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Bab 2 kls xii
- 1. Cahaya merupakan gelombang elektromagnetik yang banyak digunakan untuk kepentingan teknologi komunikasi. A. Interferensi Cahaya B. Difraksi Cahaya C. Polarisasi Cahaya Gelombang Cahaya 31 Bab 2 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?=51 C5C17 3181G1 B5A97 41 :D=@19 411= ;5894D@1 B581A981A9 %5C9;1 CDAD 8D:1 @141 B917 81A9 ;1417 ;1417 ;9C1 41@1C =59;=1C9 9418G1 @5179 G17 =57891B9 35A18G1 179C @1;18 B5251AG1 @5179 9CD 181G1 =5AD@1;1 75?=217 55;CA?=175C9; *141 12
- 2. 41 C518 =5=@51:1A9 75?=217 =5;19; =9B1G1 75?=217 19A 41 75?=217 C19 @1;18 @5A25411 1C1A1 75?=217 55;CA?=175C9; 41 75?=217 =5;19; '1DB91 C518 =5=1611C;1 75?=217 55;CA?=175C9; 411= 29417 CA1B@?AC1B9 1BCA??=9 =99C5A 41 75?7A169 ,18D;18 41 21719=11 75:11 41 39A939A9 75?=217 55;CA?=175C9; ;8DBDBG1 3181G1B5AC1@55A1@1G1$1F1211C1B@5AC1G11@5AC1G11C5AB52DC 41@1C 41 C5=D;1 @141 212 99 )58 ;1A51 9CD @51:1A9 212 99 4571 219; Sumber: Physics Today, 1995
- 3. Tes Kompetensi Awal !((.6//(/2(.%,%3+-104(2(.1/%0)%*%8%-(3,%-%0.%*41%.41%.(3+-65'%.%/6-6.%5+*%0
- 4. @1;18G1749=1;BD4457175?=21755;CA? =175C9; @1;183181G1C5A=1BD;75?=21755;CA?=175C9; C1 32 Mudah dan Aktif Belajar Fisika untuk Kelas XII A B C gelombang datang S2 S1 terang gelap terang gelap terang terang terang terang terang terang terang terang gelap gelap gelap gelap gelap gelap gelap gelap terang gelap terang gelap terang S0 Gambar 2.1 Interferensi pada celah ganda Young A. Interferensi Cahaya 14118 !! !# 1AC9G1 75?=217 =5=99;9 ! 41 # +5@5AC9 G17 C518 492181B @141 @5=2181B1 75?=217 218F1 75?=217 3181G1 25AB961C B5@5AC9 81G1 75?=217 2DG9 G19CD 41@1C 25A9C5A65A5B9 )58 ;1A51 9CD DCD; =541@1C;1 9C5A65A5B9 3181G1@D49@5AD;1BD=25A3181G1G17;?85A5G19CDBD=25A3181G1 G17 =5=99;9 6A5;D5B9 B1=1 41 2541 61B5 C5C1@ +D=25A 3181G1 G17 ;?85A5 41@1C 491=1C9 =51D9 @5A3?211 G17 491;D;1 ?58 $160) 41 3(40(.. 1. Percobaan Young dan Fresnell a. Percobaan Celah Ganda Young *5A3?21199491;D;1?58$160)4571=577D1;14D1@57 8117 *578117 @5AC1=1 =5=99;9 B1CD D217 ;539 41 @578117 ;54D1 4957;1@9 4571 4D1 D217 ;539 *5A81C9;1 %/%3 +91A=??;A?=1C9B49@5A?5841A91=@DB521719BD=25A3181G1G17 =5=131A =51D9 3518 + %5=D491 B91A 41A9 3518 + 49@131A;1 ;5 @578117 ;54D1 D1 3518 @141 @578117 ;54D1 G19CD +
- 5. 41 + G17 49@1B17 B5:1:1A 4571 + 1;1 25A6D7B9 B521719 @5=131A B91A B91A;?85A5 %54D125A;1B41A935183518+
- 6. 41+ 9925A9C5A65A5B9 @141 1G1A 1B9 9C5A65A5B9 25AD@1 71A9B C5A17 41 71A9B 751@ b. Percobaan Fresnell 571=577D1;1B52D18BD=25A3181G1+3(40(..=5=@5A?58 4D1 BD=25A 3181G1 +
- 7. 41 + G17 ;?85A5 41A9 81B9 @5=1CD1 4D1 35A=9 *5A81C9;1 %/%3 Gambar 2.2 Percobaan Fresnell untuk menunjukkan interferensi cahaya. S1 O P C2 S2 S Tokoh Augustine Fresnell (1788–1827) 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 bermacam-macam 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. Sumber: Science Encyclopedia, 1998 5A1@1;18;535@1C13181G1411=AD1781=@1 +52DC;1B961CB961C75?=2173181G1
- 8. 3 7,2 10 m 3,6 10 3m y Dari soal diketahui: m = 1; d = 10–5 m; D = 0,2 m sehingga yd 3 5 3,6 10 m 10 m 0,2m 1 3 5 3,6 10 m 10 m 0,2m 1 Gelombang Cahaya 33 S1 r1 B C P r2 S2 d y D b r2 S2 r1 S1 b 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. *141%/%3 C5AB52DC41@1C49981C218F1+14118BD=25AB91A =??;A?=1C9B +
- 9. 41 + 14118 21G171 41A9 + ?58 35A=9
- 10. 41 571 45=9;91 B91AB91A G17 41C17 @141 1G1A B5?18?18 25A1B1 41A9 +
- 11. 41 + !5?=217 3181G1 41A9 +
- 12. 41 + 99 1;1 B197 25A9C5A65A5B9@1411G1A4181B9G125A71CD7@141B59B9841A99C1B1 ;54D1 B91A 9CD *5AD 41 ;5C18D9 :9;1 ;54D1 BD=25A 3181G1 =5=99;9 1=@9CD4? G17 B1=1 @141 C5=@1C C5A:149G1 9C565A5B9 =99=D= 1;1 C5A25CD; 71A9B 751@ +5219;G1 :9;1 1=@9CD4? C941; B1=1 9C5A65A5B9 =99=D=G1 C941; 751@ B1=1 B5;19 1 2 L *5A81C9;1 %/%3 75?=217 3181G1 41C17 =5D:D 3518
- 13. 41 3518 G17 C5A5C1; @141 29417 181G1 C5AB52DC C5A496A1;B9 ?58 ;54D1 3518 C5AB52DC 41 =5781B9;1 @?1 9C5A65A5B9 @141 1G1A %9C1 41@1C =55CD;1 49 =11 B5C91@ @9C1 C5A17 1C1D @9C1 751@ C5A5C1; @141 1G1A 4571 =5=25A9;1 BD4DC 41A9 BD=2D C5718 C5A8141@ 71A9B 751@ 1C1D 71A9B C5A17 C5AB52DC -CD; =55CD;1 25B1AG1 ;9C1 81ADB=578D2D7;1G1 4571 *5A81C9;1 %/%3 ,9C9; =5AD@1;1 B52D18 C9C9; G17 C5A5C1; @141 B545=9;91 AD@1 B589771 :1A1; 2 ;5 * B1=1 4571 ;5 * 4571 45=9;91 B1=1 4571 :1A1; 41A9
- 14. ;5 D2D71 1C1A1 :1A1; 41A9
- 15. ;5 41 99 B171C AD=9C (1=D ;9C1 41@1C =5G545A811 ;1G1 4571 =571771@ 218F1 BDBD1 :1A1; 3518 C5A8141@ 1G1A :1D8529825B1A41A9@141:1A1;1C1A1;54D13518 1A971=21A C5A981C 218F1 B91A 75?=217
