Bab 2 kls xii

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Gelombang Cahaya 31 A. Interferensi Cahaya B. Difraksi Cahaya C. Polarisasi Cahaya mendeskripsikan gejala dan ciri-ciri gelombang cahaya; menerapkan konsep dan prinsip gelombang cahaya dalam teknologi. Setelah mempelajari bab ini, Anda harus mampu: menerapkan konsep dan prinsip gejala gelombang dalam menyelesaikan masalah. Hasil yang harus Anda capai: 31 Gelombang Cahaya 5>?=5>165>?=5>1 C5>C1>7 3181G1 B5A9>7 >41 :[email protected] 41<1= ;[email protected]> B581A981A9 %5C9;1 CDAD> 8D:1> @141 B91>7 81A9 ;141>7 ;141>7 ;9C1 [email protected] =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 [email protected];1> 75<?=21>7 5<5;CA?=17>5C9; *141 12 >41 C5<18 [email protected]<1:1A9 75<?=21>7 =5;1>9; =9B1<>G1 75<?=21>7 19A 41> 75<?=21>7 C1<9 @1;18 @5A25411> 1>C1A1 75<?=21>7 5<5;CA?=17>5C9; 41> 75<?=21>7 =5;1>9; '1>DB91 C5<18 =5=1>611C;1> 75<?=21>7 5<5;CA?=17>5C9; 41<1= 2941>7 CA1>[email protected]?AC1B9 1BCA?>?=9 =9<9C5A 41> 75?7A169 ,18D;18 >41 21719=1>1 75:1<1 41> 39A939A9 75<?=21>7 5<5;CA?=17>5C9; ;8DBDB>G1 3181G1 B5AC1 @5>[email protected]>>G1 $1F121> 1C1B @5AC1>G11>@5AC1>G11> C5AB52DC [email protected] >41 C5=D;1> @141 212 9>9 )<58 ;1A5>1 9CD @5<1:1A9 212 9>9 45>71> 219; Cahaya merupakan gelombang elektromagnetik yang banyak digunakan untuk kepentingan teknologi komunikasi. Bab 2 Sumber: Physics Today, 1995

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1. Cahaya merupakan gelombang elektromagnetik yang banyakdigunakan untuk kepentingan teknologi komunikasi.A. InterferensiCahayaB. Difraksi CahayaC. Polarisasi CahayaGelombang Cahaya 31Bab2Gelombang CahayaHasil 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 :[email protected] 411=;[email protected] B581A981A9 %5C9;1 CDAD 8D:1 @141 B917 81A9 ;1417;1417 ;9C1 [email protected] =59;=1C9 9418G1 @5179 G17 =57891B9 35A18G1179C @1;18 B5251AG1 @5179 9CD181G1 [email protected];1 75?=217 55;CA?=175C9; *141 12 2. 41C518 [email protected]:1A9 75?=217 =5;19; =9B1G1 75?=217 19A 4175?=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 [email protected]?AC1B9 1BCA??=9 =99C5A 41 75?7A169 ,18D;18 4121719=11 75:11 41 39A939A9 75?=217 55;CA?=175C9; ;[email protected]@[email protected]@[email protected] 41 C5=D;1 @141 212 99 )58 ;1A51 9CD @51:1A9 212 994571 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;C132 Mudah dan Aktif Belajar Fisika untuk Kelas XIIA B CgelombangdatangS2S1teranggelapteranggelapterangterangterangterangterangterangterangteranggelapgelapgelapgelapgelapgelapgelapgelapteranggelapteranggelapterangS0Gambar 2.1Interferensi pada celahganda YoungA. Interferensi Cahaya 14118 !! !# 1AC9G1 75?=217 =5=99;9 !41 # [email protected] G17 C518 492181B @141 @5=2181B1 75?=217218F1 75?=217 3181G1 25AB961C [email protected] 81G1 75?=217 2DG9 [email protected] 25A9C5A65A5B9 )58 ;1A51 9CD DCD; [email protected];1 [email protected]@5AD;1BD=25A3181G1G17;?85A5G19CDBD=25A3181G1G17 =5=99;9 6A5;D5B9 B1=1 41 2541 61B5 [email protected] +D=25A 3181G1 G17;?85A5 [email protected] 491=1C9 =51D9 @5A3?211 G17 491;D;1 ?58 $160)41 3(40(..1. Percobaan Young dan Fresnella. Percobaan Celah Ganda Young*5A3?21199491;D;1?58$160)4571=577D1;[email protected] *578117 @5AC1=1 =5=99;9 B1CD D217 ;539 41 @578117;54D1 4957;[email protected] 4571 4D1 D217 ;539 *5A81C9;1 %/%3 +91A=??;[email protected][email protected]=25A3181G1G17=5=131A =51D9 3518 + %5=D491 B91A 41A9 3518 + [email protected];1 ;[email protected] ;54D1 D1 3518 @141 @578117 ;54D1 G19CD + 5. 41 + G17 [email protected] B5:1:1A 4571 + 1;1 25A6D7B9 B521719 @5=131A B91AB91A;?85A5%54D125A;1B41A935183518+ 6. 