63ED
52A2D
5:D6?DE<2?
C63282:
36B:<ED '!  -




 
   
$C 
   
> C        0

C 
   
> 

Tantangan untuk Anda Perumusan Rayleigh-Jeans tidak cocok diterapkan untuk radiasi benda hitam pada daerah panjang gelombang pendek. Dapatkah Anda menduga, apa implikasi dari persamaan Rayleigh-Jeans pada daerah panjang gelombang pendek untuk kehidupan di Bumi?

Radiasi Benda Hitam

181

Kata Kunci • • • • • • • •

benda hitam radiasi benda hitam emisivitas benda intensitas daya radiasi spektrum energi radiasi radiasi termal

5. Aplikasi Radiasi Benda Hitam untuk Mengukur Suhu Matahari '2D292B:
>6BEA2<2?
3:?D2?8
I2?8
A2=:?8
56<2D
56?82?
E>: ;2B2
'2D292B:
>6>2?42B<2?
6?6B8:
<2=@B I2?8
C2?82D
36B8E?2
328:
<69:5EA2?
5:
E>:
?6B8:
D6BC63ED
36B2C2= 52B:
B62
:?D:
'2D292B:
2A2D<29
?52 >6>32I2?8<2?
36B2A2
36C2B
6?6B8:
I2?8
5:92C:=<2?
'2D292B:
C69:?882 C2>A2:
;2B2<
  
<>
6?6B8:
:DE
>2C:9
252
36BEA2
4292I2
52? <2=@B +25:2C:
6?6B8:
'2D292B:
C2?82D
36C2B
C6A6BD:
I2?8
D6=29
?52
A6=2;2B: #?D6?C:D2C
B25:2C:
:?:
36B32?5:?8
=EBEC
56?82?
A2?8<2D
6>A2D
CE9E?I2 '6?EBED
D6@B:
B25:2C:
I2?8
5:A2?42B<2?
:?:
>6>:=:<:
:?D6?C:D2C
I2?8 36B36523652
.?DE<
=63:9
>6>292>:
<@?C6A
:?:
A6B92D:<2?
&2'&6 
36B:<ED
:?: I(W/m2)

Gambar 7.8 Grafik intensitas radiasi Matahari yang diukur di luar angkasa.

Ingatlah 1 Å = 10–10 m

Tantangan untuk Anda Grafik pada Gambar 7.8 menunjukkan intensitas radiasi Matahari yang diukur di luar angkasa. Apakah Anda bisa menduga bagaimana grafik intensitas radiasi Matahari jika diukur di permukaan Bumi? Apakah di seluruh tempat di permukaan Bumi intensitas radiasi Matahari selalu sama? Jelaskan jawaban Anda dengan bahasa yang mudah dipahami dan diskusikanlah bersama teman sekelas.

182


















10 7 m

&2'&6

>6?E?;E<<2?
CA6
B25:2C:
'2D292B:
I2?8
5:E<EB 5:
=E2B
2?8<2C2
!B27:<
D6BC63ED
C2?82D
>:B:A
56?82?
8B27:<
:?D6?C:D2C B25:2C:
36?52
9:D2>
)=69
<2B6?2
:DE
A2B2
2CDB@?@>
>6?82CE>C:<2? '2D292B:
C63282:
36?52
9:D2>
C69:?882
3:C2
5:9:DE?8
36B2A2
CE9E
5: A6B>E<22?
'2D292B:
56?82?
>6?88E?2<2?
"E<E>
*6B86C6B2?
0:6? 2B:
8B27:<
D6B=:92D
329G2
?:=2:
:?D6?C:D2C
>2E>
D6B52A2D
A252 A2?;2?8
86=@>32?8
C636C2B
 
K
526B29
A2?;2?8
86=@>32?8
C:?2B D2>A2<
'6?EBED
"E<E>
*6B86C6B2?
0:6?
2<2?
5:52A2D<2?
?:=2:
CE9E 5:
A6B>E<22?
'2D292B:
C63282:
36B:<ED


>2%

P   >%  
P   >  

% ,6=2:?
E?DE<
>6?8E<EB
CE9E
'2D292B:
A6?6B2A2?
B25:2C:
36?52 9:D2>
:?:
52A2D
5:8E?2<2?
E?DE<
>6?8E<EB
CE9E
3:?D2?8
=2:??I2
I2:DE 56?82?
>6?82?2=:C:C
CA6
3:?D2?8
D6BC63ED

Mari Mencari Tahu ,2=29
C2DE
:CE
8=@32=
I2?8
C6B:?8
5:2?8<2D
@=69
22?2C2?
E>:
52
C6328:2?
?682B2
>2;E
I2?8
D:52<
>2E >6?2?52D2?82?:
+(-($(%
1(-(
D6?D2?8
A6>32D2C2?
<252B
6>:C:
82C
52B:
:?5ECDB: !2C
6>:C:
:?:
C6328:2?
36C2B
252=29
)
'6?EBED
A2B2
29=:
82C
)
:?:=29
I2?8 >6?82<:32D<2?
A6>2?2C2?
E>:
2D2E
5:<6?2=
56?82?
:CD:=29
$
.&!
 +'
(.,
-
2B:=29
:?7@B>2C:
52B:
36B3282:
CE>36B
3E<E
>2;2=29
:?D6B?6D 52?
=2:?=2:?
D6?D2?8
9E3E?82?
2?D2B2
676<
BE>29
<242
52?
B25:2C:
36?52
9:D2> :C<EC:<2?=29
56?82?
D6>2?
C6<6=2C
?52
D6?D2?8
92=
:?:

Mudah dan Aktif Belajar Fisika untuk Kelas XII

Tes Kompetensi Subbab

A

*6/&0&31&-
5&)&
'909
1&8.-&3







,E9E
<6B;2
7:=2>6?
DE?8CD6?
52=2>
CE2DE
=2>AE
A:;2B 252=29

%
52?
6>:C:F:D2C?I2

-6?DE<2?
=E2C A6B>E<22?
7:=2>6?
=2>AE
A:;2B

0 ,63E29
=E32?8
<64:=
A252
5:?5:?8
-'.+
>6?I6BEA2: 36?52
9:D2>
$:<2
=E2C?I2

4>
52?
CE9E?I2
 L C2>2
C6A6BD:
CE9E
5:
52=2>
D2?EB
36B2A2<29
6?6B8: I2?8
5:B25:2C:<2?
<6
=E2B
=E32?8
C6D:2A
56D:< @=2
+


4>
36BC:72D
C6A6BD:
36?52
9:D2>
36B252 52=2>
<62522?
C6:>32?8
56?82?
=:?8<E?82??I2 -6B?I2D2
3@=2
>6?I6B2A
52I2

<0
52B: =:?8<E?82??I2
6B2A2<29
CE9E
3@=2
:DE &2>AE
A:;2B
52A2D
5:2?882A
36B36?DE<
3@=2
$2B:;2B: =2>AE
A:;2B
A6BD2>2
252=29
5E2
<2=:
;2B:;2B:
=2>AE <65E2
,E9E
A:;2B
A6BD2>2
52?
<65E2
>2C:?8>2C:?8 
L
52?

L
-6?DE<2?
?:=2:
A6B32?5:?82?
2?D2B2 52I2
<2=@B
B25:2C:
=2>AE
A6BD2>2
52?
=2>AE
<65E2

 





%@?CD2?D2
6>:C:F:D2C
C63E29
36?52
9:D2>
252=29
 ":DE?8=29
:?D6?C:D2C
B25:2C:
I2?8
5:A2?42B<2?
2A23:=2 CE9E
36?52
:DE

% ,63E29
3@=2
=@82>
56?82?
;2B:;2B:

4>
5:A2?2C<2? *252
CE9E


L
3@=2
=@82>
:DE
>6>2?42B<2?
B25:2C: 56?82?
52I2
 P 
G2DD
":DE?8=29
6>:C:F:D2C
3@=2 =@82>
D6BC63ED ":DE?8=29
52I2
I2?8
5:A2?42B<2?
@=69
C63E29
<E3EC I2?8
5:A2?2C<2?
9:?882
 L
%E3EC
D6BC63ED
>6>:=:<: A2?;2?8
C:C:

4>
52?
D6B3E2D
52B:
3292?
56?82? 6>:C:F:D2C
 6?52

>6>2?42B<2?
B25:2C:
56?82?
>2H 

?> 6?52
:DE
D6BEC
>6?82=2>:
A6>2?2C2?
C69:?882 CE9E?I2

<2=:
=:A2D
CE9E
C6>E=2
-6?DE<2?=29
>2H C6D6=29
A6>2?2C2?

B. Dualisme Gelombang Partikel 1. Sifat Partikel yang Dimiliki Gelombang *252
>E=2?I2
32?I2<
7:C:<2G2?
I2?8
>6?82?882A
4292I2
C63282: 86=@>32?8
"2=
:?:
5:A6B<E2D
56?82?
252?I2
5:7B2E>
86=@>32?8
<2? D6D2A:
<6D:<2
A2B2
7:C:<2G2?
>6?6>E<2?
76?@>6?2
B25:2C:
36?52
9:D2> 676<
7@D@=:CDB:<
52?
676<
@>AD@?
2CE>C:
4292I2
C63282:
86=@>32?8
D:52< 52A2D
>6?;6=2C<2?
76?@>6?276?@>6?2
D6BC63ED
%6>E5:2?
>E?4E==29 A2?52?82?
329G2
4292I2
252=29
A2BD:<6=
,63282:
2?2=@8:
A6B92D:<29 &2'&6

*252
82>32B
D6BC63ED
?52
3:C2
>6>6B92D:<2?
A6B36522? 4292I2
;:<2
5:2CE>C:<2?
C63282:
86=@>32?8
52?
C63282:
A2BD:<6=
6?82? 2CE>C:
C63282:
86=@>32?8
4292I2
5:A2?42B<2?
C63282:
B2>32D2? 86=@>32?8
I2?8
<@?D:?E
52AE?
56?82?
2CE>C:
C63282:
A2BD:<6=
4292I2 5:A2?42B<2?
52=2>
36?DE<
A2<6DA2<6D
6?6B8:
I2?8
5:C63ED
7@D@? ,EA2I2
?52
=63:9
>6>292>:
2CE>C:
4292I2
C63282:
A2BD:<6=
?52 3:C2
>6>A6=2;2B:
676<
7@D@=:CDB:<
52?
676<
@>AD@?
36B:<ED a. Efek Fotolistrik #%
-+%),'1
%$-+('%$-+('
+"
)+&.$'
)%-
%( &
$-"$ ","'+"
' '
+$.',"
-+-'-.
",.-
$
(-(%",-+"$
=6E<22?
A=2D
=@82>
D6BC63ED
5:C63ED
6=62
<2=:
5:D6>E<2?
@=69
*68>
C6A6BD:
5:82>32B<2?
A252
&2'&6
 *6B2?8<2D
A6B4@322?
E?DE<
>6?82>2D:
676<
7@D@=:CDB:<
D6B5:B:
2D2C D23E?8
<242
92>A2
E52B2
52?
A=2D
=@82>
I2?8
5:C63ED
C63282:
<2D@56 %6D:<2
<2D@56
5:C:?2B:
56?82?
C:?2B
E=DB2F:@=6D
6=66?E;E
2?@56
C69:?882
2BEC
>6?82=:B
A252 B2?8<2:2?
2?I2A6B6>6D6B
?6B8:
<:?6D:<
I2?8
5:>:=:<: 6=66>A6B36C2B
3652
A@D6?C:2= 2?D2B2
<2D@56
52?
2?@56
C69:?882
3652
A@D6?C:2=
36BC:72D
>6?292?
=2;E

2

3 Gambar 7.9 (a) Cahaya sebagai gelombang; (b) Cahaya sebagai paket-paket energi.

tutup tabung

ruang vakum C

cahaya

T i

A i V

saklar geser – +

– +

Gambar 7.10 Skema eksperimen efek fotolistrik.