- 16. 41 14118 B5:1:1A 41 =5=25CD; BD4DC C5A8141@ BD=25A @DB1C *5A81C9;1 %/%3 ,5AG1C1 BD4DC
- 17. G17 4925CD; ?58 BD=2D @DB1C 41 BD=2D
- 18. 571 45=9;91 ;9C1 41@1C 4571 =D418 =55CD;1 218F1 B9 B9 K
- 19. (34%/%%0 9 =5AD@1;1 @5AB1=11 DCD; =55CD;1 :1A1; C5=@D8 1C1A1 B91A
- 20. 41 C5A8141@ * -CD; 9C5A65A5B9 =1;B9=D= 9C5A65A5B9 ;?BCAD;C96 C518 49;5C18D9 218F1 @1BC9 ? 1C1D 29171 751@ 41A9 @1:17 75?=217 (34%/%%0 9 41@1C 49CD9B B521719 25A9;DC Pembahasan Soal 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 2 1 2 yd m D 1 2 D m 2 2 7 3,6 10 m = 360 nm Jawaban: c
- 21. 34 Mudah dan Aktif Belajar Fisika untuk Kelas XII B9 K 4571
- 22. -CD;C5A17@DB1C;9C1=5=25A9;1 @9C1 C5A17 @5AC1=1 41 B5C5ADBG1 41@D DCD; 9C5A65A5B9 =99=D= 9C5A65A5B9 45BCAD;C96 @1BC92917171:9;19B5C571875?=217(34%/%%0 941@1C 49CD9B;1 B521719 25A9;DC
- 23. B9
- 24. K 4571
- 25. -CD; 751@ @5AC1=1 ;9C1 =5=25A9;1 @9C1 751@ ;54D1
- 26. 41 B5C5ADBG1 2. Menentukan Jarak Pita Terang ke-m atau Pita Gelap ke-m dari Terang Pusat *141 @5=2181B1 B525D=G1 C518 49B52DC;1 218F1 @?1 9C5A65A5B9 @141 1G1A 25AD@1 @9C1 C5A17 41 @9C1 751@ *5A81C9;1 %/%3 %9C1 41@1C =55CD;1 ;54D4D;1 @9C1 C5A17 ;5 1C1D @9C1 751@ ;5 @141 1G1A *5A81C9;1 ;5=219 %/%3 % )58 ;1A51 :1D8 5298 25B1A 41A9@141 BD4DC 25A919 B171C ;539 -CD; BD4DC G17 B171C ;539 1;1 25A1;D B9 C1 1A9 %/%3 % ;9C1 41@1C =55CD;1 218F1 B9 C1 # K -CD; @9C1 C5A17 =1BD;;1 (34%/%%0 9 ;5 (34%/%%0 9 B589771 49@5A?58 # K -CD; @9C1 751@ =1BD;;1 (34%/%%0 9 ;5 (34%/%%0 9 B589771 49@5A?58
- 27. #
- 28. K %5C5A171 :1A1; 1C1A3518 @141 1G1A # :1A1; C5A17 751@ ;5 41A9 @DB1C :1A1; 1G1A ;5 3518 @1:17 75?=217 3181G1 11=8199
- 29. 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.
- 30. P D i A B C d Gelombang Cahaya 35 Contoh 2.1 -CD;=55CD;1@1:1775?=217B91AG1749@131A;1?581=@D@9:1A 1CA9D=B91A99495F1C;1@1414D13518G1725A:1A1;== *141:1A1;
- 31. =5C5A 41A9351849@1B171G1A $9;181B99C5A65A5B9@1411G1A49@5A?58:1A1;71A9BC5A17 @DB1CB1=@194571;59=114118==25A1@1;18@1:1775?=217B91A1CA9D= C5AB52DC %7% 9;5C18D9 ==L
- 32. K= # ==L
- 33. K=
- 34. = k #
- 35. =
- 36. =
- 37. L
- 38. K I $149@1:1775?=217B91A1CA9D=14118 I Contoh 2.2 *141B52D18@5A3?2119C5A65A5B9497D1;14D13518B5=@9C $1A1;1C1A1;54D1 35189CD ==41495C1;;1@141:1A1; =;51G1A71A9B751@@5AC1=141A9@DB1C G17:1A1;G1== 9CD718@1:1775?=2173181G1G1 %7% == L
- 39. K=
- 40. = # ==L
- 42. =K
- 43. =
- 44. =
- 45. K
- 46. = L
- 47. K=
- 48. L
- 49. K= I $149@1:1775?=2173181G1G114118 I 3. Interferensi oleh Lapisan Tipis *5=1CD1 3181G1 =1C181A9 ?58 ! # G17 4931=@DA 4571 19A 1;1 =5=@5A981C;1 71A9B71A9B 25AF1A1 @141 =9G1; +@5;CAD= F1A1 99 =5=@5A981C;1 141G1 ?58 1@9B1=9G1;G17C9@9B9CD #C5A65A5B9C5AB52DC41@1C25AD@19C5A65A5B9 =1;B9=D==1D@D9C5A65A5B9=99=D= #C5A65A5B91C1A175?=217 G17 49@1CD;1 ?58 1@9B1 C9@9B 49CD:D;;1 @141 %/%3 +525A;1B B91A 41C17 =57519 1@9B1 C9@9B 4571 BD4DC 41C17 1;1 49291B;1 41 B521791 179 49@1CD;1 ;5=219 ;5 @5A=D;11 +91A G17 49@1CD;1 495F1C;1 @141 B52D18 5B1 @?B9C96 41 496?;DB;149C9C9;* 5A;1B3181G149C9C9;*=5AD@1;181B99C5A65A5B9 25A;1B 3181G1
- 50. 41 4571
- 51. 14118 25A;1B 3181G1 G17 49@1CD;1 17BD7 41 14118 25A;1B 3181G1 G17 =5711=9 @5=291B1 C5A5298 418DD ;5=D491 49@1CD;1 Gambar 2.4 Interferensi oleh lapisan tipis. n r lensa lapisan tipis (1) (2)
- 52. +59B98 9C1B1 ?@C9; G17 49C5=@D8 ?58 B91A 41C17 89771 =5:149 B91A @1CD ;5
- 53. 41 B91A @1CD ;5 14118 K
- 54. K K 4571 14118 945;B 291B 1@9B1 C9@9B 41 '9B1;1 C521 1@9B1 14118 =1;1 3?B B589771 3?B 41B94571 C1B589771 3?B Sumber: www.designprodygzone.com Sumber: www.funsci.com Gambar 2.6 Interferensi oleh lapisan busa sabun yang tipis. Sumber: www.instckphoto.com 36 Mudah dan Aktif Belajar Fisika untuk Kelas XII K C1B9 3?B K B9 B9 3?B 571 =577D1;1 6-6/ !0(..+64 C5C17 @5=291B1 3181G1 G1;9 B9 B9 49@5A?58 B59B98 :1A1; C5=@D8 ;54D1 B91A =5:149 3?B K B9 3?B 3?B
- 55. KB9 3?B 3?B 3?B K +D@1G1 C5A:149 9C5A65A5B9=1;B9=D=49 C9C9;* 81ADB=5AD@1 ;1 ;59@1C1 41A9 @1:17 75?=217 ;1 C5C1@9 B91A @1CD 49 =5711=9 @5AD2181 61B5
- 56. =1;1 1;1=5:149 1C1D
- 58. K 4571
- 59. (34%/%%0 9 25A1;D DCD; 945;B 291B 1@9B1 C9@9B 5298 25B1A 41A9
- 60. 1C1D
- 62. =1;1
- 67. #C5A65A5B9 =99=D= 411= 1A18 @1CD =5=5D89 @5AB1=11 3?B41 1C1D 3?B K
- 68. 4571 945;B 291B 41=5=5D89BG1A1C
- 69. 41 1C1D
- 70. 41 Gambar 2.7 Interferensi oleh lapisan minyak yang tipis. ,5CD;1C5211@9B1=99=D=G17492DCD8;1BD@1G1C5A:1499C5A65A5B9@141B52D18 1@9B1C9@9BG17=5=99;9945;B291B 4571@1:1775?=217 I Contoh 2.3 Gambar 2.5 Interferensi oleh busa sabun.