41+ [email protected] 1G1A 1B9 9C5A65A5B9 [email protected] 71A9B C5A17 41 71A9B [email protected] Percobaan Fresnell571=577D1;1B52D18BD=25A3181G1+3(40([email protected]?584D1 BD=25A 3181G1 + 7. 41 +G17 ;?85A5 41A9 81B9 @5=1CD1 4D135A=9 *5A81C9;1 %/%3 Gambar 2.2Percobaan Fresnell untukmenunjukkan interferensi cahaya.S1OPC2S2STokohAugustine Fresnell(17881827)Augustine Fresnell (17881827)adalah fisikawan Prancis yangmengembangkan teori gelombangtransversal cahaya berdasarkan hasilpenemuannya tentang lensa daninterferensi. Fresnellmemperlihatkan bahwa cahayamatahari terdiri atas bermacam-macamwarna cahaya, yang setiapwarna memiliki sudut bias tertentu.Ia juga menemukan sebuah bentuklensa yang pada keduapermukaannya berbentuk cembung.Bentuk lensa ini dikenal sebagailensa cembung. Lensa ini memilikisifat mengumpulkan cahaya.Sumber: Science Encyclopedia, 1998 [email protected];18;[email protected][email protected] +52DC;1B961CB961C75?=2173181G1 8. 37,2 10 m 3,6 10 3myDari soal diketahui:m = 1; d = 105 m; D = 0,2 msehinggayd3 5 3,6 10 m 10 m0,2m 13 5 3,6 10 m 10 m0,2m 1Gelombang Cahaya 33S1 r1B CPr2S2dyDbr2S2 r1S1bGambar 2.3(a) Sinar gelombang dari celah S1dan S2 berinterferensi di titik P.(b) Untuk Dd, r1 dan r 2 dianggapsejajar dan membentuk sudut terhadap sumbu tengah.*141%/%3 [email protected]+14118BD=25AB91A=??;A?=1C9B + 9. 41 +14118 21G171 41A9 + ?58 35A=9 10. 41571 45=9;91 B91AB91A G17 41C17 @141 1G1A B5?18?18 25A1B141A9 + 11. 41 + !5?=217 3181G1 41A9 + 12. 41 +99 1;1 [email protected]@141B59B9841A99C1B1;54D1 B91A 9CD *5AD 41 ;5C18D9 :9;1 ;54D1 BD=25A 3181G1 =5=99;[email protected]? G17 B1=1 @141 [email protected] C5A:149G1 9C565A5B9 =99=D= 1;1C5A25CD; 71A9B [email protected] +5219;G1 :9;1 [email protected]? C941; B1=1 9C5A65A5B9=99=D=G1 C941; [email protected] B1=1 B5;1912L*5A81C9;1 %/%3 75?=217 3181G1 41C17 =5D:D 3518 13. 41 3518G17 C5A5C1; @141 29417 181G1 C5AB52DC C5A496A1;B9 ?58;54D1 3518 C5AB52DC 41 =5781B9;1 @?1 9C5A65A5B9 @141 1G1A %9C1 [email protected] =55CD;1 49 =11 [email protected] @9C1 C5A17 1C1D @9C1 [email protected]; @141 1G1A 4571 =5=25A9;1 BD4DC 41A9 BD=2D [email protected] 71A9B [email protected] 1C1D 71A9B C5A17 C5AB52DC -CD; =55CD;125B1AG1 ;9C1 81ADB=578D2D7;1G1 4571 *5A81C9;1%/%3 ,9C9;[email protected];1 B52D18 C9C9; G17 C5A5C1; @141 B545=9;91 [email protected] B589771 :1A1; 2 ;5 * B1=1 4571;5 * 457145=9;91 B1=1 4571 :1A1; 41A9 14. ;5 D2D71 1C1A1 :1A1;41A9 15. ;541 99 B171C AD=9C (1=D ;9C1 [email protected] =5G545A811;1G1 4571 [email protected] 218F1 BDBD1 :1A1; 3518 [email protected] 1G1A:[email protected]:1A1;1C1A1;54D135181A971=21AC5A981C 218F1 B91A 75?=217 16. 41 14118 B5:1:1A 41 =5=25CD;BD4DC [email protected] BD=25A @DB1C *5A81C9;1 %/%3 ,5AG1C1BD4DC 17. G17 4925CD; ?58 BD=2D @DB1C 41 BD=2D 18. 57145=9;91 ;9C1 [email protected] 4571 =D418 =55CD;1 218F1B9B9 K 19. (34%/%%0 9 [email protected];1 @5AB1=11 DCD; =55CD;1 :1A1;[email protected] 1C1A1 B91A 20. 41 [email protected] * -CD; 9C5A65A5B9 =1;B9=D=9C5A65A5B9 ;?BCAD;C96 C518 49;5C18D9 218F1 @1BC9 ? 1C1D29171 [email protected] 41A9 @1:17 75?=217 (34%/%%0 9 [email protected] 49CD9BB521719 25A9;DCPembahasan SoalSeberkas cahaya monokromatisdijatuhkan pada dua celah sempitvertikal berdekatan dengan jarakd = 0,01 mm. Pola interferensiyang terjadi ditangkap layar padajarak 20 cm dari celah. Diketahuibahwa jarak antara garis gelappertama sebelah kiri ke garis gelapsebelah kanan adalah 7,2 mm.Panjang gelombang cahayatersebut adalah ....a. 180 nm d. 720 nmb. 270 nm e. 1.800 nmc. 360 nmSPMB 2003Pembahasan:Jarak pola gelap ke-1 ke pusatadalah 2 12ydmD 12D m 2 27 3,6 10 m = 360 nmJawaban: c 21. 