Radiasi Benda Hitam

183

6=622?
56?82?
<6?2:<2?
3652
A@D6?C:2=
A6?E?;E<2?
;2BE> 2>A6B6>6D6B
2<2?
>6?864:=
$:<2
A252
CE2DE
<6D:<2
;2BE>
2>A6B>6D6B >6?E?;E<
2?8<2
?@=
36C2B?I2
6?6B8:
A@D6?C:2=
C2>2
56?82?
36C2B
6?6B8: <:?6D:<
I2?8
5:>:=:<:
6=6A6B6>6D6B
I2?8
5:A2C2?8 5:
32=:<
3E<:D
>6?E?;E<<2?
2?8<2
?@=
#?:
36B2BD:
6?6B8:
<:?6D:<
6=66=6G2D:
3E<:D
A@D6?C:2=
C69:?882
36C2B?I2 A@D6?C:2=
96?D:

36BC6CE2:2?
56?82?
6?6B8:
<:?6D:<
6=6
v Ek

$




2

2D2E

I

&/ >2
intensitas tinggi intensitas rendah

–V 0

V

3 energi potensial gelombang (eV)

3

Ce Na

2

K

1 f0 0

2

4

f0

f0 6

1014 Hz

8

4 Sumber: Fisika untuk Sains dan Teknik Tipler

Gambar 7.11 (a) Apabila energi kinetik elektron melebihi energi potensial penghalang, elektron akan dapat mendaki bukit potensial; (b) Energi kinetik elektron tidak bergantung pada intensitas cahaya; (c) Setiap permukaan memiliki frekuensi (f0) berbeda.

184

M

M

6?
I2?8
5:=2<E<2?
@=69
.378*.3
56?82?
>6?8E329E329 7B6<E6?C:
4292I2
I2?8
;2DE9
A252
A=2D
=@82>
<2D@56
:?D6?C:D2C
4292I2 52?
A@D6?C:2=
A6?892=2?8
A252
676<
7@D@=:CDB:<
>6>36B:<2?
3636B2A2 <6C:>AE=2?
C63282:
36B:<ED 2 6C2B
6?6B8:
<:?6D:<
>2E>
6=6E<22?
=@82>
>6>3EDE9<2?
7B6<E6?C:
>:?:>E>
D6BD6?DE
I2?8 5:C63ED
7B6<E6?C:
2>32?8

282B
676<
7@D@=:CDB:<
52A2D
36B=2?8CE?8 A6B92D:<2?
&2'&6
( 4 =6E<22?
=@82>
92>A:B
D2?A2
C6=2?8 G232?8
D:52<
52A2D
>6?;6=2C<2?
>6?82A2
6?6B8:
<:?6D:< >2E>
6=6329
;:<2
7B6<E6?C:
4292I2
5:A6B36C2B "2C:=
66?
D6BC63ED
D:52<
52A2D
5:;6=2C<2?
@=69
D6@B:
<=2C:< D6?D2?8
86=@>32?8
6=628?6D:<
'6?EBED
D6@B:
<=2C:<
6?6B8:
I2?8 5:<2?5E?8
86=@>32?8
6=628?6D:<
36BC6CE2:2?
56?82?
:?D6?C:D2C?I2 A23:=2
:?D6?C:D2C
C6>2<:?
D:?88:
6?6B8:
86=@>32?8
6=628?6D:< 2<2?
>6?:?88:
AE=2
'652?
=:CDB:<
I2?8
5:32G2
86=@>32?8
6=628?6D:<
2<2?
>6>36B:<2?
82I2
A252
6=6E<22?
=@82> C69:?882
6=66>A6B@=69
6?6B8:
52B:
86=@>32?8
$:<2
6?6B8:
I2?8 5:>:=:<:
6=6E<22? =@82>
6B52C2B<2?
92=
:DE
C692BEC?I2
6?6B8:
<:?6D:<
6=632?8
AE?
D:52<
52A2D
5:;6=2C<2?
56?82?
D6@B: <=2C:<
6?
>6?E?;E<<2?
329G2
C6<64:=
2A2AE?
:?D6?C:D2C
4292I2 2A23:=2
7B6<E6?C:?I2
=63:9
36C2B
52B:A252
7B6<E6?C:


676<
7@D@=:CDB:< 2<2?
D6B;25:
'6?EBED
D6@B:
<=2C:<
C692BEC?I2
I2?8
252
252=29
:?D6?C:D2C 2>32?8
6?82?
<636B82?DE?82?
6?6B8:
4292I2
A252
:?D6?C:D2C?I2 9E3E?82?
=:?62B
2?D2B2
6?6B8:
<:?6D:<
6=632?8 6=628?6D:<
C2>2
C6<2=:
D:52<
52A2D
5:;6=2C<2? '6?EBED
1'*68
.378*.3
C6>E2
6?6B8:
7@D@?
5:36B:<2?
<6A252 6=66?I6323<2?
7@D@?
=6?I2A
#2
36BA2?52?82?
329G2 <E2?DE>
>6BEA2<2?
C:72D
86=@>32?8
:DE
C6?5:B:
'6?EBED?I2
4292I2

Mudah dan Aktif Belajar Fisika untuk Kelas XII

D6B5:B:
2D2C
A2<6DA2<6D
I2?8
5:C63ED
C63282:
7@D@?
@D@?7@D@?
>6BEA2<2? A2BD:<6=
I2?8
D:52<
36B>2CC2
I2?8
5:C6B2A
52?
5:A2?42B<2?
?6B8:
I2?8
5:>:=:<: @=69
7@D@?
C2>2
56?82?
<E2?D2
I2?8
5:<6>E<2<2?
@=69
*=2?4<
I2:DE


!
 56?82?
!
252=29
<@?CD2?D2
*=2?4<
52?

252=29
7B6<E6?C:
86=@>32?8 6=628?6D:< :?CD6:?
52A2D
>6?;6=2C<2?
C642B2
C6>AEB?2
92C:=
676<
7@D@=:CDB:< 292I2
I2?8
;2DE9
5:
2D2C
=@82>
5:=E<:C<2?
C63282:
B6?D6D2?
7@D@?
I2?8 >6?E>3E<
A252
=@82>
C6D:2A
7@D@?
>6?E>3E<
6=6
=@82> *252
A6B:CD:G2
DE>3E<2?
D6BC63ED
7@D@?
>6>36B:<2?
C6=EBE9
6?6B8:?I2 <6A252
6=66?;25:
=6?I2A ?6B8:
I2?8
5:A6B=E<2?
E?DE<
>6=6A2C<2?
5:B:
52B:
A6B>E<22?
=@82> 5:C63ED
7E?8C:
<6B;2

.?DE<
;6?:C
=@82>
I2?8
36B3652
D6?DE?I2
7E?8C: <6B;2
2<2?
36B3652
AE=2
<2B6?2
A6B36522?
6?6B8:
:<2D
2?D2B2
6=6
=@82>
$25:
E?DE<
=@82>
D6BD6?DE
36B=2<E 

!


M

$:<2
7B6<E6?C:
7@D@?
=63:9
36C2B
52B:A252
7B6<E6?C:
2>32?8
<6=63:92? 6?6B8:
I2?8
5:D6B:>2
6=66?;25:
6?6B8:
<:?6D:<
6=6
M 

!


!


$ !




$

M 

Keterangan: ! 
<@?CD2?D2
*=2?4<


P
M
$C  
7B6<E6?C:
86=@>32?8
4292I2
"J  
7B6<E6?C:
2>32?8
"J $:<2
5:3E2D
52=2>
C63E29
D236=
7E?8C:
<6B;2
7@D@=:CDB:<
3636B2A2
;6?:C =@82>
3:C2
5:=:92D
52=2>
"&'*1
 "&'*1
 E?8C:
%6B;2
=6 4,&2 6C:E> %2=:E> (2DB:E> >2C -6>3282 *6B2< *=2D:?2

&2'&3, C % (2 E E 8 *D

93,7.
*6/&
%
*$

 



    

Ingatlah 1 eV = 1,6 × 10–19 J

Informasi untuk Anda Efek fotolistrik selain memiliki peranan penting dalam teori foton pada cahaya, juga memiliki banyak sekali aplikasi dalam kehidupan sehari-hari. Sebagai contoh adalah alarm pencuri dan pembuka otomatis yang juga menggunakan sirkuit fotolistrik. Ketika seseorang menghalangi sinar yang dipancarkan, penurunan arus listrik dapat mengaktifkan saklar sehingga bel berbunyi atau pintu terbuka. Biasanya sinar yang digunakan adalah sinar ultraviolet dan inframerah karena kedua sinar ini tidak terlihat.

Information for You The photoelectric effect, besides playing an important historical role in confirming the photon theory of light, also has many practical applications. Burglar alarms and automatic door openers often make use of the photocell circuit. When a person interrupts the beam of light, the sudden drop in current in the circuit activates a switch which operates a bell or opens the door. UV or IR light is sometimes used in burglar alarms because of its invisibility. Sumber: Physics for Scientist & Engineers

Sumber: Physics, 1993

Contoh 7.6 B6<E6?C:
2>32?8
CE2DE
3292?

P
 
"J
$:<2
3292?
D6BC63ED
5:C:?2B:
56?82?
C:?2B I2?8
>6>:=:<:
7B6<E6?C:

P
 
"J
D6?DE<2?
36C2B
6?6B8:
<:?6D:<
6=6E<22?
=@82>
D6BC63ED &;&' $

!
2
!

!

2
 $


P
M
$C

P
 
"J
M

P
 

"J $


P
M
$C

P
 
"J



P
M 
;@E=6

Radiasi Benda Hitam

185

Ingatlah Foton merupakan paket-paket energi pada cahaya. Konsep ini merupakan kelanjutan dari asumsi cahaya merupakan partikel.

)=69
<2B6?2

;@E=6


P
M 
6/
36C2B
6?6B8:
<:?6D:<
6=6
      6/


6/     

Contoh 7.7 %2=:E>
5:C:?2B:
56?82?
4292I2
E=DB2F:@=6D
I2?8
A2?;2?8
86=@>32?8?I2

K
$:<2
7E?8C: <6B;2
<2=:E>

6/
D6?DE<2?=29
?:=2:
A@D6?C:2=
96?D:
A252
A6B:CD:G2
676<
7@D@=:CDB:< D6BC63ED
!


P
M
$C &;&' :<6D29E:
 

K


P
M
>  

6/
 
6/
 
P
M 
$C 

P
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$ ?6B8:
<:?6D:<
>2E>
6=6

!