- 71. 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. Gelombang Cahaya 37 %7% #C5A65A5B9=1;B9=D=@1411@9B1C9@9B=5=5D89@5AB1=11 3?B
- 73. 413?B
- 74. B58977149@5A?58
- 76. I
- 77. I I $149C5211@9B1C9@9BG17492DCD8;114118 I Contoh 2.4 ,5CD;1@1:1775?=217B91AG17497D1;1:9;1C5A:1499C5A65A5B9=99=D= ?A45;54D141A91@9B1C9@9B49D41A14571;5C5211
- 80. =3?BJ
- 82. = Kata Kunci • interferensi • sinar monokromatis • interferensi maksimum • interferensi minimum Tes Kompetensi Subbab A (3,%-%0.%*'%.%/6-6.%5+*%0
- 83. @1G1749=1;BD49C5A65B93181G1 +52D183518714125A:1A1;==49251;173518 41@141:1A1; =5C5A495C1;;1B52D181G1A 518 49B91A9 4D1 B91A =??;A?=1C9B 4571 @1:17 75?=217=41= 5A1@1;18:1A1;71A9B C5A17?A45;55=@1C;54D1B91A@1411G1A +525A;1B3181G1=55F1C94D13518B5=@9CG17B1CD B1=11925A:1A1;== $1A1;3518;51G1A
- 85. L
- 86. K 3= 5A1@1;18@1:1775?=2173181G1 G17497D1;1 -CD; =57D;DA @1:17 B91A =5A18 491;D;1 @5A3?211B52171925A9;DC +91A 29AD 4571 @1:17 75?=217 = 49:1CD8;1 C571; DADB @141 3518 7141 *?1 9C5A65A5B9C5A:149@1411G1AG1725A:1A1; =41A9 3518 !1A9BC5A17?A45@5AC1=125A:1A1;==41A9 71A9BC5A17@DB1C +5C5189CDB91A=5A1849:1CD8;1 @1413518 ,5AG1C171A9BC5A17?A45@5AC1=125A:1A1; ==41A971A9BC5A17@DB1C ,5CD;118@1:17 75?=217B91A=5A189CD +52D181@9B1C9@9B=5=99;9945;B291B 497D1;1 DCD; =5981C 75:11 9C5A65A5B9 $9;1 @1:17 75?=217G1 I C5CD;1 C521 =99=D= 1@9B1C5AB52DCBD@1G1C5A:1499C5A65A5B9 B. Difraksi Cahaya *141@51:1A175C1A14175?=21749%51B/C518492181B218F1 75?=217 19A G17 =55F1C9 B52D18 @578117 4571 B52D18 3518 B5=@9C1;1=5711=95CDA1 !5?=217G1741C1741@1C25A25?; B5C518 =51D9 3518 C5AB52DC *5=25?;1 75?=217 G17 49B5212;1 ?58 141G1 @578117 25AD@1 3518 49B52DC 496A1;B9 75?=217 +1=1 81G1 4571 75?=217 3181G1 G17 495F1C;1 @141 B52D18 3518 B5=@9C :D71 1;1 =5711=9 5CDA1
- 87. 96A1;B93181G1C5A:149:D71@1413518B5=@9CG17C5A@9B18B5:1:1A B1CD B1=1 19 @141 :1A1; G17 B1=1 518 B5=@9C G17 45=9;91 49B52DC %9B9 14118 ;5@971 ;131 G17 497?A5B =5DADC 71A9B B5:1:1A 41 21G1; :D=18G1 $1A1; 1C1A1 4D1 3518 49B52DC C5C1@1 ;9B9 1. Difraksi Celah Tunggal 96A1;B9 @141 3518 CD771 1;1 =5781B9;1 @?1 71A9B C5A17 41 751@ @141 1G1A 518 CD771 41@1C 491771@ C5A49A9 1C1B 2525A1@1 3518 B5=@9C G17 4921C1B9 C9C9;C9C9; 41 B5C91@ 3518 9CD =5AD@1;1 BD=25A 3181G1 B589771 B1CD B1=1 19G1 41@1C 25A9C5A65A5B9 *5A81C9;1 %/%3 -CD; =57119B9B @?1 496A1;B9 3518 @141 %/%3 492179 4D1 21791 *5A81C9;1 75?=217
- 88. 41 !5?=217
- 89. =55=@D8 9C1B1 G17 5298 :1D8 B525B1A B9 d 1 2 d 2 maksimum utama Contoh 2.5 d 5 4 3 2 d sin 2 38 Mudah dan Aktif Belajar Fisika untuk Kelas XII 41A9@141 75?=217 +1=1 81G1 4571 75?=217 41 G17 =5=99;9 2541 9C1B1 B525B1A B9 #C5A65A5B9 =99=D= G17 =5781B9;1 71A9B 751@ C5A:149 :9;1 ;54D1 75?=217 25A2541 61B5
- 90. J 1C1D 2541 9C1B1G1 B1=1 4571
- 91. @1:17 75?=217 B9 B9 $9;1 3518 492179 5=@1C 21791 4941@1C 71A9B 751@ ;5C9;1 B9 B9 1 B5AD@1 4571 9CD :9;1 C518 492179 51= 21791 4941@1C 71A9B 751@ ;5C9;1 B9 B9 +531A1 D=D= 41@1C 49G1C1;1 218F1 @9C1 751@ ;5 C5A:149 :9;1 K
- 93. B9 %5C5A171 521A 3518 BD4DC B9=@17 45E91B9
- 94. -CD; 1C1D C5A:149 =1;B9=D= DC1=1 @9C1 C5A17 C5718 B5@5AC9 49@5A981C;1 @141 %/%3 Gambar 2.9 Maksimum utama terjadi untuk k = 0 atau = 0. 571=577D1;1@5781173518CD771@1411G1AC1=@1;@?1496A1;B9 71A9BC5A17@DB1C4171A9B751@;55=@1C=5=25CD;BD4DCJC5A8141@71A9B?A=1 $9;13181G1G17497D1;1=5=99;9@1:1775?=217 IC5CD;1521A 3518G17497D1;1 Gambar 2.8 Difraksi cahaya pada celah tunggal.