34 Mudah dan Aktif Belajar Fisika untuk Kelas XIIB9 K 4571 22. -CD;[email protected];9C1=5=25A9;[email protected] C5A17 @5AC1=1 41 [email protected] DCD; 9C5A65A5B9 =99=D= 9C5A65A5B9 45BCAD;[email protected]:9;19B5C571875?=217(34%/%%[email protected];1 B521719 25A9;DC 23. B9 24. K4571 25. -CD; [email protected] @5AC1=1 ;9C1 =5=25A9;1 @9C1 [email protected] ;54D1 26. 41 B5C5ADBG12. Menentukan Jarak Pita Terang ke-m atau Pita Gelap ke-mdari Terang Pusat*141 @5=2181B1 B525D=G1 C518 49B52DC;1 218F1 @?19C5A65A5B9 @141 1G1A [email protected] @9C1 C5A17 41 @9C1 [email protected] *5A81C9;1%/%3 %9C1 [email protected] =55CD;1 ;54D4D;1 @9C1 C5A17 ;5 [email protected] [email protected] ;5 @141 1G1A *5A81C9;1 ;5=219 %/%3 % )58;1A51:1D8 5298 25B1A [email protected] BD4DC 25A919 B171C;539 -CD; BD4DC G17 B171C ;539 1;1 25A1;D B9 C1 1A9%/%3 % ;9C1 [email protected] =55CD;1 218F1B9 C1 # K-CD; @9C1 C5A17 =1BD;;1 (34%/%%0 9 ;5 (34%/%%0 9B589771 [email protected]?58# K-CD; @9C1 [email protected] =1BD;;1 (34%/%%0 9 ;5 (34%/%%0 9 B589771 [email protected]?58 27. # 28. K%5C5A171:1A1; 1C1A3518 @141 1G1A#:1A1; C5A17 [email protected] ;5 41A9 @DB1C:1A1; 1G1A ;5 [email protected]:17 75?=217 3181G111=8199 29. Tantanganuntuk AndaSeberkas cahaya monokromatisdijatuhkan pada dua celah sempitvertikal berdekatan dengan jarakd = 0,01 mm. Pola interferensi yangterjadi ditangkap pada jarak 20 cmdari celah. Diketahui bahwa jarakantara garis gelap pertama disebelah kiri ke garis gelap pertamadi sebelah kanan adalah 7,2 mm.Hitunglah panjang gelombangberkas cahaya tersebut. 30. PDiABCdGelombang Cahaya 35Contoh 2.1-CD;=55CD;[email protected]:[email protected];[email protected]@9:1A1CA9D=B91A99495F1C;[email protected]:1A1;==*141:1A1; 31. [email protected]$9;[email protected]@5A?58:1A1;[email protected][email protected];[email protected];[email protected]1:1775?=217B91A1CA9D=C5AB52DC%7%9;5C18D9==L 32. K= # ==L 33. K= 34. =k# 35. = 36. = 37. L 38. [email protected]:1775?=217B91A1CA9D=14118IContoh 2.2*[email protected]?2119C5A65A5B9497D1;[email protected]$1A1;1C1A1;54D135189CD ==41495C1;;[email protected]:1A1; =;[email protected]@[email protected]:1A1;[email protected]:1775?=2173181G1G1%7%== L 39. K= 40. =# ==L 41. [email protected](34%/%%09 [email protected]?58# 42. =K 43. = 44. = 45. K 46. =L 47. K= 48. L 49. [email protected]:1775?=2173181G1G114118I3. Interferensi oleh Lapisan Tipis*5=1CD1 3181G1 =1C181A9 ?58 ! # G17 [email protected] 19A 1;1 [email protected];1 71A9B71A9B 25AF1A1 @141 =9G1;[email protected];CAD= F1A1 99 [email protected];1 141G1 [email protected]=9G1;[email protected]#[email protected]@19C5A65A5B9=1;[email protected]=99=D=#C5A65A5B91C1A175?=217G17 [email protected];1 ?58 [email protected] [email protected] 49CD:D;;1 @141 %/%3 +525A;1B B91A 41C17 =57519 [email protected] [email protected] 4571 BD4DC 41C17 1;1 49291B;1 41 B521791 179 [email protected];1 ;5=219 ;5 @5A=D;11+91A G17 [email protected];1 495F1C;1 @141 B52D18 5B1 @?B9C96 41496?;DB;149C9C9;*5A;1B3181G149C9C9;*[email protected];181B99C5A65A5B925A;1B 3181G1 50. 41 4571 51. 14118 25A;1B 3181G1 [email protected];1 17BD7 41 14118 25A;1B 3181G1 G17 [email protected]=291B1 C5A5298 418DD ;5=D491 [email protected];1Gambar 2.4Interferensi oleh lapisan tipis.nrlensalapisantipis(1)(2) 52. +59B98 9C1B1 [email protected]; G17 [email protected] ?58 B91A 41C17 89771=5:149 B91A @1CD ;5 53. 41 B91A @1CD ;514118 K 54. K K457114118 945;B 291B [email protected] [email protected] 41 '9B1;1 [email protected] 14118=1;13?BB5897713?B41B94571 C1B589771 3?B Sumber: www.designprodygzone.comSumber: www.funsci.comGambar 2.6Interferensi oleh lapisanbusa sabun yang tipis.Sumber: www.instckphoto.com36 Mudah dan Aktif Belajar Fisika untuk Kelas XIIK C1B93?BK B9 B93?B571 =577D1;1 6-6/ !0(..