2

!  2
   $>


    $C    
> C      
$

   
> $>


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M 
$ %6>E5:2?
8E?2<2?
*67&2&&3
?
C69:?882

$>


2D2E




     
$ $>





/     
 

$25:
A@D6?C:2=
96?D:
A252
676<
7@D@=:CDB:<
D6BC63ED
252=29

/

Tugas Anda 7.2 Proses terjadinya sinar-X merupakan kebalikan dari proses fotolistrik. Carilah informasi dari berbagai sumber tentang sinar-X dan diskusikanlah bersama teman sekelas Anda.

b. Efek Compton 2=2>
A6B:CD:G2
676<
7@D@=:CDB:<
4292I2
I2?8
5:;2DE9<2?
A252
<6A:?8 =@82>
5:A6B=2<E<2?
C63282:
A2<6D
6?6B8:
I2?8
5:C63ED
7@D@?
@D@?
:DE >6?82=2>:
A6B:CD:G2
DE>3E<2?
56?82?
6=63E<2? C6=2=E
5:9E3E?8<2?
56?82?
>@>6?DE>
*252
A6B:CD:G2
D6BC63ED
2<2? 36B=2<E
"E<E>
%6<6<2=2?
'@>6?DE>
52?
"E<E>
%6<6<2=2?
?6B8: *6?6=:D:2?
92>3EB2?
C:?2B1
I2?8
5:=2<E<2?
68-96

425843   M  
>6?892C:=<2?
76?@>6?2
32BE
I2:DE
A6B86C6B2?
A2?;2?8 86=@>32?8
2D2E
A6BE3292?
7B6<E6?C:
C636=E>
52?
C6CE529
DE>3E<2? !6;2=2
:?:
5:;6=2C<2?
@=69
425843
56?82?
>6?82?882A
329G2
I2?8 D6B;25:
252=29
DE>3E<2?
2?D2B2
<E2?DE>
4292I2
7@D@?
52?
6=66?E>3E<
6=66>:=:<:
6?6B8:
<:?6D:< 52AE?
6?6B8:
7@D@?
C6D6=29
DE>3E<2?
2<2?
36B<EB2?8
'6?EBED
D6@B: <=2C:<
A6?8EB2?82?
6?6B8:
D:52<
2<2?
5::<ED:
@=69
A6BE3292?
7B6<E6?C: 2D2E
A2?;2?8
86=@>32?8
(2>E?
>6?EBED
D6@B:
<E2?DE>
A6BE3292? 6?6B8:
36B2BD:
2<2?
D6B;25:
A6BE3292?
7B6<E6?C:
52?
A6BE3292?
A2?;2?8 86=@>32?8
#?:
5:3E2D2?
I2?8
>6?E?;E<<2? 329G2
C6D6=29
DE>3E<2?
A2?;2?8
86=@>32?8
7@D@?
36BD2>329
36C2B      
)=69
<2B6?2
6?6B8:
7@D@?
5:BE>EC<2?
C63282:
!  
;6=2C=29  329G2
6?6B8:
7@D@?
C6D6=29
DE>3E<2?
2<2?
36B<EB2?8  42*3892
+4843 %6C6D2B22?
>2CC2
52?
6?6B8:
>6?EBED
.378*.3
252=29


&
"2= D6BC63ED
2<2?
?52
A6=2;2B:
52=2>
A@<@<
3292C2?
36B:<ED?I2
C652?8<2?

186

Mudah dan Aktif Belajar Fisika untuk Kelas XII

>6?EBED
*=2?4<
4292I2
36B36?DE<
A2<6DA2<6D
6?6B8:
56?82?
6?6B8: 7@D@?
C636C2B


!


6?82?
56>:<:2?
A6?56<2D2?
C642B2
B6=2D:F:CD:< 52A2D
5:DE=:C &  ! 

!  & 

M 

Tokoh Arthur Compton (Holly) (1892–1962)

2B:
*67&2&&3
? 
?52
52A2D
>6>A6B@=69
A6BC2>22?
>@>6?DE> B6=2D:F:CD:<
52B:
C63E29
7@D@?
I2:DE )  & 

! 

M 

 :<6D29E:
  
52?
 
>2<2
*67&2&&3
? 
52A2D
5:DE=:C    )  !

2D2E

  ! ) 

M 

%6D6B2?82? ) 
>@>6?DE>
C63E29
7@D@?
<8
> C  
46A2D
B2>32D
4292I2
> C  
A2?;2?8
86=@>32?8
4292I2
> ! 
D6D2A2?
*=2?4<

P M
$C  
7B6<E6?C:
4292I2
"J  *69'&-&3
&3/&3,
*142'&3,
5&)&
&2'96&3
425843 *6B92D:<2?
DE>3E<2?
2?D2B2
C63E29
7@D@?
52?
C63E29
6=6
DE>3E<2?
7@D@?
>6>:=:<: A2?;2?8
86=@>32?8
 
52?
6=6
<62522?
5:2>
,6D6=29 DE>3E<2?
7@D@?
D6B92>3EB
56?82?
CE5ED

52?
>6?82=2>:
A6?EBE?2? 6?6B8:
>6?;25:

:
=2:?
A:92<
C6328:2?
6=66>36?DE<
CE5ED
56?82?
2B29
86B2<
7@D@?
C636=E>
DE>3E<2?
"2>3EB2?
I2?8
5:2=2>: @=69
7@D@?
5:C63ED
92>3EB2?
@>AD@?
56?82?
4:B:4:B:
<92C?I2
D6B;25: A6BE3292?

52?
A6BE3292?

.?DE<
>6?86D29E:
A6BE3292?
6?6B8:
2D2E A2?;2?8
86=@>32?8
7@D@?
C6D6=29
D6B92>3EB
5:8E?2<2?
2?2=:C:C
56?82? "E<E>
%6<6<2=2?
'@>6?DE>
52?
"E<E>
%6<6<2=2?
?6B8:
)=69 <2B6?2
6?6B8:
I2?8
D6B=:32D
C2?82D
D:?88:
C6BD2
<6>E?8<:?2?
36C2B <646A2D2?
6=63E<2?
C2?82D
36C2B
A6BC2>22?
5:?2>:<2 I2?8
5:8E?2<2?
252=29
5:?2>:<2
B6=2D:F:CD:< 6?82?
>6?88E?2<2?
"E<E>
%6<6<2=2?
?6B8:
52?
"E<E> %6<6<2=2?
'@>6?DE>
52=2>
5:?2>:<2
+6=2D:F:CD:<
5:52A2D<2? A6BC2>22?
92>3EB2?
@>AD@?
C63282:
36B:<ED

    !
 4@C  &  Keterangan:  
A2?;2?8
86=@>32?8
7@D@?
C636=E>
DE>3E<2?
> ’ 
A2?;2?8
86=@>32?8
7@D@?
C6CE529
DE>3E<2?
> ! 
D6D2A2?
*=2?4<


P
M
$C  
46A2D
B2>32D
4292I2


P

> C  
CE5ED
A6?I:>A2?82?
7@D@? &  
>2CC2
5:2>
6=6
Arthur Compton adalah seorang ahli Fisika yang dilahirkan di Wooster, Ohio, USA. Dia belajar di Universitas Princetion dan lulus menjadi profesor Fisika di Chicago (1923). Dia meneliti dan menjelaskan Efek Compton, penambahan panjang gelombang sinar-X yang dihamburkan ketika bertumbukan dengan elektron. Untuk sumbangan pemikirannya tersebut, dia mendapat penghargaan Nobel Fisika pada 1927. Sumber: www.allbiographies.com

elektron diam

foton dengan  , p

2

fo

to

n

t

a erh



M 

mb

ur

 ', p' elektron terdorong pe

3

v

Gambar 7.12 Tumbukan foton dan elektron (a) sebelum tumbukan (b) sesudah tumbukan

Radiasi Benda Hitam

187

Pembahasan Soal Ek (eV)

*252
*67&2&&3
? 
CE<E
!
5:C63ED
;E82
A2?;2?8
86=@>32?8 &  @>AD@?
  
?:=2:?I2
252=29    
$C   !      
> &  
   
<8 
   
> C 

Contoh 7.8

0,2 f0 f (Hz) 3,7 Grafik di atas menunjukkan hubungan antara energi kinetik maksimum elektron terhadap frekuensi foton pada efek fotolistrik. Jika h = 6,6 × 10–34 Js dan 1 eV = 1,6 × 10–19 J, besar f adalah .... a. 48 × 1014 Hz b. 21 × 1014 Hz c. 14 × 1014 Hz d. 9,5 × 1014 Hz e. 8,9 × 1014 Hz UMPTN 1996 Pembahasan Menurut persamaan efek fotolistrik Ek = hf – W maka

Ek  W h Dari grafik, nilai Ek = 0,2 eV = 0,2 (1,6 × 10–19) J Pada saat f = 0, berlaku Ek = – W atau W = –Ek Dari grafik, saat f = 0, nilai Ek = –37 eV, maka W = 3,7 eV = 3,7 (1,6 × 10–19) J Jadi,
0,2  3,7 1,6 10 19 J f = Ek h W = 6,6 10 34 Js = 9,5 × 1014 Hz

,:?2B1
56?82?
A2?;2?8
86=@>32?8
 
 
?>
5:92>3EB<2?
@=69
C63E29
32=@<
<2B3@? ,:?2B1
I2?8
5:92>3EB<2?
5:2>2D:
>6?I:>A2?8
L
D6B9252A
2B29?I2
C6>E=2
":DE?8 A2?;2?8
86=@>32?8
C:?2B1
I2?8
5:92>3EB<2?
:?: &;&' 652
A2?;2?8
86=@>32?8
7@D@?
C636=E>
52?
C6CE529
5:92>3EB<2?
52A2D
5:9:DE?8
56?82? *67&2&&3
? 

      !
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?>
!



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86=@>32?8
7@D@?
C:?2B1
I2?8
5:92>3EB<2?
252=29

’ 
  

 
?>


?>


 
?>

f=

Jawaban: d

2. Sifat Gelombang pada Partikel *252
A6>3292C2?
C636=E>?I2
D6=29
5:;6=2C<2?
329G2
86=@>32?8 >6>:=:<:
C:72D
A2BD:<6=
,6<2B2?8
>E?4E=
A6BD2?I22?
2A2<29
A2BD:<6= >6>:=:<:
C:72D
86=@>32?8
2=2>
A6B:CD:G2
:?D6B76B6?C:
52?
5:7B232?8
4292I2
=63:9
>6?@?;@=
52B:A252
C:72D
A2BD:<6= 4292I2
52AE?
52=2>
A6B:CD:G2
676<
7@D@=:CDB:<
52?
676<
@>AD@?
C:72D A2BD:<6=
4292I2
=63:9
>6?@?;@=
52B:A252
C:72D
86=@>32?8
4292I2 *252
 
:C:<2G2?
36B<632?8C22?
*B2?4:C
49.7
)*
64,1.*
252=29 @B2?8
A6BD2>2
I2?8
>6?82;E<2?
9:A@D6C:C
329G2
A2BD:<6=
C6A6BD:
6=632?8 a. Panjang Gelombang de Broglie 49.7
)*
64,1.*
>6?I2D2<2?
329G2
6=6E?8<:?2?
>6>:=:<:
C:72D
86=@>32?8
56?82?
A2?;2?8
86=@>32?8
I2?8 C6CE2:
*2BD:<6=
I2?8
36B86B2<
56?82?
<646A2D2?
/
>6>:=:<:
>@>6?DE> )

&
/
2D2E
C642B2
B6=2D:F:CD:<
5:DE=:C<2?
C63282:
36B:<ED )

! 

)=69
<2B6?2




>2<2

)

! !   

C69:?882
A2?;2?8
86=@>32?8
56
B@8=:6
252=29

! ! ) &/

188

Mudah dan Aktif Belajar Fisika untuk Kelas XII

M 

%6D6B2?82?  
A2?;2?8
86=@>32?8
A2BD:<6=
>6?EBED
56
B@8=:6
> ! 
D6D2A2?
*=2?4<


P
M
$C & 
>2CC2
5:2>
A2BD:<6=
<8 / 
=2;E
A2BD:<6=
> C 2=2>
A6?82>2D2??I2
)*
64,1.*
>6=:92D
329G2
36B<2C
6=63EB<2?
@=69
2D@>2D@>
A6B>E<22?
?:<6=
D6A2D
C6A6BD:
I2?8 5:A6B<:B2<2?
E?DE<
5:7B232?8
>6?EBED
BE>EC
B288
52?
>6>:=:<: A2?;2?8
86=@>32?8
I2?8
C6CE2:
56?82?
I2?8
5:36B:<2?
@=69
A6BC2>22? 56
B@8=:6
"2C:=
A6B4@322?
:?:
>6I2<:?<2?
)*
64,1.*
329G2
A2BD:<6= 6=66>:=:<:
C:72D
86=@>32?8
I2?8
5:4:B:<2? @=69
A2?;2?8
86=@>32?8?I2

Tokoh Louis Victor de Broglie (1892–1987)

Contoh 7.9 -6?DE<2?
A2?;2?8
86=@>32?8
56
B@8=:6
E?DE<
36?52
36B:<ED
:?: 2 @=2
56?82?
>2CC2

<8
I2?8
36B86B2<
56?82?
<646A2D2?