- 95. dp B B1 A d C D E F G Gelombang Cahaya 39 garis gelap terang %7% 571=577D1;1(34%/%%0 949@5A?58 B9 B9J I=1;1 d P Q pusat
- 96. I $149521A3518G114118 I -CD;=541@1C;1@?1496A1;B9=1;B9=D=25419C1B141A99C5A 65A5B9=99=D=81ADB49;DA1794571
- 97. )58;1A51;54D13181G1 B561B5 2541 61B5 ;54D1G1 =5:149 J D1 75?=217 4571 2541 61B5
- 98. 1C1D2541BD4DC61B5J49B52DC:D71B561B5 *5AB1=119C5A65A5B9 =1;B9=D= 41A9 @?1 496A1;B9G1 1;1 =5:149 B9
- 99. B9
- 100. 1C1D B9
- 101. K
- 102. K
- 103. 141182917171:9
- 104. ,5CD;1521A3518=99=D=G17492DCD8;1@141496A1;B93518CD771:9;1 49979;1BD4DC496A1;B9G1J41@1:1775?=217G17497D1;1= DCD;@?1496A1;B9=1;B9=D= %7% 571=577D1;1(34%/%%0 9 49@5A?58 B9
- 107. = = = = B9
- 108. $149521A3518=99=D=14118 = Pembahasan Soal 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 m 3 3 0,4 10 m 0,5 10 m 1 0,4 m = 5,6 × 10–7 m Jawaban: d Contoh 2.6 2. Difraksi pada Kisi 96A1;B9 3181G1 C5A:149 @D1 @141 3181G1 G17 =51D9 21G1; 3518 B5=@9C 4571 :1A1; 3518 B1=1 518 B5=@9C G17 45=9;91 49B52DC ;9B9 496A1;B9 1C1D 49B97;1C ;9B9 +5=1;9 21G1; 3518 @141 B52D18 ;9B9 B5=1;9 C1:1= @?1 496A1;B9 G17 4981B9;1 @141 1G1A '9B1G1 @141 415A18 B5521A 3= C5A41@1C 3518 AC9G1 ;9B9 C5AB52DC C5A49A9 1C1B 3518 3= 1C1D 3518 3= 571 45=9;91 :1A1; Gambar 2.10 Difraksi pada kisi T M P K 1C1A3518 14118
- 109. 3= L
- 110. K 3= *5A81C9;1 %/%3
- 111. intensitas m = 1 m = 2 2 1 1 3 1 2 3 dsin Gambar 2.12 Difraksi minimum kedua untuk N banyak celah. biru spektrum orde ke 2 merah biru biru merah spektrum orde ke-1 spektrum orde ke-0 (putih) spektrum orde ke-1 spektrum orde ke-2 Gambar 2.13 merah merah biru Difraksi cahaya putih akan menghasilkan pola berupa pita-pita spektrum. cahaya putih kisi difraksi %/%3 =5=@5A981C;1 B525A;1B B91A =??;A?=1C9B G17 495F1C;1 @141 B52D18 ;9B9 41 =5781B9;1 @?1 496A1;B9 @141 1G1A *?1 496A1;B9 25AD@1 71A9B C5A17 41 71A9B 751@ B531A1 25A71C91 96A1;B9=1;B9=D=C5A:149:9;1@1411G1AC1=@1;71A9B71A9BC5A17 541 9C1B1 G17 495F1C9 3181G1 G17 41C17 41A9 4D1 3518 25A45;1C1 14118
- 112. 1C1D2917131318;19@1:1775?=217G1 *?1 496A1;B9 =1;B9=D= DC1=1 @141 ;9B9 14118 B9 49=1114118?A45496A1;B94114118:1A1;1C1A35181C1DC5C1@1 ;9B9 96A1;B9=99=D=491C1A1 =1;B9=D=C5A:149:9;1@1411G1AC1=@1; 71A9B71A9B 751@ 41 =99=D= @5AC1=1 B5BD418 =1;B9=D= ;5 C5A:149 :9;1 B9
- 113. :D71=99=D=;54D1 :9;1 B9 +52D18;9B94571 71A9B 3=495F1C;13181G1C571;DADB4571@1:17 75?=217 !1A9BC5A17496A1;B9?A45@5AC1=1=5=25CD;BD4DCJC5A8141@71A9B ?A=1=1;B9=D=DC1=1 ,5CD;118@1:1775?=217 %7% 9;5C18D9
- 114. 71A9B 3= L
- 115. K3=B9J
- 116. Mari Mencari Tahu 40 Mudah dan Aktif Belajar Fisika untuk Kelas XII
- 118. K 3=
- 119. L
- 120. K3=
- 122. I +5217193?C?8DCD; 351849@5A?58%/%3 B5417 ;1 DCD; 21G1; 3518 49@5A?58 %/%3 %/%3 =5=@5A981C;1 3181G1 @?9;A?=1C9; @141 3518 G17 =5AD@1;1 3181G1 @DC98 +91A @DC98 @?9;A?=1C9; C5A49A9 1C1B 25A21719 F1A1 4571 @1:17 75?=217 C5A;539 F1A1 D7D 41 C5A25B1A F1A1 =5A18 571 45=9;1 F1A1 G17 C5A45;1C 4571 14118F1A1D7D41G17C5A:1D814118F1A1=5A18G17=5AD@1;1 B@5;CAD= F1A1 57;1@ G19CD D7D 29AD 89:1D ;D97 :9771 41 =5A18 +5C91@ ?A45 496A1;B9 =5D:D;;1 B@5;CAD= F1A1 Contoh 2.7 *5A18;1841=5981C@5179*5179=5AD@1;1B118B1CD75:1111=B521719 81B9 496A1;B9 ,D71B 41 ;D=@D;1 96?A=1B9 =57519 C5A:149G1 @5179 ;5=D491@A5B5C1B9;1CD9B1414945@1;51B Gambar 2.11 Difraksi minimum kedua untuk N = 2 celah. m = –1 m = 0 m = 1 m = 2 0 dsin 2 Tantangan untuk Anda Tentukan daya urai sebuah celah dengan diameter 2 mm, jarak celah ke layar 1 meter dengan panjang gelombang cahayanya 590 nm.