+64 C5C17 @5=291B1 3181G1G1;9B9 [email protected]?58 B59B98 :1A1; [email protected] ;54D1 B91A =5:1493?BKB9 3?B 3?B 55. KB9 3?B3?B [email protected] C5A:149 9C5A65A5B9=1;B9=D=49 C9C9;* [email protected];1 ;[email protected] 41A9 @1:17 75?=217 ;1 [email protected] B91A @1CD 49 =5711=9 @5AD2181 61B5 56. =1;11;1=5:149 1C1D 57. K#C5A65A5B9=1;[email protected]@[email protected]@9B1;[email protected]=11 3?B 58. K4571 59. (34%/%%0 9 25A1;D DCD; 945;B 291B [email protected] [email protected] 5298 25B1A41A9 60. 1C1D 61. [email protected];[email protected]?589C5A65A5B9=99=D=;[email protected] =5=99;9 2541 61B5 62. =1;1 63. #C5A65A5B9 =99=D= 411= 1A18 @1CD =5=5D89 @5AB1=113?B411C1D3?BK 64. 4571 945;B 291B41=5=5D89BG1A1C 65. 41 1C1D 66. 41 Gambar 2.7Interferensi oleh lapisan minyakyang tipis.,5CD;[email protected]=99=D=G17492DCD8;[email protected]:[email protected]@[email protected]=5=99;9945;B291B [email protected]:1775?=217IContoh 2.3Gambar 2.5Interferensi oleh busa sabun. 67. Tugas Anda 2.1Coba Anda perhatikan kembaliGambar 2.5. Gelembung tersebutsebenarnya berwarna-warni.Mengapa demikian? Anda dapatmencari jawabannya dari bukureferensi atau internet.Gelombang Cahaya 37%7%#C5A65A5B9=1;[email protected]@[email protected][email protected]=11 3?B 68. [email protected]@[email protected]@9BG1=1;1 69. 413?B 70. [email protected]?58 71. I 72. I [email protected]@9BG17492DCD8;114118 IContoh 2.4,5CD;[email protected]:1775?=217B91AG17497D1;1:9;1C5A:1499C5A65A5B9=99=D=?A45;[email protected]@9B49D41A14571;5C5211 73. =BD4DC291BJ41945;[email protected] 74. %7%571=577D1;1(34%/%%[email protected]?58 3?B 75. =3?BJ 76. [email protected]:1775?=217G17497D1;114118 77. =Kata Kunci interferensi sinar monokromatis interferensi maksimum interferensi minimumTes Kompetensi Subbab A(3,%-%0.%*'%.%/6-6.%5+*%0 78. @1G1749=1;BD49C5A65B93181G1 +52D183518714125A:1A1;==49251;[email protected]:1A1; =5C5A495C1;;1B52D181G1A51849B91A9 4D1 B91A =??;A?=1C9B 4571 @1:[email protected];18:1A1;71A9BC5A17?A45;[email protected];[email protected] +525A;[email protected]=11925A:1A1;==$1A1;3518;51G1A 79. =5C5A41:1A1;[email protected] 80. L 81. K [email protected];[email protected]:1775?=2173181G1G17497D1;1 -CD; =57D;DA @1:17 B91A =5A18 491;D;[email protected]?211B52171925A9;DC+91A 29AD 4571 @1:17 75?=217=49:1CD8;1 C571; DADB @141 3518 7141 *?19C5A65A5B9C5A:[email protected]:1A1; [email protected]=125A:1A1;[email protected]+5C5189CDB91A=5A1849:1CD8;[email protected],[email protected]=125A:1A1;[email protected],5CD;[email protected]:1775?=217B91A=5A189CD [email protected]@9B=5=99;9945;B291B 497D1;1DCD; =5981C 75:11 9C5A65A5B9 $9;1 @1:1775?=217G1 I C5CD;1 C521 [email protected]@1G1C5A:1499C5A65A5B9B. Difraksi Cahaya*[email protected]:1A175C1A14175?=21749%51B/C518492181B218F175?=217 19A G17 =55F1C9 B52D18 @578117 4571 B52D18 [email protected];[email protected]?;B5C518 =51D9 3518 C5AB52DC *5=25?;1 75?=217 G17 49B5212;1?58 141G1 @578117 [email protected] 3518 49B52DC 496A1;B9 75?=217 +1=181G1 4571 75?=217 3181G1 G17 495F1C;1 @141 B52D18 [email protected] :D71 1;1 =5711=9 5CDA1 82. 96A1;B93181G1C5A:149:[email protected][email protected]@9B18B5:1:1AB1CD B1=1 19 @141 :1A1; G17 B1=1 518 [email protected] G17 45=9;9149B52DC %9B9 14118 ;[email protected] ;131 G17 497?A5B =5DADC71A9B B5:1:1A 41 21G1; :D=18G1 $1A1; 1C1A1 4D1 3518 [email protected] ;9B9 1. Difraksi Celah Tunggal96A1;B9 @141 3518 CD771 1;1 =5781B9;1 @?1 71A9B C5A1741 [email protected] @141 1G1A 518 CD771 [email protected] [email protected] C5A49A9 1C1B [email protected] [email protected] G17 4921C1B9 C9C9;C9C9; 41 [email protected] 3518 9CD [email protected];1BD=25A 3181G1 B589771 B1CD B1=1 19G1 [email protected] 25A9C5A65A5B9*5A81C9;1 %/%3 -CD; =57119B9B @?1 496A1;B9 3518 @141 %/%3 4921794D1 21791 *5A81C9;1 75?