> C 3 =6 C    
$C 2   !      
> & /

<8 

> C  3 $

=
6/


P
M 
$

)  & $ 

   
<8 
    
$             
<8
> C

   
$C      
> ! )      
<8
> C

Sumber: Conceptual of Physics, 2000

Louis Victor De Broglie (1892– 1987) lahir di Dieppe, Prancis, dan bersekolah di Sorbonne. Ia melanjutkan kuliahnya di Universitas Paris tahun 1926. Pada 1929, ia menerima hadiah Nobel dalam bidang Fisika atas teorinya mengenai sifat gelombang pada partikel. Teori ini menjadi dasar dari mekanika kuantum, cabang dari ilmu Fisika. Pada awal 1800, para fisikawan percaya bahwa cahaya terdiri atas gelombang energi. Pada 1900, Max Planck menunjukkan bahwa cahaya dapat bersifat partikel. Pada 1924, de Broglie menyatakan bahwa elektron dapat bersifat sebagai partikel dan gelombang.

Contoh 7.10 ,63E29
6=6E=2>E=2
5:2>
5:A6B46A2D
@=69
3652
A@D6?C:2=
=:CDB:<
 $:<2
>2CC2
6=6E2D2?
6=632?8
6=632?8
 

K 36B2A2
36C2B

I2?8
92BEC
5:36B:<2?
&

 P M
<8



P
M 

52? !


P
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$C &;&' ,63E29
6=6
2<2? >6?6B:>2
)
=:CDB:<

52?
>6?8E329?I2
>6?;25:
$
6=6
 &/   2D2E
/  &

*2?;2?8
86=@>32?8
56
B@8=:6
6=6
  !

&  &/

&  &  & & 82B
 

K


P
M 
>
>2<2

52A2D
5:9:DE?8
C63282:
36B:<ED    
$C      
>


   
<8 
    
  / P M 
>
 P M
<8
 P M 



P M
$C 




F@=D

Tantangan untuk Anda Efek Compton diaplikasikan dalam dunia kedokteran untuk mendeteksi kepadatan tulang. Dengan bahasa Anda sendiri, dapatkah Anda menjelaskan cara kerjanya?

Radiasi Benda Hitam

189


filamen berkas elektron

detektor

kristal nikel

Gambar 7.13 Skema alat percobaan DavissonGermer. Berkas elektron dijatuhkan pada kristal tunggal nikel. Interferensi yang terjadi pada sudut diamati pada detektor.

sinar datang



b. Eksperimen Davisson-Germer *252
 

&:.7743
52?

*62*6
>6=2<E<2?
A6?6=:D:2? >6?86?2:
92>3EB2?
6=66?;2DE9<2?
36B<2C
6=66>:=:<:
6?6B8:
<:?6D:<
D6BD6?DE
<6
2D2C
6?82>2D:
329G2
A252
CE5ED

D6BD6?DE
D6B;25:
D:D:<
>2E>
I2?8 36B9E3E?82?
56?82?
A@=2
:?D6B76B6?C:
6B52C2B<2?
52D2
I2?8
5:A6B@=69 A2?;2?8
86=@>32?8

I2?8
5:52A2D<2?
D6B?I2D2
252
<6C6CE2:2?
2?D2B2 92C:=
I2?8
>6B6<2
A6B@=69
52?
A2?;2?8
86=@>32?8

I2?8
5:52A2D<2? >6?EBED
9:A@D6C:C
56
B@8=:6
,<6>2
2=2D
A6B4@322?
I2?8
5:=2<E<2? >6B6<2
D2>A2<
A252
&2'&6
 =66?
A252
D23E?8
92>A2
E52B2 5:=6G2D<2?
>6=2=E:
>652?
=:CDB:<
56?82?
A@D6?C:2=
I2?8
52A2D
5:2DEB
,6D6=29 5:9:DE?8
A2?;2?8
86=@>32?8
I2?8
5:>:=:<:
@=69
6=62 36C2B
56?82?
;2B2<
2D@>2D@>
52=2>
6?82>2D:
CE5EDCE5ED
A2?DE=
6=66?86?2:
6>A6B@=69
A@=2A@=2
5:7B22
56?82?
A@=2
5:7B26?
&:.7743
52? *62*6
>6?E?;E<<2?
329G2
A2BD:<6=
C6A6BD:
6=632?8 *252
66?
&:.7743*62*6
36B<2C
6=62?DE=
2?DE=
I2?8 >6>:=:<:
C6=2?8
;2B2<
C636C2B

2<2?
>6?892>3EB<2?
6=63EB

"E3E?82?
;2B2<
2?D2B2
2D@>

52?
;2B2<
2?D2B2
3:52?8 A6>2?DE=

>6>6?E9:
A6BC2>22?
36B:<ED

 2 2

   4@C

a

a

d a

2

d d

Gambar 7.14 Gambaran mengenai hamburan elektron oleh bidang kristal.

190




M 

2B:
&2'&6

D6B=:92D
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Mudah dan Aktif Belajar Fisika untuk Kelas XII

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36B:<ED pengamat

Kata Kunci • • • • • • • •

gelombang partikel efek fotolistrik foton fungsi kerja efek Compton panjang gelombang de Broglie jarak antara bidang prinsip ketidakpastian Heisenberg

pengamat

pengamat

foton datang

momentum elektron semula

2

Gambar 7.15 elektron bertumbukan

3

momentum elektron akhir

Tumbukan foton dan elektron ketika mengamati elektron.

4 Sumber: Konsep Fisika Modern, 1983

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191

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Refleksi Setelah mempelajari materi pada bab ini, Anda tentu dapat memahami mengenai benda hitam. Anda juga paham bahwa cahaya juga dapat bersifat seperti partikel dan partikel bisa bersifat seperti gelombang. Dari keseluruhan materi yang ada pada bab ini, bagian

manakah yang Anda anggap sulit untuk dipahami? Coba Anda diskusikan dengan teman atau guru Fisika Anda. Coba Anda tuliskan juga manfaat mempelajari bab ini. Diskusikan konsep yang ada di dalamnya bersama teman Anda.

Radiasi Benda Hitam

193

Tes Kompetensi Bab 7  .1.-1&-
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4



194

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196

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6/

Bab

8 Sumber: Chemistry: Matter and Its Changes, 2003

Atom terdiri atas proton dan neutron yang terletak pada inti serta elektron yang mengorbit pada kulit.

Fisika Atom Hasil yang harus Anda capai: menganalisis berbagai besaran fisis pada gejala kuantum dan batas-batas berlakunya relativitas Einstein dalam paradigma Fisika Modern. Setelah mempelajari bab ini, Anda harus mampu: membuat ulasan tentang perkembangan teori atom

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A. Evolusi Model Atom B. Atom Berelektron Banyak

Fisika Atom

197

Tes Kompetensi Awal #,),3;4
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Tokoh John Dalton (1766–1844)

 

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1. Model Atom Demokritus

John Dalton adalah seorang ilmuwan Inggris. Dia orang pertama yang mengemukakan tentang teori atom berdasarkan hasil eksperimen yang dilakukan Joseph Proust. Dalton menyatakan bahwa atom adalah bagian terkecil dari benda yang tidak dapat dibagi-bagi lagi. Sumber: www.allbiographies.com

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198

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Tokoh Sir J.J. Thomson (1856–1940)

Thomson adalah seorang ahli Fisika yang lahir di Checham Hill, Greater Manchester, Inggris. Dia belajar di Cambridge dan meraih gelar profesor pada 1884. Dia meraih hadiah Nobel Fisika tahun 1906. Eksperimennya yang terkenal tentang tabung sinar katode. Dari eksperimennya, dia menemukan nilai nisbah e/m. Sumber: www.allbiographies.com

Fisika Atom

199

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Gambar 8.1 Percobaan Thomson

Ingatlah Jika ada elektron yang bergerak dengan kecepatan v dalam suatu medan magnet B, arah gaya Lorentznya mengikuti kaidah tangan kanan. F

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Gambar 8.2 Model Atom Thomson

minyak disemprot lubang kecil radiasi sinar-X

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Gambar 8.3

tetes minyak Sumber: Chemistry, 2001

Percobaan tetes minyak Millikan

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Gambar 8.4 Gaya berat, gaya Archimedes, dan gaya stokes bekerja pada tetesan minyak pada percobaan Millikan.

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Dengan menggunakan logika dan pemahaman Anda tentang konsep gerak lurus, apakah Anda dapat menduga bagaimana Millikan mengukur kecepatan (v) dari tetesan minyak tersebut? Jelaskan dengan alasan-alasan ilmiah menurut pendapat Anda.

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Sumber: Science Encyclopedia, 1998

Ernest Rutherford lahir di New Zealand pada 1871. Ia mendapat beasiswa di Universitas Cambridge, Inggris pada 1894. Di sana, ia mulai melakukan penelitian tentang radiasi elektromagnetik, tetapi kemudian perhatiannya beralih ke sinar-X dan radioaktivitas. Ia menunjukkan bahwa ada perbedaan tipe radioaktivitas, yang ia namakan radiasi alfa, beta, dan gamma. Ia mendapat hadiah Nobel untuk bidang kimia pada 1908. Pada 1911, Rutherford menjadi seorang profesor di Universitas Manchester, ia menemukan bahwa hampir seluruh massa atom berkumpul di inti atom.

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Gambar 8.6 Percobaan Rutherford

Fisika Atom

203

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Gambar 8.7 Skema penyimpangan sinar alfa.

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204

Mudah dan Aktif Belajar Fisika untuk Kelas XII

merah

biru

ungu

warna

Gambar 8.9 Spektrum hidrogen pada daerah tampak

frekuensi (Hz) 4 × 1014

5 × 1014

6 × 1014

7 × 1014

8 × 1014

Sumber: Konsep Fisika Modern, 1983

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Tantangan untuk Anda Diskusikanlah bersama teman Anda, apakah Anda dan teman Anda dapat menduga mengapa dalam eksperimennya Rutherford menggunakan sinar  dan lempeng emas?

Keterangan: 

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Tantangan untuk Anda Hitunglah panjang gelombang terpendek dan terpanjang untuk deret Balmer, deret Lyman, deret Pascha, deret Bracket, dan deret Pfund.

Fisika Atom

205

Contoh 8.1 '141
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K         =           = K 



















       
= 

6. Model Atom Bohr

v elektron m F r

+Ze inti

Gambar 8.10 Model atom Bohr (Z = 1) untuk hidrogen

)5D5<18
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=1;1 * #
    *

206

Mudah dan Aktif Belajar Fisika untuk Kelas XII

M

41@E>
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Tokoh Niels Bohr (1885–1962)

  #   (        *  *     *

"5D5B1>71>  

P
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 %=  
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a. Jari-Jari Lintasan Stasioner 9C9;1
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#   

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%*




  



Sumber: Jendela Iptek, 1997

Pada awal 1900-an, struktur atom menjadi lebih jelas, namun hukum Fisika pada saat itu tidak dapat menjelaskan mengapa elektron tidak cepat menghasilkan spiral menuju inti. Niels Bohr, murid Rutherford, membantu memecahkan misteri itu dengan menunjukkan bahwa elektron-elektron hanya diberi energi tertentu. Ia menemukan bahwa elektronelektron yang memiliki energi terendah mengorbit paling dekat ke inti, dan elektron dengan energi tinggi mengorbit paling jauh. Pada 1913, Niels Bohr juga menjelaskan hubungan antara materi dan cahaya. Ia menunjukkan bahwa jika elektron berpindah dari satu tingkat energi ke tingkat energi lainnya, elektronelektron itu mengeluarkan atau menyerap paket radiasi dalam bentuk cahaya. Paket ini dinamakan foton.