- 123. celah bulat pola difraksi Informasi untuk Anda Information for You Gelombang Cahaya 41 3. Daya Urai Optik # ! 14118 ;5=1=@D1 B52D18 5B1 DCD; =5=9B18;1 21G17141A94D1C9C9;G17C5A@9B18B1CDB1=119@141:1A1;=99=D= %5=1=@D1 @5A25B1A1 11C11C ?@C9; =9B1G1 D@ =9;A?B;?@ 41 C5A?@?74921C1B9?5841G1DA195B141:D714921C1B9?58@?1496A1;B9 G17 C5A25CD; @141 21G171 2541 9CD %/%3 =5=@5A981C;1 218F1 21G171 G17 C5A:149 =5 AD@1;1 @?1 496A1;B9 G17 49B5212;1 ?58 ! D1B B9BC5= 5B1 ?58 11C11C ?@C9; C5AB52DC *?1496A1;B9G174925CD;?58B52D1835182D1CC5A49A91C1B25CD; C5A17 @DB1C G17 49;59979 3939 C5A17 41 751@ +118 B5?A17 9=DF1 13' %8.(+)* =5G9=@D;1 218F1 4D1 2D18 C9C9; BD=25A G17 C5A17G1 B1=1 1;1 C5A981C C5A@9B18 :9;1 =1;B9=D= B5CA1 1C1D @DB1C@DB1C 41A9 @?1 496A1;B9 G17 B1CD 25AC5@1C1 5C1;G1 4571 =99=D= @5AC1=1 41A9 @?1 496A1;B9 C9C9; G17 19 *?1 C5AB52DC 41@1C 49:51B;1 4571 =577D1;1 %/%3
- 124. $1A9:1A9 97;1A1 C5A17 G17 C5A25CD; 41@1C 491AC9;1 41G1 @9B18 @?1 496A1;B9 G17 C5A21C1B $9;1 3181G1 =51D9 AD17 81=@1 1C1D D41A141G1DA1941A9351897;1A141@1C49C5CD;14571@5AB1=11
- 125. K
- 126. '5DADC %8.(+)*41(%04;A9C5A91:1A1;1C1A1;54D1=1;B9=D= C5AB52DC @197 ;539 B1=1 4571 :1A9:1A9 97;1A1 C5A17 '1;B9=D= G17 ;54D1 :1CD8 @141 =99=D= G17 @5AC1=1 1C1D :1A1; BD4DC 1C1A1 ;54D1 @DB1C 21G171 G19CD B9
- 127. K
- 128. -CD; BD4DC G17 ;539
- 129. K
- 130. %5C5A171 41G1 DA19 = :1A1; 2541 41A9 5B1 = @1:17 75?=217 3181G1 = 491=5C5A 2D;11 5B1 = BD4DC 45E91B9 Gambar 2.14 Bayangan dari optik fisis dua benda yang berdekatan karena (a) beririsan dan (b) terpisah dengan baik. D Gambar 2.15 Lukisan sinar dari sumber cahaya dari sebuah celah bulat. r cahaya sumber datang 2 Contoh 2.8 %5C9;1491=5C5A=1C149@5A25B1AB1=@19==25A1@1:1A1;=99=D=1C1A14D1 BD=25AC9C9;G17=1B9841@1C492541;1?58=1C1@141:1A1;3=41A9=1C1 *1:1775?=2173181G149D41A1=41945;B291B=1C1 s1 D s2 L *5A81C9;171=21A25A9;DC 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. 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
- 131. = =
- 132. 518 CD771 B5521A == 49B91A9 3181G1 G17 @1:1775?=217G1 I *?1496A1;B949C17;1@ @1411G1AG17:1A1;G13=41A93518 ,5CD;1 :1A1;1C1A171A9B751@;5C9714171A9BC5A17@DB1C DCD;BD4DC G17;539 B9 C1 5A1@1 521A 3518 CD771 G17 49@5AD;1 BD@1G1 41@1CC5A:1499C5A65A5B9=1;B9=D=?A45;5C9714571 BD4DC496A1;B9J41A9B525A;1BB91A=??;A?=1C9BG17 =5=99;9@1:1775?=217 I ,5CD;11841G1DA19B52D1835184571491=5C5A ==41:1A1;3518;51G1A
- 133. =5C5A4571@1:17 75?=2173181G1G1= cahaya alami yang datang getaran horizontal diserap sempurna oleh polaroid getaran vertikal diserap sebagian cahaya diteruskan terpolarisasi linear 42 Mudah dan Aktif Belajar Fisika untuk Kelas XII $1A1;4D11=@D=?2914118
- 134. =5C5A41491=5C5A @D@9=1C1B5?A1711; == $9;1@1:1775?=217 3181G1G1749@131A;1;54D11=@DC5AB52DCA1C1 A1C1 I25A1@1;18:1A1;=?29=1;B9=D=BD@1G1 G111=@D9CD=1B9841@1C49@9B18;1?58=1C1 +5?A17 11; =5=2D;1 =1C1 521A521A B589771 491=5C5A9A9BG13= $9;1497D1;1B91A;D97 4571@1:1775?=217 I25A1@1;1841G1 @9B18=1C111;C5AB52DC411=A14914145A1:1C Tes Kompetensi Subbab B (3,%-%0.%*'%.%/6-6.%5+*%0 C. Polarisasi Cahaya 1. Polarisasi pada Kristal 181G1 G17 81G1 =5=99;9 1A18 75C1A1 C5AC5CD 49B52DC 3181G1 C5A@?1A9B1B9 9417C1=@1;;54D4D;11A1875C1A1C5AC5CD41A93181G1 C5A@?1A9B1B9 49B52DC 29417 @?1A9B1B9 $9;1 B52D18 3181G1 11=918 495F1C;1@141B52D18;A9BC11A183181G1G17;5D1A41A9;A9BC181G1 411= B1CD 1A18 B1:1 B589771 49B52DC 3181G1 C5A@?1A9B1B9 951A $9;1 ;A9BC1 =5G5A1@ B521791 1A18 75C1AG1 ;A9BC1 9CD 49B52DC *5A81C9;1 %/%3 3181G1 41C17 =9B1G1 B91A 11=9 =1C181A9 C941; C5A@?1A9B1B9 495F1C;1 @141 B52D18 ;A9BC1 %?=@?5 E5AC9;1 49B5A1@ ?58 ;A9BC1 41 3181G1 G17 49C5ADB;1 C5A@?1A9B1B9 951A +519 ;A9BC1 @?1A?94 @D 41@1C =5=2D1C 3181G1 C5A@?1A9B1B9 %/%3 =5=@5A981C;1 BDBD1 4D1 ;5@97 @?1A?94 B5:1:1A *?1A?94 @5AC1=1 49B52DC 41 ;5@97 G17 ;54D1 49B52DC 181G1 G17 ;5D1A 41A9 @?1A?94 81G1 =5=99;9 B1CD 1A18 75C1A1 C5AC5CD 1C1D 3181G1 C5A@?1A9B1B9 ;1A51 1A18 75C1A1 19G1 49B5A1@ %/%3 %=5=@5A981C;1@?1A9B1C?A41119B1C?AG1749@1B17 Kata Kunci • difraksi gelombang • kisi difraksi • sudut simpang (deviasi) • difraksi maksimum • difraksi celah tunggal • difraksi pada kisi • daya urai optik • apertur %7% 9;5C18D9491=5C5A5B1=1C1==L
- 135. K=@1:1775?=2173181G1 =L
- 136. K=945;B291B=1C1 *1:1775?=2173181G1;5C9;1=5=1BD;9=1C149@5A?584571 =1C1
- 137. =
- 138. = ! $1A1;C9C9;2541;55B1 3== 1G1DA195B1=1C141@1C4989CD7
- 139. = L
- 140. K= Gambar 2.16 Cahaya tak terpolarisasi dilewatkan pada sebuah kristal.