=217 83. 41 !5?=217 84. [email protected] G17 5298 :1D8 B525B1A B9d 1 2d2maksimumutamaContoh 2.5d5432dsin238 Mudah dan Aktif Belajar Fisika untuk Kelas XII [email protected] 75?=217 +1=181G1 4571 75?=217 41G17 =5=99;9 2541 9C1B1 B525B1AB9 #C5A65A5B9 =99=D= G17 =5781B9;1 71A9B [email protected] C5A:149 :9;1;54D1 75?=217 25A2541 61B5 85. J 1C1D 2541 9C1B1G1 B1=1 4571 86. @1:17 75?=217B9 B9$9;1 3518 492179 [email protected] 21791 [email protected] 71A9B [email protected] ;5C9;1B9 B91 [email protected] 4571 9CD :9;1 C518 492179 51= 21791 [email protected] [email protected] ;5C9;1B9 B9 +531A1 D=D= [email protected] 49G1C1;1 218F1 @9C1 [email protected] ;5 C5A:149 :9;1K 87. B9 %5C5A171521A 3518BD4DC [email protected] 45E91B9 88. -CD;1C1DC5A:149 =1;B9=D= DC1=1 @9C1 C5A17C5718 [email protected] [email protected];1 @141 %/%3 Gambar 2.9Maksimum utama terjadiuntuk k = 0 atau= 0.571=577D1;[email protected]@[email protected];@?1496A1;[email protected]@;[email protected]=5=25CD;[email protected]?A=1$9;13181G1G17497D1;1=5=99;[email protected]:1775?=217IC5CD;1521A3518G17497D1;1Gambar 2.8Difraksi cahaya pada celah tunggal. 89. dpBB1AdCDEFGGelombang Cahaya 39garis gelapterang%7%571=577D1;1(34%/%%[email protected]?58 B9B9JI=1;1dPQpusat 90. I$149521A3518G114118I-CD;[email protected];[email protected]?1496A1;B9=1;B9=D=25419C1B141A99C5A65A5B9=99=D=81ADB49;DA1794571 91. )58;1A51;54D13181G1B561B5 2541 61B5 ;54D1G1 =5:149 J D1 75?=217 4571 254161B5 92. 1C1D2541BD4DC61B5J49B52DC:D71B561B5*5AB1=119C5A65A5B9=1;B9=D= 41A9 @?1 496A1;B9G1 1;1 =5:149B9 93. B9 94. 1C1DB9 95. K 96. K 97. 141182917171:9 98. ,5CD;1521A3518=99=D=G17492DCD8;[email protected];B93518CD771:9;149979;1BD4DC496A1;[email protected]:1775?=217G17497D1;1=DCD;@?1496A1;B9=1;B9=D=%7%571=577D1;1(34%/%%[email protected]?58B9 99. = == =B9 100. $149521A3518=99=D=14118=Pembahasan SoalSuatu berkas sinar sejajar mengenaicelah yang lebarnya 0,4 mm secarategak lurus. Di belakang celah diberilensa positif dengan jarak titik api40 cm. Garis terang pusat (orde nol)dengan garis gelap pertama padalayar di bidang titik api lensaberjarak 0,56 mm. Panjanggelombang sinar adalah ....a. 6,4 107 mb. 6,0 107 mc. 5,2 107 md. 5,6 107 me. 0,4 107 mPPI 1983Pembahasan:Jarak titik api = jarak celah kelayar == 40 cm.Gelap pertama m = 1 m 3 3 0,4 10 m 0,5 10 m10,4 m = 5,6 107 mJawaban: dContoh 2.62. Difraksi pada Kisi96A1;B9 3181G1 C5A:149 @D1 @141 3181G1 G17 =51D9 21G1; [email protected] 4571 :1A1; 3518 B1=1 518 [email protected] G17 45=9;91 49B52DC;9B9 496A1;B9 1C1D 49B97;1C ;9B9 +5=1;9 21G1; 3518 @141 B52D18 ;9B9B5=1;9 C1:1= @?1 496A1;B9 G17 4981B9;1 @141 1G1A '9B1G1 @141415A18 B5521A 3= [email protected] 3518 AC9G1 ;9B9 C5AB52DC C5A49A91C1B 3518 3= 1C1D 3518 3= 571 45=9;91 :1A1;Gambar 2.10Difraksi pada kisiTMPK1C1A3518 14118 101. 3= L 102. K 3= *5A81C9;1 %/%3 103. intensitasm = 1 m = 2 21 131 23dsinGambar 2.12Difraksi minimum keduauntuk N banyak celah.biruspektrumorde ke 2merahbirubirumerahspektrumorde ke-1spektrumorde ke-0(putih)spektrumorde ke-1spektrumorde ke-2Gambar 2.13merahmerahbiruDifraksi cahaya putih akanmenghasilkan pola berupapita-pita spektrum.cahayaputihkisidifraksi%/%3 [email protected];1 B525A;1B B91A =??;A?=1C9B G17495F1C;1 @141 B52D18 ;9B9 41 =5781B9;1 @?1 496A1;B9 @141 1G1A *?1 496A1;B9 [email protected] 71A9B C5A17 41 71A9B [email protected] B531A1 25A71C9196A1;B9=1;B9=D=C5A:149:9;[email protected][email protected];71A9B71A9BC5A175419C1B1 G17 495F1C9 3181G1 G17 41C17 41A9 4D1 3518 25A45;1C114118 104. 