 *  *  *



 &
   %*  *

* !9;1
 


%

M 



4941@1D;1>
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Fisika Atom

207

1B9
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1D1E
? n


E=0

n=4 n=3

E = –0,8 eV E = –1,5 eV

n=2

E = –3,4 eV

n=1

E = –13,6 eV

Gambar 8.11

b. Energi Elektron Berdasarkan Teori Bohr 5C1B
5>5B79
5<5;DB?>
41@1D
49@5B?<58
45>71>
=5>7712E>7;1> !,89(4((5
@ 
41>
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@ 
I19DE &  

Dalam model atom Bohr, energi yang dibutuhkan oleh elektron hidrogen untuk pindah dari orbit dengan bilangan kuantum 1 ke 3 adalah .... a. 1,50 eV b. 1,90 eV c. 2,35 eV d. 12,10 eV e. 13,60 eV UMPTN 1992 Pembahasan: Energi yang dibutuhkan  E = E 3 – E1 Diketahui bahwa energi dasar = E1 = –13,6 eV Maks, E E3 = 12 n 13,6 eV = 32 Jadi, 13,6 eV – (– 13,6 eV) 9 = 12,10 eV

E =

Jawaban: d

208

M 

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4941@1D
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C5@5BD9
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Tingkat-tingkat energi atom hidrogen.

Pembahasan Soal

%  

 
&

&  

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Fisika Atom

211

Informasi

Contoh 8.5

untuk Anda

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Atom adalah partikel yang sangat kecil sehingga tidak dapat diamati dengan cahaya tampak. Menggambarkan struktur atom merupakan pekerjaan yang sangat menarik. Untuk mempelajari atom, para ahli mempelajarinya dari cahaya yang diesensikan oleh atom-atom dari unsur murni. Spektrum cahaya tersebut mengandung informasi tentang struktur, atom, energi ionisasi, dan lain-lain.

Information for You Atom are tiny, and not observable using visible light. Trying to figure out the structure of atoms is a facinating task. A major source of information about atoms comes from a study of the light emitted by the atoms of a pure material. The spectrum of the light give information about atom structure, ionization energy, etc.

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1

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5<5;DB?>
CE2;E<9D 
141<18

5<5;DB?>
41>
C5D5BEC>I1
C5@5BD9
$(),3
 c. Bilangan Kuantum Magnetik (m  ) &<58
;1B5>1
5<5;DB?>
25B=E1D1>
<9CDB9;
75B1;
5<5;DB?>
41@1D 49@5>71BE89
?<58
=541>
=17>5D9;
;921D>I1
5<5;DB?>
I1>7
25B75B1; 49
41<1=
?B29D>I1
=5=9<9;9
=?=5>
=17>5D9;
$?=5>
=17>5D9;
41B9 5<5;DB?>
9>9
49>I1D1;1>
45>71>
29<1>71>
;E1>DE=
=17>5D9;
=   9<1>71>
;E1>DE=
=17>5D9;
=5>I1D1;1>
1B18
=?=5>DE=
CE4ED 5<5;DB?>
!9;1
49D5D1@;1>
1B18
=541>
=17>5D9;
<E1B
C5:1:1B
45>71>
CE=2E 0
;5=E>7;9>1>

41<1=
1B18
0
49>I1D1;1>
45>71>
@5BC1=11> 0  %



M 

9<1>71>
;E1>DE=
=17>5D9;
=  
=5=9<9;9
>9<19
1>D1B1
M  41>
   $9C1<>I1 E>DE;



81B71
= 


1D1E
D5B41@1D

?B29D1< E>DE;



81B71
= 

M

41>

1D1E
D5B41@1D

?B29D1< E>DE;



81B71
= 

M 
M



1D1E
D5B41@1D

?B29D1< 5C1B
J
41@1D
49<981D
@141
(4)(8
 

)E2;E<9D

+ (   



 

1>I1;
%5

   

 



1>I1;
5<5;DB?>     

P
21>I1;
= 

Lz

= 2 m= 2 m= 1 m= 0 m= 1

2



0 –


m= 2

–2


Gambar 8.13 Arah dan besar momentum sudut Lz untuk  = 2

Fisika Atom

213

z

z

z Z  h 2

Z  h  Z  h 2

Z  0

Z  0

Z  0

Z  h 2 Z  h  2

Z  h 2

0

 1

1

2

3 Gambar 8.14

Arah momentum sudutsudut orbital s, p, dan d (a) orbital hanya memiliki 1 kemungkinan arah; (b) orbital p memiliki 3 kemungkinan; dan (c) orbital d memiliki 5 kemungkinan.

s px

py

'141
(4)(8
 
49@5B<981D;1>
F5;D?BF5;D?B
=?=5>DE=
CE4ED 45>71>
29<1>71>
;E1>DE=
?B29D1<
D5BD5>DE
9<1>71>
;E1>DE=
=17>5D9; :E71
=5>771=21B;1>
;535>45BE>71>
;54E4E;1>
?B29D1<
49
41<1=
BE1>7 C5@5BD9
D5B<981D
@141
(4)(8
  d. Bilangan Kuantum Spin (ms) 9<1>71>
;E1>DE=
C@9>
%C
=5>I1D1;1>
1B18
@5B@ED1B1>
5<5;DB?> D5B8141@
CE=2E>I1
C@9>
I1>7
41@1D
=5>9=2E<;1>
=?=5>
=17>5D9;  %9<19
29<1>71>
;E1>DE=
C@9>
141
4E1
I19DE
%C


E>DE;
1B18
@ED1B1>  5<5;DB?>
;5
;1>1>
41>
%C


E>DE;
1B18
@ED1B1>
5<5;DB?>
;5
;9B9 $1C9>7=1C9>7
49DE<9C
45>71>
>?D1C9
1B18
@1>18
;5
1D1C
41>
;5
21G18 )E2;E<9D
+
=5=9<9;9
C52E18
?B29D1<
41>
@1<9>7
21>I1;
499C9
4E1
5<5;DB?> )E2;E<9D
(
=5=9<9;9
D971
?B29D1<
(H
(I
(J
41>
@1<9>7
21>I1;
499C9
?<58
P



5<5;DB?>
<5;DB?>
41<1=
CE2;E<9D
+
(
41>

41@1D
4971=21B;1> 45>71>
4917B1=
?B29D1<
C5@5BD9
(4)(8
  5>41
I1>7
25B@ED1B
=5=9<9;9
=?=5>DE=
CE4ED
,5;D?B
=?=5>DE= CE4ED
I1>7
25B;19D1>
45>71>
29<1>71>
;E1>DE=
+
9>9
141<18

I1>7 @1>:1>7>I1

pz

        




d xy dxz dyz d x2y2 d x2y2

Gambar 8.15 Elektron dalam subkulit s, p, dan d.

1

M 

B18
=?=5>DE=
CE4ED
49D5>DE;1>
?<58
;?=@?>5>
F5;D?B
@141
CE=2E0 I19DE
CE=2E
D571;
@141
C9CD5=
;??B49>1D
D971
49=5>C9

2



C






45>71>
%C

 1D1E
 

M 

1=21B
C@9>
41>
F5;D?B
=?=5>DE=
CE4ED
+
D1=@1;
@141
(4)(8
  1 2

1 2

3 SZ  h

4p

S

SZ   h 4p

Gambar 8.16 Arah perputaran dan vektor momentum sudut S.

$(),3

 "5C9=@E<1>
"55=@1D
9<1>71>
"E1>DE= (4(

6:(90

03(0

?(5.
07,82,5(52(5

9<1>71>
;E1>DE=
ED1=1

&






9<1>71>
;E1>DE=
?B29D1<







>M

9<1>71>
;E1>DE=
=17>5D9;

%

M  

M 
M







9<1>71>
;E1>DE=
C@9>

%C

 

Contoh 8.7 *E>:E;;1><18
218G1
;E<9D

=1;C9=E=
49D5=@1D9
?<58

5<5;DB?> (=() "E<9D

=5=9<9;9
29<1>71>
;E1>DE=
&

 +>DE;
&


>9<19
29<1>71>
;E1>DE=
?B29D1<>I1



41>


 +>DE;



D5B41@1D
 




P




2E18
>9<19
%  +>DE;



D5B41@1D
 




P




2E18
>9<19
%  !E=<18 


2E18
>9<19
% 

214

Mudah dan Aktif Belajar Fisika untuk Kelas XII

+>DE;


2E18
% 
49D5=@1D9
?<58

2E18
5<5;DB?>
!149
E>DE;

2E18

% 
49D5=@1D9 ?<58

P



2E18
5<5;DB?> (:(:(5 M E>DE;
&


29<1>71>
;E1>DE=
?B29D1<>I1  


41>
M E>DE;
&


29<1>71>
;E1>DE=
?B29D1<>I1  



41>


41>
C5D5BEC>I1

*5<18
>41
;5D18E9
218G1
29<1>71>
;E1>DE=
=17>5D9;
5B1D 8E2E>71>>I1
45>71>
=541>
=17>5D9;
I1>7
49D9=2E<;1>
?<58
75B1; 5<5;DB?>
21719=1>1
:9;1
1D?=
894B?75>
49D5=@1D;1>
41<1=
=541> =17>5D
8?=?75>

'5B81D9;1>
75B1;
5<5;DB?>
41<1=
=541>
=17>5D @141
(4)(8
 
!9;1
C52E18
5<5;DB?>
25B=1CC1
%
25B75B1;
41<1= ?B29D
I1>7
25B:1B9:1B9
*
45>71>
6B5;E5>C9

=?=5>DE=
CE4ED
I1>7
49=9<9;9 5<5;DB?>
141<18


%5.*

%5 * 
C541>7;1>
1B18
=?=5>DE=>I1 25BCE4ED

D5B8141@
CE=2E0
5C1B>I1
1BEC
I1>7
D9=2E<
1;921D
75B1;1> 5<5;DB?>
494569>9C9;1>
C521719
21>I1;>I1
=E1D1>
I1>7
=5>71<9B
D91@ ) 45D9;
!149
1BEC
I1>7
49D9=2E<;1>
75B1;1>
5<5;DB?>


 )     >41
:E71
=5=181=9
218G1
75B1;1>
5<5;DB?>
:E71
=5>9=2E<;1>
=541> =17>5D
9>71D
=541>
=17>5D
49
C5;9D1B
1BEC
<9CDB9;
"5D9;1
49
D5=@1D 5<5;DB?>
21>I1;
4925B9;1>
=541>
=17>5D
I19DE
=541>
=17>5D
41B9
<E1B =1;1
1;1>
D5B:149
9>D5B1;C9
1>D1B1
4E1
=17>5D
49
D5=@1D
9>9
=@<9;1C9>I1 141<18
41@1D
=5=25B9;1>
D?BC9

I1>7
=5>I5212;1>
75B1;1>
5<5;DB?> 25BE218
5C1B>I1
D?BC9
1;921D
=541>
=17>5D

141<18 

 P


M 





M 

Tantangan untuk Anda Tentukan besar momentum sudut untuk subkulit: a. f; b. g; c h.