- 141. cahaya yang diteruskan terpolarisasi polarisator analisator sumber cahaya 1 sumber cahaya polarisator analisator tidak ada cahaya 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. ip cermin 1 terpolarisasi 2 ip r terpolarisasi Gelombang Cahaya 43 B5:1:1A 1C1D B1=1 ;54D4D;1G1 3181G1 G17 49C5ADB;1 C5A@?1A9B1B9 *141 %/%3 1A18 CA1B=9B9 119B1C?A C571; DADB @141 1A18 CA1B=9B9 @?1A9B1C?A C941; 141 75C1A1 G17 41@1C 49C5ADB;1 119B1C?A B589771 41 C941; 41@1C =5981C 3181G1 $9;1 25A;1B 3181G1 11=918 4571 9C5B9C1B 495F1C;1 @141 B52D18 @?1A9B1C?A 9451 9C5B9C1B 3181G1 G17 495F1C;1 14118 1C1D
- 142. ;1 C5C1@9 :9;1 ;54D1G1 49@1B17 25AB9171 C941; 141 9C5B9C1B 3181G1 G17 41@1C =55F1C9 119B1C?A 19B1C?A 25A6D7B9 411= =57119B9B B91A G17 495F1C;1 ?58 @?1A9B1C?A $9;1119B1C?A49@DC1A@141B11C9CD=1C1=5981CB91AC5A17 +91A =5A54D@ @141 B11C @?1A9B1C?A 41 119B1C?A B197 C571; DADB 41 C1; 141 3181G1 G17 49C5ADB;1 '5DADC 5+(00( 16+4 %.64
- 143. K
- 144. 9C5B9C1B 3181G1 G17 495F1C;1 @?1A9B1C?A 14118
- 145. $9;1 BD4DC BD=2D @?1A9B1C?A 41 119B1C?A 14118 =1;1 9C5B9C1B3181G1B5C518=51D9119B1C?A141183?B B589771
- 146. 3?B K
- 147. *5AB1=119949B52DC:D716-6/%.644571 141189C5B9C1B 3181G1 G17 5F1C 119B1C?A 41 14118 BD4DC 1C1A1 @?1A9B1C?A 41 119B1C?A +1CD1 9C5B9C1B 3181G1 14118 F1CC = Contoh 2.9 +525A;1B3181G111=918495F1C;1@1414D1;5@97;131@?1A?94G171A18@?1A9B1B9 B1CDB1=119=5=25CD;BD4DCJ $9;19C5B9C1B3181G111=918G114118
- 148. . = C5CD;1 9C5B9C1B 3181G1 G17 C518 =55F1C9 ;54D1 ;131 @?1A?94 C5AB52DC %7% 571=577D1;1(34%/%%0 9 49@5A?58
- 149. 3?B
- 150. . = 3?B . =
- 151. . = $1499C5B9C1B3181G1G17495F1C;114118. = 2. Polarisasi pada Pemantulan dan Pembiasan *5A81C9;1 %/%3 +525A;1B B91A 41C17 G17 495F1C;1 @141 @5A=D;11 29417 21C1B 4D1 =549D= G17 945;B 291BG1 25A2541 =9B1G1
- 152. 41 B521791B91A49@1CD;141B52179117949291B;1 $9;1 B91A @1CD 41 B91A 291B B197 C571; DADB =5=25CD; BD4DC J B91A @1CD 25AD@1 B91A C5A@?1A9B1B9 951A @?1A9B1B9 B5=@DA1 *141 B11C 9CD BD4DC 41C17 49B52DC BD4DC @?1A9B1B9 +D4DC 41C17 99 49B52DC BD4DC @?1A9B1B9 1C1D BD4DC %/%3 =5=@5A981C;1 B91A 41C17 @141 29417 21C1B +521791 49@1CD;141B52179117949291B;1 +5BD194571D;D=+59DB
- 153. B9 B9 4571 J 1C1D K =1;1 41@1C 49CD9B;1 @5AB1=11G1 G19CD
- 154. B9 B9K
- 155. B9 3?B
- 156. B9 3?B n1 n2 Gambar 2.18 Polarisasi pada (a) pemantulan dan (b) pembiasan.
- 157. n1 n2 (1) (2) Gambar 2.19 Polarisasi pembiasan ganda. partikel-partikel gas normal gelombang datang tak terpolarisasi gelombang hamburan terpolarisasi cahaya alami tak terpolarisasi cahaya terpolarisasi E0 sumbu analisator sumbu polarisator 0 E cos 44 Mudah dan Aktif Belajar Fisika untuk Kelas XII
- 158. C1 K
- 160. B5AC1 14118 945;B 291B =549D= B1CD 41 =549D= 4D1 3. Polarisasi pada Pembiasan Ganda *141 ;A9BC1 ;1B9C 1) ;D1AB1 +9) =9;1 C?@1B 41 5B 3181G1 41@1C =5711=9 @5=291B1 7141 ;1A51 =5=99;9 4D1 919 945;B 291B *141 %/%3 C1=@1; 141 4D1 21791 B91A G17 49291B;1 +91A
- 161. C941;=579;DC9@5=291B1=5DADCD;D=+59DB 1C1D49B52DCB91A9BC9=5F1B91A =579;DC98D;D=@5=291B1+59DB 1C1D 49B52DC B91A 291B 4. Polarisasi karena Hamburan 181G1G1741C17@141H1C71B1;1=5711=9@?1A9B1B9B521791 5;CA?55;CA? 411= @1AC9;5 1;1 =5G5A1@ 41 =5=131A;1 ;5=219 B521791 41A9 3181G1 %/%3
- 162. # # ! ! 179C @141 B917 81A9 C1=@1; 25AF1A1 29AD ;1A51 @5A9BC9F1 81=2DA1 *1AC9;5@1AC9;5 D41A1 =5G5A1@ B91A =1C181A9 41 =5=131A;1G1 ;5=219 C5ADC1=1 B91A 29ADG1 *141 @179 41 B?A5 81A9 @1AC9;5@1AC9;5 D41A1 1;1 =5781=2DA;1 5298 21G1; 3181G1 29AD B589771 G17 C5AB9B1 41A9 3181G1 =1C181A9 14118 3181G1 =5A18 D1 C941; =5=99;9 1C=?B65A B589771 C941; 41@1C =5781=2DA;1 3181G1 =1C181A9 )58 ;1A51 9CD 1C=?B65A D1 C5A981C 751@ $9;1 3181G1 C941; C5A@?1A9B1B9 41C17 @141 BD1CD =549D= 71B 3181G1 G17 4981=2DA;1 41@1C C5A@?1A9B1B9 B521791 1C1D B5DAD8G1 A18 @?1A9B1B9 B545=9;91 AD@1 B589771 C571; DADB C5A8141@ 29417 G17 4925CD; ?58 71A9B B91A 41C17 41 71A9B @57981C1 5. Pemutaran Bidang Polarisasi %/%3 =5=@5A981C;1 B52D18 G17 C5A49A9 1C1B 4D12D18@?1A?94G1749@1B17@141B52D1811CG174957;1@94571 B;11 45A1:1C 41 ;?C1; 1ADC1 *?1A?94 G17 45;1C 4571 BD=25A 3181G1 49B52DC @?1A9B1C?A 41 G17 19G1 14118 119B1C?A 'D1=D1 =1C1 49 251;17 119B1C?A C941; =5981C 3181G1 G17 41C17 751@ *5D:D; 119B1C?A =5D:D;;1 BD4DC
- 163. ;5=D491 49 1C1A1 @?1A9B1C?A 41 119B1C?A 495C1;;1 25:11 ;131 G17 25A9B9 1ADC1 7D1 181G1 G17 =51D9 @?1A9B1C?A 1;1 =55F1C9 1ADC1 99 B525D= B1=@19 ;5 119B1C?A +5C518 491=1C9 C5AG1C1 B5;1A17 =1C1 =5981C 141G1 3181G1 C5A17 1ADC1 7D1 411= 81 99 25A6D7B9 B521719 @5=DC1A 29417 75C1A 71A =5:149 5298 751@ 179 119B1C?A 49@DC1A B589771 =5D:D;;1 BD4DC $149 25B1AG1 BD4DC @DC1A1 29417 75C1A 3181G1 G17 491;D;1 ?58 1ADC1 7D1 14118
- 164. K
- 165. 1ADC1 7D1 C5AB52DC 49B52DC 1ADC1 1ADC1 C5AB52DC 141 G17 41@1C =5=DC1A 29417 75C1A @?1A9B1B9 ;5 ;9A9 41 141 :D71 G17 ;5 ;11 571 11C B5=131= 99 ?A17 41@1C =55CD;1 Gambar 2.20 Polarisasi karena hamburan. Gambar 2.21 Pemutaran bidang polarisasi.