1C1D2917131318;[email protected]:1775?=217G1*?1 496A1;B9 =1;B9=D= DC1=1 @141 ;9B9 14118B949=1114118?A45496A1;B94114118:1A1;[email protected];9B996A1;B9=99=D=491C1A1 =1;B9=D=C5A:149:9;[email protected][email protected];71A9B71A9B [email protected] 41 =99=D= @5AC1=1 B5BD418 =1;B9=D= ;5 C5A:149:9;1B9 105. :D71=99=D=;54D1 :9;1B9 +52D18;9B9457171A9B 3=495F1C;13181G1C571;[email protected]:1775?=217 !1A9BC5A17496A1;[email protected]=1=5=25CD;[email protected]?A=1=1;B9=D=DC1=1,5CD;[email protected]:1775?=217%7%9;5C18D9 106. 71A9B 3= L 107. K3=B9J 108. Mari Mencari Tahu40 Mudah dan Aktif Belajar Fisika untuk Kelas XII 109. 571=577D1;1(34%/%[email protected]?58 B9 L 110. K3= 111. L 112. K3= 113. [email protected]:1775?=217G17497D1;114118 114. I+5217193?C?8DCD; [email protected]?58%/%3B5417;1 DCD;21G1; 3518 [email protected]?58 %/%3 %/%3 [email protected];1 3181G1 @?9;A?=1C9; @141 3518 G17 [email protected];1 3181G1 @DC98 +91A @DC98 @?9;A?=1C9; C5A49A9 1C1B25A21719 F1A1 4571 @1:17 75?=217 C5A;539 F1A1 D7D 41C5A25B1A F1A1 =5A18 571 45=9;1 F1A1 G17 C5A45;1C 4571 14118F1A1D7D41G17C5A:[email protected];[email protected];CAD= F1A1 57;[email protected] G19CD D7D 29AD 89:1D ;D97 :9771 41=5A18 [email protected] ?A45 496A1;B9 =5D:D;;1 [email protected];CAD= F1A1Contoh 2.7*5A18;[email protected]*[email protected];1B118B1CD75:1111=B52171981B9 496A1;B9 ,D71B 41 ;[email protected];1 96?A=1B9 =57519 C5A:149G1 @5179;[email protected];[email protected];51BGambar 2.11Difraksi minimum keduauntuk N = 2 celah.m = 1 m = 0 m = 1 m = 20dsin 2Tantanganuntuk AndaTentukan daya urai sebuah celahdengan diameter 2 mm, jarak celahke layar 1 meter dengan panjanggelombang cahayanya 590 nm. 115. celah bulat pola difraksiInformasiuntuk AndaInformation for YouGelombang Cahaya 413. Daya Urai Optik# ! 14118 ;[email protected] B52D18 5B1 DCD; =5=9B18;121G17141A94D1C9C9;[email protected][email protected]:1A1;=99=D=%[email protected] @5A25B1A1 11C11C [email protected]; =9B1G1 [email protected] =9;A?B;[email protected] [email protected]?74921C1B9?5841G1DA195B141:[email protected]?1496A1;B9G17 C5A25CD; @141 21G171 2541 9CD%/%3 [email protected];1 218F1 21G171 G17 C5A:149 [email protected];1 @?1 496A1;B9 G17 49B5212;1 ?58! D1B B9BC5= 5B1?58 11C11C [email protected]; C5AB52DC*?1496A1;B9G174925CD;?58B52D1835182D1CC5A49A91C1B25CD;C5A17 @DB1C G17 49;59979 3939 C5A17 41 [email protected] +118 B5?A179=DF1 13'%8.(+)* [email protected];1 218F1 4D1 2D18 C9C9; BD=25AG17 C5A17G1 B1=1 1;1 C5A981C [email protected] :9;1 =1;B9=D= B5CA1 [email protected]@DB1C 41A9 @?1 496A1;B9 G17 B1CD [email protected] 5C1;G1 4571=99=D= @5AC1=1 41A9 @?1 496A1;B9 C9C9; G17 19 *?1 C5AB52DC [email protected]:51B;1 4571 =577D1;1 %/%3 116. $1A9:1A9 97;1A1 C5A17G17 C5A25CD; [email protected] 491AC9;1 [email protected] @?1 496A1;B9 G17 C5A21C1B $9;1 3181G1 =51D9 AD17 [email protected] 1C1DD41A141G1DA1941A9351897;[email protected];[email protected]=11 117. K 118. '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 119. K 120. -CD; BD4DC G17 ;539 121. K 122. %5C5A17141G1 DA19 =:1A1; 2541 41A9 5B1 [email protected]:17 75?=217 3181G1 =491=5C5A 2D;11 5B1 =BD4DC 45E91B9Gambar 2.14Bayangan dari optik fisis dua bendayang berdekatan karena (a) beririsandan (b) terpisah dengan baik.DGambar 2.15Lukisan sinar dari sumber cahayadari sebuah celah bulat.rcahaya sumber datang2Contoh 2.8%5C9;[email protected][email protected][email protected]:1A1;=99=D=1C1A14D1BD=25AC9C9;[email protected];[email protected]:1A1;3=41A9=1C1*1:1775?