Z L



Gambar 8.17 Gerak elektron dalam medan magnet B.

45>71> "5D5B1>71> 

=?=5>
=17>5D9; 


<E1C
41B9
2941>7
5<5;DB?> &<58
;1B5>1


M

41>


 B 
>41
41@1D
=5>E<9C;1>

      *   



 *

M
    %     %  * 

M



*1>41
>571D96
=5>E>:E;;1>
1B18
%
25B<1G1>1>
45>71>
1B18
=?=5>DE=
CE4ED 
!149
25C1B>I1
D?BC9
I1>7
49C5212;1>
?<58
=541>
=17>5D

141<18

    
C9>
     % 

Z L Lz = me 

M 

*?BC9
9>9<18
I1>7
=5=ED1B
?B29D
5<5;DB?>
C589>771
25BE218
1B18 '5B81D9;1>
(4)(8
 
I1>7
=5<E;9C;1>
?B29D?B29D
5<5;DB?>
41> @5B81D9;1>
@5BE2181>
1B18
@ED1B1>>I1
E>DE;
=5>75D18E9
25B1@1 ;5=E>7;9>1>
@5BE2181>
75B1;1>
5<5;DB?>
41@1D
;9D1
<981D
=5<1<E9
5>5B79 =17>5D9;
41B9
@5>71BE8
75B1;1>
5<5;DB?>
41B9
=541>
=17>5D
<E1B
 5C1B
5>5B79
=17>5D
=5=5>E89
@5BC1=11>



Gambar 8.18 Arah momentum sudut elektron.

  

   3?C     3?C %  

Fisika Atom

215

 5C1B1>
% 49>1=1;1>
@5=21>49>7
79B?=17>5D9;
/*'%&,!  *,!' '5B81D9;1>
(4)(8
 
1=21B
D5BC52ED
=5>E>:E;;1>
218G1 % 3?C



C589>771
;9D1
@5B?<58    %  

   3?C         %  %     

  

 %  

  % 

M 

&<58
;1B5>1
%5
=5=9<9;9
2525B1@1
>9<19
41>
25BC961D
49C;B9D
5>5B79 9>D5B1;C9
=541>
=17>5D9;
9>9
:E71
=5=9<9;9
2525B1@1
>9<19

   %  

   %  )521719
3?>D?8
:9;1
29<1>71>
;E1>DE=
 

5>5B79
5<5;DB?>
1;1> =5>71<1=9
@5BE2181>
C521719
25B9;ED    "5D9;1
%5


5>5B79
5<5;DB?>
=5>:149
D?D

5<5;DB?>
1G1<
M     %   "5D9;1
%5


5>5B79
5<5;DB?>
D5D1@
D941;
25BE218

Gambar 8.19 Terpecahnya garis spektrum akibat medan magnet.

   "5D9;1
%5

M
5>5B79
5<5;DB?>
=5>:149
D?D

5<5;DB?>
1G1<
   %    !149
D9>7;1D
5>5B79
5<5;DB?>
@5318
=5>:149

21791>
+>DE;


5>5B79
5<5;DB?>
1;1>
@5318
=5>:149

21791>
C5CE19
45>71>
>9<19
% 9>9<18
@5>D9>7>I1
29<1>71>
;E1>DE=
=17>5D9;
41<1=
=5>5>DE;1>
75B1; ?B29D
5<5;DB?> '5318>I1
D9>7;1DD9>7;1D
5>5B79
9>9
D5<18
491=1D9
?<58
',,4(5 =5<1<E9
5;C@5B9=5>>I1
1
=5>71=1D9
C@5;DBE=
1D?=
894B?75>
;5D9;1 4925B9
=541>
=17>5D9;
;E1D
D5B>I1D1
71B9C71B9C
C@5;DBE=
I1>7
491=1D9 @5318
=5>:149
2525B1@1
21791> '5318>I1
71B9C
C@5;DBE=
9>9
D5B>I1D1
1;921D
DB1>C9C9
41B9
D9>7;1D
5>5B79 I1>7
D5B@5318
9DE
41@1D
>41
@181=9
@141
(4)(8
 
'5BE2181> ;539<
5>5B79
=5=5>E89
@5BC1=11>
  
C589>771
41@1D
=5=5>71BE89 @1>:1>7
757
=5<1<E9
=5D?45
4965B5>C91C9


 
1;1>
49@5B?<58  

 D1E
45>71>
=5>71=29<
>9<19
=ED<1;>I1
49@5B?<58

    

216

Mudah dan Aktif Belajar Fisika untuk Kelas XII

M 

Contoh 8.8 9DE>7<18
@5BE2181>
@1>:1>7
757
6?D?>
I1>7
D5B:149
1;921D
DB1>C9C9
41B9
(
;5 +
1@129<1
1D?=
894B?75>
49D5=@1D;1>
41<1=
=541>
=17>5D
C525C1B

* (=()             M


 
    &  &  !149
@1>:1>7
757
I1>7
49@1>31B;1>
141<18




M





  P    !C 
P  = C  

P
 
=   
  5, 
P   ! 5,  '5BE2181>
5>5B79
    
 
P
 M 
! * 
*
P
 M
! !149
25B41C1B;1>
!,89(4((5
@ 
49@5B?<58


 P   =  
P
 M
!

 
P
 M
=     
P
   !C 
P   = C 

3. Sifat Atom )961DC961D
CE1DE
E>CEB
41@1D
491=1D9
45>71>
=5<981D
CDBE;DEB 1D?=>I1
)ECE>1>
5<5;DB?>
41<1=
CE1DE
1D?=
41@1D
49@1;19
C521719
41C1B E>DE;
=5>75D18E9
C961DC961D
1D?=
D5BD5>DE
'141
C521791>
E>CEB
;E<9D ;E<9D
1D?=
141
I1>7
D5B9C9
@5>E8
5<5;DB?>
41>
141
I1>7
;?C?>7
"E<9D I1>7
D941;
@5>E8
D5B9C9
5<5;DB?>
25B141
@141
;E<9D
@1<9>7
<E1B
<5;DB?> @141
;E<9D
D5B<E1B
9>9
49>1=1;1>
$#,*'&
.$&+! '5>79C91>
5<5;DB?>
49=E<19
41B9
D9>7;1D
5>5B79
D5B5>418
"?>697EB1C9 I1>7
=1>D1@
D5B41@1D
@141
CE2;E<9D
I1>7
D5B9C9
@5>E8
!9;1
CE2;E<9D
D5<18 D5B9C9
@5>E8
C9C1
5<5;DB?>
1;1>
=5>79C9
CE2;E<9D
C5<1>:ED>I1
$9C1<>I1 141
C1DE
5<5;DB?>
I1>7
D5B9C9
@141
CE2;E<9D
D5B<E1B
=1;1
5<5;DB?>
9>9 35>45BE>7
=E418
<5@1C
CE@1I1
1D?=
=5>:149
CD129<
'5<5@1C1>
5<5;DB?> 9>9
41@1D
D5B:149
45>71>
31B1
25B9;1D1>
45>71>
1D?=
<19> )521719
3?>D?8
1D?=
%1
I1>7
=5=9<9;9
;?>697EB1C9

+     

+ ( +  

"E<9D

D5B49B9
1D1C
CE2;E<9D
+
(
41>

1B9
;?>697EB1C9
1D?=
%1
@141 (4)(8
 
D5B<981D
81>I1
141
C1DE
5<5;DB?>
I1>7
=5>79C9
;E<9D
 I19DE
@141
?B29D
+
+>DE;
=5>:149
CD129<
1D?=
%1
35>45BE>7
=5<5@1C;1> C1DE
5<5;DB?>
D5B<E1B>I1
I1>7
D5B41@1D
@141
;E<9D

)521<9;>I1
:9;1 @141
CE2;E<9D
D5B<E1B
;5;EB1>71>
C1DE
5<5;DB?>
E>DE;
=5>:149
CD129< 1D?=
9>9
35>45BE>7
=E418
=5>5B9=1
C1DE
5<5;DB?>
41B9
<E1B '5B81D9;1>
:E71
(4)(8
 
'141
1D?=
$>
;E<9D
D5B<E1B>I1
I19DE ;E<9D

D5B9C9

5<5;DB?>
"E<9D

D5B9C9
=1;C9=1<

5<5;DB?>
25B1BD9 D5B41@1D
;5;EB1>71>

5<5;DB?>
C589>771
1D?=
$>
35>45BE>7
=5>5B9=1 C1DE
5<5;DB?>
41B9
1D?=
<19>
CE@1I1
=5>:149
CD129<
D?=
%1
41>
$> 41@1D
25B712E>7
=5>:149
=?<5;E<
I1>7
CD129<
;1B5>1
@141
C5D91@
1D?= 1;1>
=5=9<9;9
;E<9D
1D1E
CE2;E<9D
I1>7
D5B9C9
@5>E8
?<58
5<5;DB?>

elektron dilepas

Na

1

Na

2

Gambar 8.20 (a) Atom Na, kulit terluarnya terisi satu elektron. (b) Atom Na yang stabil setelah melepas elektron terluarnya.

Mn

Na

1

2

e

Gambar 8.21 (a) Atom Mn menerima elektron dari atom Na. (b) Atom Na yang tidak stabil.

Fisika Atom

217

1B9
EB191>
D5BC52ED
41@1D
491=29<
;5C9=@E<1>
218G1
1D?=1D?= 35>45BE>7
E>DE;
=5=9<9;9
;?>697EB1C9
C5@5BD9
71C
=E<91
I19DE
=5=9<9;9 ;51411>
I1>7
CD129<
45>71>
31B1
=5<5@1C
1D1E
=5>5B9=1
5<5;DB?>

4. Sistem Periodik

Tantangan untuk Anda Jelaskan sifat-sifat yang dimiliki atom berikut ini. 1.

40 20

Ca

2.

30 31

Ga

3.

80 35

Br

1s 2s

2p

3s

3p

3d

4s

4p

4d

5s

5p

5d

6s

6p

6d

4f

7s

Gambar 8.22 Pengisian elektron menurut aturan Aufbau.

218

)9CD5=
@5B9?49;
E>CEBE>CEB
49CECE>
25B41C1B;1>
;5>19;1>
>?=?B 1D?=
1D1E
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Fisika Atom

219

Gambar 8.23 Berdasarkan asas larangan Pauli, satu orbital maksimum ditempati dua elektron dengan spin yang berlawanan.

benar

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5. Spektrum Emisi dan Spektrum Absorpsi

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kisi



sumber



spektrum

Gambar 8.24 Proses untuk menghasilkan spektrum emisi

220

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222

Mudah dan Aktif Belajar Fisika untuk Kelas XII

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M M M  M 

M  M

Sumber: General Chemistry, 1990

Kata Kunci • • • • • • • • • • • • • •

teori atom Bohr bilangan kuantum utama bilangan kuantum orbital bilangan kuantum magnetik bilangan kuantum spin prinsip Pauli aturan Aufbau kaidah Hund spektrum emisi spektrum absorsi garis-garis Praunhoffer energi ionisasi afinitas elektron elektron valensi

Fisika Atom

223

Tes Kompetensi Subbab

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Refleksi Setelah mempelajari bab ini, tentu Anda telah memahami bagaimana perkembangan model atom sampai terakhir digunakan model atom Bohr. Anda juga tentu telah paham bagaimana kedudukan dan sifat elektron dalam bab inti. Dari semua materi yang ada dalam bab ini, bagian manakah yang menurut Anda sulit dipahami? Coba Anda diskusikan bersama teman atau guru Fisika Anda.