- 166. Gelombang Cahaya 45 ;?B5CA1B9 1ADC1 ?@C9; 1;C96 *?1A9=5C5A G17 ;8DBDB DCD; =55 CD;1 ;?B5CA1B9 1ADC1 7D1 49B52DC 1A9 25A21719 @5A3?211 49B9=@D;1 218F1 61;C?A61;C?A G17 =5=571AD89BD4DC@DC1A2941775C1A14118:59B1ADC1C5211ADC1 @1:17 41 ;?B5CA1B9 1ADC1 +531A1 =1C5=1C9B 41@1C 49CD9B;1 B521719 25A9;DC
- 167. K
- 168. %5C5A171
- 169. BD4DC@DC1A 29417 75C1A ;?B5CA1B9 1ADC1 @1:17 1ADC1 C521 BD4DC @DC1A1 :59B 1ADC1 Contoh 2.10 $9;1945;B291BD41A114118
- 171. 571=577D1;1(34%/%%0 949@5A?58 C1
- 172. C1
- 173. C1J
- 175. Contoh 2.11 +52D18B1381A9=5C5AG17@1:17G13=25A9B91ADC17D1@1B9A4571@DC1A1 :59B1ADC1G1J $9;1497D1;1B91A1CA9D=@5=DC1A129417@?1A9B1B9G1 J89CD718;?B5CA1B91ADC19CD %7% 9;5C18D9 3== J
- 177. =
- 179. L
- 181. C5CD;118 945;B291B21?;5BC5AB52DC @12919C5B9C1B3181G1G17;5D1A41A94D1;131 @?1A?94G1749@1B17=5=25CD;BD4DC
- 182. JB1CDB1=1 1941A93181G1=D1=D1 ,5CD;125B1AG1BD4DC G174925CD;?58;54D1;131@?1A?94C5AB52DC Kata Kunci • bidang polarisasi • dichroic • polarisator • analisator • sudut polarisasi/sudut Brewster • hamburan • polarimeter • larutan optik aktif • sacharimeter Tes Kompetensi Subbab C (3,%-%0.%*'%.%/6-6.%5+*%0 +52D18B1381A9=5C5A=5=99;9C12D7G17@1:17G1 3=25A9B91ADC17D14571;5@5;1C1
- 184. #C5A65A5B9 Refleksi (.1/%0).(-531/%)0(5+- 3?C?8G1 181G1 *?1A9B1B9@141 *5=291B1!141 Setelah mempelajari bab ini, tentu Anda menjadi tahu bahwa cahaya merupakan gelombang elektro-magnetik yang dapat mengalami proses interferensi, difraksi, dan polarisasi. Dari semua materi pada bab ini, 46 Mudah dan Aktif Belajar Fisika untuk Kelas XII
- 185. 3?B 4571 BD4DC G17 *5=DC1A1 9417 *?1A9B1B9 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. Rangkuman
- 186. 181G1 C5A=1BD; 75?=217 55;CA?=175C9; B589771@5A1=21C1G1C941;=5=5AD;1=549D= #C5A65A5B9 14118 @5A9BC9F1 @57712D71 4D1 75?=2171C1D529841A975?=217G17;?85A5 181G141@1C25A9C5A65A5B9:9;1BD=25A3181G1G1 ;?85A51AC9G1=5=99;96A5;D5B9B1=1412541 61B5C5C1@ +D=25A3181G1G17;?85A541@1C491=1C9 =51D9@5A3?211$160)413(40(.. #C5A65A5B93181G1=5781B9;1@?1751@C5A17 *?1751@4981B9;141A99C5A65A5B945BCAD;C96B197 =55=18;11;921C@57712D714D175?=217 G17=5=99;961B525A1F11 *5AB1=11B59B98:1A1; G1749C5=@D875?=21714118 B9
- 187. 4571
- 188. *?1C5A174981B9;141A99C5A65A5B9;?BCAD;C96 B197 =57D1C;1 1;921C @57712D71 4D1 75?=217G17=5=99;961B5B1=1 *5AB1=11B59B98 :1A1;G1749C5=@D875?=21714118 B9 4571
- 189. $1A1;41A9C5A17@DB1C;5@?1C5A17;514118 # $1A1;41A9C5A17@DB1C;5@?1751@;514118
- 190. # 96A1;B9 75?=217 14118 @A?B5B @5=25?;1 75?=217G1749B5212;1?58141G1@578117 25AD@1 3518 1C1D BD4DC @578117 G17 =5781179 B521791 =D;1 75?=217 518 45=9;9149B52DC;9B9496A1;B9 $1A1;1C1A3518411= ;9B949B52DCC5C1@1;9B9 *141496A1;B93518CD771@9C1751@;5C5A:149 :9;1 B9 4571
- 191. B5417;1 @9C1C5A17;5C5A:149:9;1 B9
- 192. 4571 K
- 193. 141182917171:9
- 194. 96A1;B9@141;9B9C5A:149:9;13181G1495F1C;1@141 35183518G17=5=99;9:1A1;G17B1=1 1G1 DA19 ?@C9; 14118 ;5=1=@D1 B52D18 5B1 DCD; =5=9B18;1 21G171 41A9 4D1 C9C9; G17 C5A@9B18B1CDB1=119@141:1A1;=99=D=
- 195. *?1A9B1B914118@A?B5B@5G1A9711A1875C1ABD1CD 75?=217 1C DCD; =5G1A97 1A18 75C1A 99 49B52DC@?1A?94B118B1CD3?C?8G114118;A9BC1
- 198. #C5B9C1B3181G1G17495F1C;1@141119B1C?A 14118 4925CD; 1C1A1 BD=2D @?1A9B1C?A 41 BD=2D 119B1C?A 96A1;B9 =5=2181B 96A1;B9 518,D771 96A1;B9 @141%9B9 *?1A9B1B9 =5=2181B *?1A9B1B9 @141%A9BC1 *?1A9B1B9@141*5=1CD1 41*5=291B1 C5A49A9 1C1B #C5A65A5B9 '99=D= #C5A65A5B9 '1;B9=D= *?1A9B1B9;1A51 1=2DA1 Peta Konsep 41@1C =5711=9