=2173181G149D41A1=41945;B291B=1C1 s1D s2L *5A81C9;171=21A25A9;DCMerak jantan dengan bulu-buluekornya yang berwarna-warni danberukuran lebar menyebabkanlebih kelihatan menarik dibandingmerak betina. Keindahan bulumerak tersebut merupakan contohefek difraksi gelombang cahayaoleh bulu merak.Male with the largest or most colorfuladornments are often the mostattractive to females. The extraordinaryfeathers of a peacocks tail are anexample of diffraction effect ofpeacocks tail light wakes.Sumber: Biology ConceptsConnections,2006 123. = = 124. 518 CD771 B5521A == 49B91A9 3181G1 [email protected]:1775?=217G1I*?1496A1;B949C17;[email protected]@1411G1AG17:1A1;G13=41A93518,5CD;1:1A1;[email protected];[email protected];BD4DC G17;539 B9C1 [email protected] 521A 3518 CD771 G17 [email protected];1 [email protected]@1CC5A:1499C5A65A5B9=1;B9=D=?A45;5C9714571BD4DC496A1;B9J41A9B525A;1BB91A=??;A?=1C9BG17=5=99;[email protected]:1775?=217I ,5CD;11841G1DA19B52D1835184571491=5C5A ==41:1A1;3518;51G1A 125. [email protected]:1775?=2173181G1G1=cahaya alamiyang datanggetaranhorizontaldiserapsempurnaoleh polaroidgetaran vertikaldiserapsebagiancahaya diteruskanterpolarisasi linear42 Mudah dan Aktif Belajar Fisika untuk Kelas XII $1A1;[email protected]=?2914118 126. [email protected]@9=1C1B5?A1711; ==$9;[email protected]:[email protected];1;[email protected]@1;18:1A1;=?29=1;[email protected][email protected][email protected]@9B18;1?58=1C1 +5?A17 11; =5=2D;1 =1C1 521A521A B589771491=5C5A9A9BG13=$9;1497D1;1B91A;[email protected]:[email protected];[email protected]=1C111;C5AB52DC411=A14914145A1:1CTes Kompetensi Subbab B(3,%-%0.%*'%.%/6-6.%5+*%0C. Polarisasi Cahaya1. Polarisasi pada Kristal181G1 G17 81G1 =5=99;9 1A18 75C1A1 C5AC5CD 49B52DC [email protected][email protected];;54D4D;[email protected]?1A9B1B9 49B52DC 29417 @?1A9B1B9 $9;1 B52D18 3181G1 11=918495F1C;[email protected];A9BC11A183181G1G17;5D1A41A9;A9BC181G1411= B1CD 1A18 B1:1 B589771 49B52DC 3181G1 [email protected]?1A9B1B9 951A $9;1;A9BC1 [email protected] B521791 1A18 75C1AG1 ;A9BC1 9CD 49B52DC *5A81C9;1 %/%3 3181G1 41C17 =9B1G1 B91A 11=9 =1C181A9C941; [email protected]?1A9B1B9 495F1C;1 @141 B52D18 ;A9BC1 %[email protected]?5 E5AC9;[email protected] ?58 ;A9BC1 41 3181G1 G17 49C5ADB;1 [email protected]?1A9B1B9 951A+519 ;A9BC1 @?1A?94 @D [email protected] =5=2D1C 3181G1 [email protected]?1A9B1B9%/%3 [email protected];1 BDBD1 4D1 ;[email protected] @?1A?94 B5:1:1A*?1A?94 @5AC1=1 49B52DC 41 ;[email protected] G17 ;54D1 49B52DC 181G1 G17 ;5D1A 41A9 @?1A?94 81G1 =5=99;9 B1CD 1A18 75C1A1C5AC5CD 1C1D 3181G1 [email protected]?1A9B1B9 ;1A51 1A18 75C1A1 19G1 [email protected]%/%3%[email protected];[email protected][email protected] 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 127. [email protected]:1775?=2173181G1=L 128. K=945;B291B=1C1*1:1775?=2173181G1;5C9;1=5=1BD;[email protected]?584571 =1C1 129. = 130. =!$1A1;C9C9;2541;55B1 [email protected] 131. =L 132. K=Gambar 2.16Cahaya tak terpolarisasidilewatkan pada sebuah kristal. 133. cahaya yangditeruskanterpolarisasipolarisator analisatorsumbercahaya1sumbercahaya polarisator analisatortidak adacahaya2Gambar 2.17(a) Polarisator dan analisator dipasangsejajar sehingga cahaya yangditeruskan di belakang analisator akanterpolarisasi linear.(b) Polarisator dan analisator dipasangtegak lurus sehingga cahaya tidakditeruskan oleh analisator.ipcermin1terpolarisasi2iprterpolarisasiGelombang Cahaya 43B5:1:1A 1C1D B1=1 ;54D4D;1G1 3181G1 G17 49C5ADB;1 [email protected]?1A9B1B9*141 %/%3 1A18 CA1B=9B9 119B1C?A C571; DADB @141 1A18CA1B=9B9 @?1A9B1C?A C941; 141 75C1A1 G17 [email protected] 49C5ADB;1 119B1C?AB589771 41 C941; [email protected] =5981C 3181G1$9;1 25A;1B 3181G1 11=918 4571 9C5B9C1B495F1C;1 @141B52D18 @?1A9B1C?A 9451 9C5B9C1B 3181G1 G17 495F1C;1 14118 1C1D 134. ;1 [email protected] :9;1 ;54D1G1 [email protected] 25AB9171 C941; 1419C5B9C1B 3181G1 G17 [email protected] =55F1C9 119B1C?A19B1C?A 25A6D7B9 411= =57119B9B B91A G17 495F1C;1 [email protected]?1A9B1C?A$9;[email protected]@141B11C9CD=1C1=5981CB91AC5A17+91A [email protected] @141 B11C @?1A9B1C?A 41 119B1C?A B197 C571; DADB41 C1; 141 3181G1 G17 49C5ADB;1 '5DADC 5+(00( 16+4 %.64 135. K 136. 