Sekarang Anda menjadi tahu bahwa rekayasa atom untuk dimanfaatkan dalam kehidupan bergantung sekali pada elektron valensi atom. Coba Anda cari manfaat lain yang diperoleh setelah mempelajari bab ini, kemudian diskusikan konsep Fisika apa yang terkandung di dalamnya bersama teman Anda.

Fisika Atom

225

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Bab

9 Sumber: Jendela IPTEK, Energi

Pesawat ruang angkasa Starship Enterprise dalam serial televisi Star Trek seringkali berjalan dengan kecepatan cahaya.

Teori Relativitas Khusus Hasil yang harus Anda capai: menganalisis berbagai besaran fisis pada gejala kuantum dan batas-batas berlakunya relativitas Einstein dalam paradigma Fisika Modern. Setelah mempelajari bab ini, Anda harus mampu: memformulasikan teori relativitas khusus untuk waktu, panjang, dan massa, serta kesetaraan massa dengan energi yang diterapkan dalam teknologi

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Sumber: Science Encyclopedia, 1998

Hendrik Antoon Lorentz (1853– 1928) adalah seorang fisikawan Belanda yang kali pertama mengemukakan tentang pemantulan dan pembiasan cahaya, ditinjau dari teori bahwa cahaya termasuk gelombang elektromagnetik. Teori ini dikembangkan oleh James Clerk Maxwell yang menyatakan bahwa cahaya terdiri atas gelombang listrik dan gelombang magnetik dengan sudut yang berbeda. Lorentz juga memprediksi keberadaan elektron. Beliau menerima hadiah Nobel tahun 1902 dalam bidang fisika atas penelitiannya mengenai medan magnet. Dalam bab teori relativitas, Lorentz berjasa dalam teorinya mengenai "Transformasi Lorentz".

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248

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10 Sumber: Energy: Technology and Development, 1995

Pembangunan PLTN merupakan salah satu solusi timbulnya krisis energi yang suatu saat akan dialami penduduk Bumi.

Fisika Inti dan Radioaktivitas Hasil yang harus Anda capai: menunjukkan penerapan konsep Fisika inti dan radioaktivitas dalam teknologi dan kehidupan sehari-hari. Setelah mempelajari bab ini, Anda harus mampu: • •

menganalisis karakteristik inti atom dan radioaktivitas; mendeskripsikan pemanfaatan radiokativitas dalam kehidupan sehari-hari dan teknologi untuk kesejahteraan manusia.

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Fisika Inti dan Radioaktivitas

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Informasi untuk Anda Eksperimen Rutherford menyatakan bahwa inti atom memiliki diameter sekitar 10–15 sampai 10–14 m. Dari teori kinetik dan khususnya analisis Einstein tentang gerak Browman, diameter atom memiliki ukuran sekitar 10–10 m. Itu artinya elektron akan terlihat berjarak sekitar 10.000 sampai 100.000 kali lipat ukuran inti diukur dari inti atom tersebut (jika inti diibaratkan bola baseball, atom akan memiliki diameter seukuran kota yang berdiameter beberapa kilometer). Jadi, sebuah atom sebagian besar berupa ruang kosong.

Information for You Rutherford’s experiments suggested that the nucleus must have a radius of about 10–5 to 10–14 m. From kinetic theory, and especially Einstein’s analysis of Browman movement, the radius of atoms was estimated to be about 10–10 m. Thus the electrons would seem to be at a distance from the nucleus of about 10,000 to 100,000 times the radius of the nucleus itself (if the nucleus were the size of a baseball, the atom would have the diameter of a big city several kilometers across). So an atom would be mostly empty space.

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Fisika Inti dan Radioaktivitas

259

hidrogen biasa

deuterium

tritium

proton neutron elektron

Gambar 10.1 Isotop hidrogen

Nomor neutron (N) 140 130 120 100 90 80 70 60 50 40 30 20 10 0

Z=N

0 10 20 30 40 50 60 70 80 90 Nomor proton (Z) Sumber: Konsep Fisika Modern, 1983

Gambar 10.2 Diagram kestabilan inti (N-Z)

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260

Mudah dan Aktif Belajar Fisika untuk Kelas XII

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Sumber: Konsep Fisika Modern, 1983

Gambar 10.3 Tingkat energi beberapa isotop

Fisika Inti dan Radioaktivitas

261

#'(+2
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DKAB;ED

    

Contoh 10.2



'7II7
?IEJEF
" 7:7B7>
  
IC7
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;D;H=?
?A7J
?DJ?
"
@?A7
C7II7
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IC7
:7D
C7II7
D;KJHED


IC7

Sumber: Conceptual Physics, 1998

262

 P 

$?A7
$

$'

$
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F;HI7C77D
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Mudah dan Aktif Belajar Fisika untuk Kelas XII

'='( ?A;J7>K?

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7J7I

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';0

B/A, MeV per nukleon 10 56 35Cl Fe 20 Ne 12C 19 F 14 N 4 He 11 B 9 Be 5

89 Y 110 Cd

Pr 180Hf

75

As

125Te

160Dy

209

Bi

197

Au

238 U

6Li

2H

0

100 nomor massa A

6. Gaya Inti !7N7
?DJ?
8;HI?<7J
I7B?D=
C;DEB7A
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<;HC? : =7N7
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8K7>
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Tes Kompetensi Subbab

141

200

Gambar 10.4 Grafik energi ikat per nukleon terhadap nomor massa berbagai inti.

Kata Kunci • • • • • • • •

isotop hipotesis proton-elektron hipotesis proton-neutron nomor atom nomor massa nuklida defek massa energi ikat inti

A

+80'1'42'.
6'*'
(;1;
2':/.'4





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8

 

C 8 ; -? B 9 '7II7
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;
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DKAB;ED
KDJKA
7JEC
 

Fisika Inti dan Radioaktivitas

263

B. Radioaktivitas 1. Sejarah Penemuan Radioaktif !;@7B7
H7:?E7AJ?<
?D?
:?J;CKA7D
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+48>
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:?I;8KJ =;@7B7
H7:?E7AJ?<
7J7K
H7:?E7AJ?L?J7I

2. Jenis-Jenis Radioaktif

magnet kuat P Q

sumber 

ratemeter Sumber: Physics for ‘O’ Level, 1990

Gambar 10.5 Pengujian pembiasan sinar  dalam medan magnet.







radium

sumber radioaktif

Sumber: Conceptual Physics, 1998

Gambar 10.6 Dalam medan magnet, hanya sinar  yang tidak dibelokkan.

264

7H?
F;D;CK7D
+)7;+8+2
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I7B7>
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   a. Pemancaran Sinar Alfa (  ) '7II7
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7JEC
O7J
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Mudah dan Aktif Belajar Fisika untuk Kelas XII

7A7D
8;HAKH7D=

:7H?
DECEH
7JEC
?D:KADN7
:7D
DECEH
C7II7DN7
7A7D 8;HAKH7D=

-;97H7

KCKC
H;7AI?
F;C7D97H7D
7B<7
  
:7F7J
:?JKB?IA7D I;87=7?
8;H?AKJ  

EDJE>

 

–

– 1,015 MeV

2   2   27 12

 /    2  

b. Pemancaran Sinar Beta (  ) 'K7J7D
:7D
C7II7
I?D7H

I7C7
:;D=7D
;B;AJHED
"7B
?D?
A7H;D7 I?D7H
8;J7
J?:7A
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7:7B7>
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:;D=7D
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N7D=
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C;C7D97HA7D
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7D7A
N7D=
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H;7AI?
F;C7D97H7D

:?JKB?I
I;87=7?
8;H?AKJ  

2

  



Al*

0,834 MeV 27 12

Al*

Gambar 10.7



Pemancaran  dan peluruhan

27 12

Mg dan

dalam

27 13

Al .

 

3  ,

EDJE>
    (     , :;D=7D
,
7:7B7>
D;KJH?DE c. Pemancaran Sinar Gamma (  ) ;H8;:7
>7BDN7
:;D=7D
F;C7D97H7D
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:7D
I?D7H

N7D= C;C7D97HA7D
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8;HCK7J7D
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=;BEC87D=
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8;H7:7
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8?7I7DN7
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8;H?AKJ

Tugas Anda 10.1 Carilah informasi mengenai pemanfaatan sinar  ,  , dan

.

sumber radioaktif

 kertas

 aluminium



2    3     

timbal

EDJE>           %;I?CFKB7D
:7H?
A7H7AJ;H?IJ?A
I?D7H
H7:?E7AJ?<
N7?JK
I?D7H  
I?D7H  
:7D I?D7H

:?JKD@KAA7D
F7:7
#'(+2

7N7
J;C8KI
I?D7H
 
I?D7H   :7D
I?D7H
 C7I?D=C7I?D=
:?JKD@KAA7D
F7:7
'3('8
 

Sumber: Conceptual Physics, 1998

 

Gambar 10.8 Daya tembus partikel

 ,  , dan  .

#'(+2
 %7H7AJ;H?IJ?A
-?D7H
,7:?E7AJ?< +4/9
"/4'8 !'*/5'1:/, B<7
  

!+'19/
2'3/'. #DJ?
>;B?KC

B;AJHED ;J7
   !7CC7
   ,7:?7I?
;B;AJHEC7=D;J?A

;':'4 *EI?J?<

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(;=7J?< ,;D:7> .?:7A
8;HCK7J7D .?:7A
8;HC7II7

+3(/'9'4
*'2'3 +*'4
'-4+: %;9?B ;I7H .?:7A
:?8?7IA7D

+3'36;'4 '>'
#+3(;9 %;HJ7I
J?F?I %KB?J
C7DKI?7 BKC?D?KC

CC .?C87B
J?F?I

Sumber: Physics for ‘O’ Level, 1990

Fisika Inti dan Radioaktivitas

265

3. Pelemahan Intensitas Sinar Radioaktif $?A7
I?D7H
H7:?E7AJ?<
:?B;M7JA7D
F7:7
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J;HDN7J7
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F;HI7C77D

    .

X

Gambar 10.9 Pelemahan intensitas oleh bahan dengan ketebalan X.

 P

%;J;H7D=7D  
AE;7D
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 C  
8?B7D=7D
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C7A7

  +89'3''4
 ?
:7F7J
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C;D@7:?


   .





BD  BD 
P


Tokoh Marie Curie (1867–1934)


  


.

.


  . 
  .

I;>?D==7
:?F;HEB;> .

 

 P

"7H=7
.
F7:7
+89'3''4
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Contoh 10.3 Sumber: Science Encyclopedia, 1998

Marie Curie dilahirkan di Warsawa, Polandia. Nama aslinya adalah Marie Sklodowska. Ia belajar Matematika, Fisika, dan Kimia di Paris. Marie bersuamikan Pierre Curie seorang ahli fisika juga. Mereka bekerja sama dalam meneliti radiasi yang dihasilkan oleh bahan radioaktif dan mereka berhasil menemukan unsur thorium yang bersifat radioaktif. Pada 1898, mereka menemukan dua unsur radioaktif baru, yaitu polonium dan radium. Untuk keberhasilan ini, Marie dan Pierre Curie serta Henri Becquerel mendapat hadiah Nobel untuk bidang Fisika pada 1903. Pada 1910, Marie berhasil memisahkan radium murni dan mempelajari sifat-sifat kimianya. Untuk hasil kerja kerasnya ini, Marie kembali mendapatkan hadiah Nobel yang kedua dalam bidang kimia pada 1911. Marie meninggal tahun 1934 akibat kanker, yang mungkin disebabkan ia terlalu lama berinteraksi dengan bahanbahan radioaktif.