- 199. Gelombang Cahaya 47 Tes Kompetensi Bab 2 +.+*.%*4%.%*4%56,%7%%08%0)2%.+0)5(2%5'%0-(3,%-%0.%*2%'%6-6.%5+*%0
- 200. +D4DC@9C1C5A17C571841A9496A1;B9?A45;54D1G17 4981B9;1?58;9B94571 3518 3=B525B1AJ '1;1@1:1775?=2173181G1G1749@5A7D1;1 14118 1 I 2 I 3 I 4 I 5 I $9;1B91AG17:1CD8C571;DADB491C1B@5A=D;11 =9G1;491C1B19AC5211@9B1=9G1;945;B291BG1 41 @1:17 75?=217 41A9 3181G1 G17 =57 81B9;19C5A65A5B9=1;B9=D= =1;1@5AB1=11 G17=5=5D8914118 1 2
- 201. 3 4
- 202. 5 -CD;=55CD;1@1:1775?=217=??;A?=1C9B 497D1;1 @5A3?211 0?D7 G17 41C1G1 B521719 25A9;DC $1A1;1C1A1;54D13518G1==:1A1;3518;5 1G1A3=B5AC1:1A1;1C1A171A9B751@;5 41 71A9B751@;5@1411G1A
- 203. == *1:17 75?=217 B91A =??;A?=1C9B C5AB52DC 14118 1 = 4 = 2 = 5 = 3 = *141@5A3?2110?D74D1351825A:1A1;
- 204. ==49 5C1;;1@141:1A1;
- 205. =41A9B52D181G1A $9;1:1A1; C5A45;1C1C1A1@?19C5A65A5B971A9BC5A17@5AC1=1 41 71A9B C5A17 ;5B5251B 14118 == @1:17 75?=2173181G1G17=5G91A914118 1
- 206. I 4 I 2 I 5 I 3 I *A9B9@41B1A41A94D1BD=25A3181G1;?85A514118 1 ;54D1G1B171C25A45;1C1 2 1=@9CD4?G1B1=1 3 B9=@171G1B51DB1=1 4 254161B5;54D1G114118C5C1@ 5 ;54D1G1=5=131A;13181G1G1725A@1@1B1 939(5FC?C5A:149;1A5175:11 1 496A1;B9 2 @?1A9B1B9 3 49B@5AB9 4 9C5A65A5B9 5 A56A1;B9 *141@5A3?2110?D735187141:9;1:1A1;1C1A1 4D13518G149:149;14D1;19B5=D1=1;1:1A1; 1C1A14D171A9B751@G1725ADADC1 1 ;19B5=D1 2 ;19B5=D1 3 M;19B5=D1 4 N;19B5=D1 5 C5C1@C941;25AD218 @12919C5B9C1B3181G1=D1=D1G17=55F1C9 4D1 @1CA5C1A41B9 14118 1C1A1 ;54D1 @1C =5=25CD; BD4DC J 9C5B9C1B 3181G1 G17 495F1C;1?58;54D1@1C14118 1
- 207. 4
- 208. 2
- 209. 5
- 211. 3 +52D18@?1A?94=5=99;9@?1A9B1C?A41119B1C?AG17 49@1B17=5=25CD;BD4DCJ9C5B9C1B3181G1G17 49C5ADB;11;1B5214974571 1
- 212. 4
- 213. 2 5
- 214. 3
- 216. = 41A9 3518 *1:17 75?=217 B91A G17 49@1;19 14118 1 L
- 217. K== 2 L
- 218. K== 3
- 219. L
- 220. K== 4 L
- 221. K== 5 L
- 222. K==
- 224. +525A;1B B91A =??;A?=1C9B 4571 @1:17 75?=217L
- 225. K=41C17C571;DADB@141;9B9 $9;1B@5;CAD=?A45;54D1=5=2D1CBD4DCJ4571 71A9B?A=1@141;9B9=1;1:D=1871A9B@5A3=;9B9 14118 1 L
- 226. 4 L
- 227. 2 L
- 228. 5 L
- 229. 3 L
- 231. J 3 J
- 233. D135184571:1A1; ==49B91A9C571;DADB !1A9BC5A17;5C971C5A5C1;==41A971A9BC5A17;5 ?@1411G1AG1725A:1A1;
- 234. =41A93518 5A1@1 @1:1775?=217B91AG1749@1;19 181G1BD1CDBD=25A=51D94D13518B5=@9CG17 C5A@9B18
- 236. 3= 41 :1A1; 1C1A1 71A9B 751@ @5AC1=14171A9BC5A17@5AC1=114118 == 25A1@1;18 @1:17 75?=217 3181G1 G17 497D1;1 48 Mudah dan Aktif Belajar Fisika untuk Kelas XII $9;1@5A3?211351871410?D74935D@;1;5411= 19A21719=11;18@5AD2181@1419C5A65A5B9G17 C5A:149 +52D18B1381A9=5C5A=5=99;9C12D7G17499B91ADC1 7D1 @1:17G1 3= %?B5CA1B9 7D1 G17 497D1;1
- 238. 3= $9;1497D1;1B91A1CA9D= C5CD;1 @5=DC1A1 29417 @?1A9B1B9 3181G1 ?58 1ADC1
- 239. +525A;1B 3181G1 =??;A?=1C9B 49:1CD8;1 @141 4D13518B5=@9CE5AC9;125A45;1C14571:1A1;
- 240. == *?19C5A65A5B9G17C5A:14949C17;1@ @141:1A1; 3=41A93518 9;5C18D9218F1:1A1; 1C1A171A9B751@@5AC1=149B52518;9A9;571A9B751@ @5AC1=1 49 B52518 ;11 14118 == *1:17 75?=21725A;1B3181G114118 1
- 241. = 2 = 3 = 4 = 5
- 242. = ! ()+10%.
- 244. 3=B545=9;9189771C5A25CD; @?1751@C5A17@1411G1AG1725A:1A1;3= %5C9;1 @5=9B1811C1A@?1C5A17141183==1;1 @1:1775?=2173181G1G17497D1;1C5AB52DC 14118 1 = 4 = 2 = 5 = 3 = # %810
- 245. .1A1D7D41A9B@5;CAD=?A45;5C97125A9=@9C4571 F1A1=5A18?A45;54D141A9BD1CD@5A9BC9F1496A1;B9 G17=5=@5A7D1;1;9B99925A1AC9@5A214971 1C1A1B91A@1:1775?=217B91AD7D41B91A =5A1814118 1 2 3 4 5 B5=D1:1F121B118