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Sumber: Konsep Fisika Modern, 1983

Gambar 10.13 Deret uranium

Fisika Inti dan Radioaktivitas

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7A7D C;D=?EDA7D
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:?
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:?J7H?A
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Sumber: Conceptual Physics, 1998

Gambar 10.15 Pencacah Geiger-Muller

jendela tipis gas kawat elektrode (anode) isolator tabung dari logam (katode)

– +

+ – 103 V

Sumber: Physics for Scientist and Engineers, 2000

Gambar 10.16 Bagian-bagian alat pencacah Geiger-Muller

sumber jejak ion radioaktif (R) keping gelas uap

cahaya

pengisap (P)

Gambar 10.17 Kamar kabut Wilson

Fisika Inti dan Radioaktivitas

271

partikel datang kristal sintilasi fotokatode foton +200 V +400 V +600 V +800 V +1000 V +1200 V +1400 V +1600 V

pulsa keluaran

fotomultiplier

Sumber: Physics for Scientist and Engineers, 2000

Gambar 10.18 Diagram sintilator yang dilengkapi dengan fotomultiplier

Tes Kompetensi Subbab

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272

Mudah dan Aktif Belajar Fisika untuk Kelas XII

Pada setiap reaksi inti selalu berlaku hukum kekekalan berikut ini. "KAKC
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 &?  ";

";

Tokoh



Enrico Fermi (1901–1954)

1. Energi yang Dihasilkan dari Reaksi Inti

Sumber: Science Encyclopedia, 1998

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Contoh 10.7

Fermi adalah seorang fisikawan kelahiran Itali. Setelah tahun 1939, ia pergi ke Amerika Serikat untuk bekerja dan hidup disana. Ia dikenal karena menggunakan neutron untuk menembak atom yang dapat menghasilkan isotop. Enrico Fermi mendapat hadiah Nobel pada 1938. Pada 2 Desember 1942, reaksi rantai nuklir pertama dihasilkan oleh sebuah tim ilmuwan yang dipimpinnya. Percobaan yang sukses ini bertempat di lapangan Squash Universitas Chicago, Amerika Serikat. Reaksi rantai bekerja menyerupai penyalaan korek api. Panas yang dihasilkan oleh gesekan pada waktu mematik korek api menyebabkan sebagian atom terbakar dan melepaskan lebih banyak panas sehingga menyebabkan reaksi terusmenerus berlangsung.

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Kata Kunci • • • • • •

energi reaksi inti reaksi eksotermik reaksi endotermik reaksi berantai reaksi fisi reaksi fusi

Fisika Inti dan Radioaktivitas

273

2. Pembuatan Isotop Radioaktif dari Reaksi Inti ,7:?E7AJ?<
:7F7J
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3. Reaksi Berantai *;D;C87A7D
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    D
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+

inti bulatan n U

Gambar 10.19 Diagram reaksi berantai

 #IEJEF
  2;
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7:7B7>
?IEJEF?IEJEF
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4. Reaksi Fisi reaksi fisi kedua menghasilkan 4 neutron inti U-235

neutron pertama

reaksi fisi U-235 memunculkan 2 neutron Sumber: Physics for ‘O’ Level, 1990

Gambar 10.20 Pada reaksi pembelahan inti, jumlah neutron bertambah pada setiap tingkatan.

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274

Mudah dan Aktif Belajar Fisika untuk Kelas XII

'E

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1 1H

1 1H

1 1H

1 1H







1 1H

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3 2 Hc





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Gambar 10.21 Penggabungan empat inti hidrogen membentuk inti helium disertai pelepasan energi.

5. Reaksi Fusi ")
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Gambar 10.22 Reaksi fusi di Matahari

medan magnetik toroid

ruang vakum toroid arus listrik luar medan magnetik total

medan magnetik poloid

arus plasma plasma

Sumber: Physics for Scientist and Engineers, 2000

Gambar 10.23 Tokamak

Fisika Inti dan Radioaktivitas

275

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D. Reaktor Nuklir, Bom Nuklir, dan Radioisotop

Informasi

1. Reaktor nuklir

untuk Anda

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sistem kedua air panas

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Beberapa jenis limbah reaktor nuklir bersifat radioaktif. Limbah ini menghasilkan radiasi nuklir mematikan yang merusak sel-sel hidup. Beberapa jenis limbah radioaktif dapat bertahan ribuan tahun. Oleh karena itu, limbah yang disimpan rapat dalam peti kemas tertutup harus dipendam di bawah tanah. Untuk menangani limbah nuklir maka orang harus menggunakan pakaian seperti terlihat pada gambar berikut.

Information for You Some kind of the nuclear reactor waste is radioactive material. This waste produce deadly nuclear radiation that damage living cells. Some kind of radioactive waste can last for thousand years. So, nuclear reactor waste placed in shielded box must be buried underground. Overcoming nuclear waste, people must wear suit like you have seen this picture.

generator

kondensor batang pengendali

pelindung

pompa

Gambar 10.24 Sumber: Physics for Scientist and Engineers, 2000

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Perancangan dasar reaktor Nuklir air-tekan.

Gambar 10.25 Reaktor nuklir dengan moderator air

Fisika Inti dan Radioaktivitas

277

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Gambar 10.26

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(6)

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Komponen-komponen dasar sebuah reaktor atom: (1) teras reaktor; (2) elemen bahan bakar; (3) moderator; (4) batang kendali; (5) shielding; (6) pendingin.

aliran dingin

pompa

Informasi untuk Anda Sebuah reaktor nuklir tidak dapat meledak seperti bom atom. Akan tetapi, suatu kecelakaan pada pusat tenaga nuklir dapat menimbulkan akibat yang merusak sejumlah besar wilayah sekitarnya. Pada 1986, terjadi kecelakaan nuklir terbesar di pusat tenaga nuklir Chernobyl di Ukraina. Kecelakaan ini melepaskan zat radioaktif ke dalam atmosfer dan mengakibatkan gangguan kesehatan yang hebat pada penduduk di sekitar Chernobyl. Penduduk di Eropa pun mendapatkan dampak yang tidak kecil.

Information for You A nuclear reactor can not explode like atom bomb. But, an accident in centre of the nuclear power can cause a large destruction. In 1986, the biggest nuclear accident happened in the Chernobyl nuclear power plant, Ukraine. This accident released radioactive material to the atmosphere and caused severe health disorder to civilians near by some European also got the serious effect.

278

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Sumber: Physics for ‘O’ Level, 1990

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Fisika Inti dan Radioaktivitas

279

Tantangan untuk Anda Mengapa molekul dari uranium (U–235) lebih cepat bergerak daripada molekul U–238 pada suhu yang sama?

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Gambar 10.28 Bagan PLTN dengan sistem BWR pompa

kondensator Sumber: Conceptual Physics, 1998

2. Bom Nuklir

Kata Kunci • • • • • •

reaktor nuklir reaktor daya reaktor produksi isotop reaktor penelitian sistem PWR sistem BWR

Tugas Anda 10.2 Apakah Anda pernah mendengar tragedi Chernobyl? Chernobyl merupakan pembangkit listrik tenaga nuklir di Ukraina yang mengalami kebocoran. Carilah informasi dari berbagai sumber, mengapa kebocoran ini bisa terjadi?

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Informasi untuk Anda

Kedua tangan ini memegang sebongkah uranium–235. Isotop dari uranium yang digunakan sebagai bahan dalam reaktor nuklir. Uranium merupakan salah satu logam berat, sebongkah kecil uranium ini massanya 4,5 kg dan harganya lebih dari 200.000 dolar. Sebelum memegang uranium, tangan harus dilindungi dengan sarung tangan khusus.

Information for You This two hand hold a slab of Uranium-235. It is uranium isotop used as material in nuclear reactor. Uranium is one of the heavy metal, that small uranium slab weights 4,5 kg and the price is more than 200.000 dollars. Before holding uranium, hands must be covered by a special handgloves.

Tugas Anda 10.3 Carilah foto-foto tentang kondisi kota dan warga Hiroshima dan Nagasaki setelah pengeboman oleh Amerika Serikat. Diskusikanlah bersama teman sekelas Anda tentang dampak negatif bom atom. Diskusikan juga apakah ada dampak positif dari bom atom.

Fisika Inti dan Radioaktivitas

281

Gambar 10.29 Magnetic Resonance Imaging

Sumber: Physics for Scientist and Engineers, 2000

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Gambar 10.30 Karbon dapat digunakan untuk menentukan usia sebuah gulungan perkamen yang ditemukan di sebuah gua di Laut Mati.

Sumber: Fundamental of Physics, 2001

Mari Mencari Tahu Tantangan untuk Anda Apakah radioisotop yang disuntikkan ke dalam tubuh pasien untuk mendiagnosis penyakit, membahayakan pasien itu sendiri?

282

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Refleksi Setelah mempelajari bab ini, tentu Anda telah memahami bagian-bagian penyusunan inti dan bagaimana pengaruh rekayasa penyusun inti tersebut. Anda juga tentu telah memahami mengenai radioaktivitas dan peluruhan serta perhitunganperhitungan yang ada di dalamnya. Dari keseluruhan materi yang ada pada bab ini, bagian mana yang menurut

284

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Anda sulit dipahami? Coba Anda diskusikan dengan teman atau guru Fisika Anda. Banyak sekali manfaat mempelajari Fisika Inti dan Radioaktivitas. Salah satunya adalah dibangunnya PLTN yang hemat bahan bakar. Coba Anda cari manfaat lain mempelajari bab ini. Diskusikan hasilnya bersama temanteman Anda.

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23

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Indeks

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Daftar Pustaka Allonso, M. dan Finn. 1980. Fundamental Physics, Vol 1and 2. New York: Addision-Wesley Publishing Company Inc. Beiser, Arthur. 1983. Konsep Fisika Modern, Edisi Ketiga. Jakarta Pusat: Erlangga. Biryam, M. 1992. Hukum-Hukum Kekekalan dalam Mekanika. Jakarta: Departemen Pendidikan dan Kebudayaan. Bueche, Fredrick. 1982. Introduction to Physics for Scientist and Insights. New York: Grow Hill Book Company Inc. Dorling Kindsley. 1995. Jendela IPTEK, seri 1–4. Jakarta: Balai Pustaka. Giancoli, Douglas C. 2000. Physics, 3th Edition. USA: PrenticeHall International. Grolier International Inc. 1995. Oxford Ensiklopedi Pelajar. Jakarta: Widyadara. Haliday, D, R. Resnick, J. Waker. 2001. Fundamental of Physics Sixth Edition. USA: John Willey and Sons Inc. Harsanto. 1980. Motor Bakar. Jakarta: Djambatan. Hermawan Edi. 2004. Radar Atmosfer Khatulistiwa. Surakarta: Pabelan. Hewwit, Paul G. 1993. Conceptual Physics, 6th Edition. USA: Harper Collins College Publisher. Hewwit, Paul G. 1998. Conceptual Physics, 8th Edition. USA: Addison-Wesley Publishing Company Inc. Jones, Edwin R. dan Richard L. Childers. 1993. Contemporary College Physics. USA: Addison-Wesley Publishing Company, Inc. Sears, F.W. et.al. 1993. University Physics. USA: Addison-Wesley Publishing Company Inc. Tipler, Paul. 1998. Fisika untuk Sains dan Teknik, Jilid 1 (alih bahasa: Prasetyo dan Rahmad W. Adi). Jakarta: Erlangga. Tipler, Paul. 2001. Fisika untuk Sains dan Teknik, Jilid 2 (alih bahasa: Bambang Soegijono). Jakarta: Erlangga. Warlijo, Liek. 2003. Kamus Fisika, Cetakan Kedua. Jakarta: Balai Pustaka.

306

Mudah dan Aktif Belajar Fisika untuk Kelas XII

ISBN 978-979-068-816-2 (No. Jilid lengkap) ISBN 978-979-068-931-2

Harga Eceran Tertinggi (HET) Rp.19.502,-