9C5B9C1B 3181G1 G17 495F1C;1 @?1A9B1C?A 14118 137. $9;1 BD4DC BD=2D @?1A9B1C?A 41 119B1C?A 14118 =1;19C5B9C1B3181G1B5C518=51D9119B1C?A141183?B B589771 138. 3?B K 139. *5AB1=119949B52DC:D716-6/%.644571 141189C5B9C1B3181G1 G17 5F1C 119B1C?A 41 14118 BD4DC 1C1A1 @?1A9B1C?A 41119B1C?A +1CD1 9C5B9C1B 3181G1 14118 F1CC = Contoh 2.9+525A;1B3181G111=918495F1C;[email protected];[email protected];[email protected][email protected]?1A9B1B9B1CDB1=119=5=25CD;BD4DCJ$9;19C5B9C1B3181G111=918G114118 140. . =C5CD;1 9C5B9C1B 3181G1 G17 C518 =55F1C9 ;54D1 ;131 @?1A?94C5AB52DC%7%571=577D1;1(34%/%%09 [email protected]?58 141. 3?B 142. . = 3?B . = 143. . = $1499C5B9C1B3181G1G17495F1C;114118. = 2. Polarisasi pada Pemantulan dan Pembiasan*5A81C9;1 %/%3 +525A;1B B91A 41C17 G17 495F1C;[email protected] @5A=D;11 29417 21C1B 4D1 =549D= G17 945;B 291BG1 25A2541=9B1G1 144. 41 [email protected];141B52179117949291B;1$9;1 B91A @1CD 41 B91A 291B B197 C571; DADB =5=25CD; BD4DCJ B91A @1CD [email protected] B91A [email protected]?1A9B1B9 951A @?1A9B1B9 [email protected]*141 B11C 9CD BD4DC 41C17 49B52DC BD4DC @?1A9B1B9 +D4DC 41C17 99 49B52DC BD4DC @?1A9B1B9 1C1D BD4DC %/%3 [email protected];1 B91A 41C17 @141 29417 21C1B [email protected];141B52179117949291B;1+5BD194571D;D=+59DB 145. B9 B94571 J 1C1DK=1;1 [email protected];1 @5AB1=11G1 G19CD 146. B9 B9K 147. B9 3?B 148. B93?Bn1n2Gambar 2.18Polarisasi pada (a) pemantulan dan(b) pembiasan. 149. n1n2(1)(2)Gambar 2.19Polarisasi pembiasan ganda.partikel-partikelgasnormalgelombang datangtak terpolarisasigelombang hamburanterpolarisasicahaya alamitak terpolarisasicahayaterpolarisasiE0sumbuanalisatorsumbupolarisator 0 E cos44 Mudah dan Aktif Belajar Fisika untuk Kelas XII 150. C1 K 151. [email protected]?1A9B1B91C1DBD4DC 41 152. B5AC1 14118 945;B 291B =549D= B1CD 41 =549D= 4D13. Polarisasi pada Pembiasan Ganda*141 ;A9BC1 ;1B9C 1) ;D1AB1 +9)=9;1 [email protected] 41 5B3181G1 [email protected] =5711=9 @5=291B1 7141 ;1A51 =5=99;9 4D1 919945;B 291B *141 %/%3 [email protected]; 141 4D1 21791 B91A G1749291B;1+91A 153. C941;=579;[email protected]=291B1=5DADCD;D=+59DB1C1D49B52DCB91A9BC9=5F1B91A =579;DC98D;[email protected]=291B1+59DB1C1D 49B52DC B91A 291B4. Polarisasi karena [email protected];[email protected]?1A9B1B9B5217915;CA?55;CA? 411= @1AC9;5 1;1 [email protected] 41 =5=131A;1;5=219 B521791 41A9 3181G1 %/%3 154. ##!! 179C @141 B91781A9 [email protected]; 25AF1A1 29AD ;1A51 @5A9BC9F1 81=2DA1 *1AC9;[email protected];5D41A1 [email protected] B91A =1C181A9 41 =5=131A;1G1 ;5=219 C5ADC1=1B91A 29ADG1 *141 @179 41 B?A5 81A9 @1AC9;[email protected];5 D41A1 1;1=5781=2DA;1 5298 21G1; 3181G1 29AD B589771 G17 C5AB9B1 41A93181G1 =1C181A9 14118 3181G1 =5A18 D1 C941; =5=99;9 1C=?B65AB589771 C941; [email protected] =5781=2DA;1 3181G1 =1C181A9 )58 ;1A519CD 1C=?B65A D1 C5A981C [email protected]$9;1 3181G1 C941; [email protected]?1A9B1B9 41C17 @141 BD1CD =549D= 71B3181G1 G17 4981=2DA;1 [email protected] [email protected]?1A9B1B9 B521791 1C1D B5DAD8G1A18 @?1A9B1B9 B545=9;91 [email protected] B589771 C571; DADB [email protected] 29417G17 4925CD; ?58 71A9B B91A 41C17 41 71A9B @57981C15. Pemutaran Bidang Polarisasi%/%3 [email protected];1 B52D18 G17 C5A49A9 [email protected][email protected]@141B52D1811CG174957;[email protected];11 45A1:1C 41 ;?C1; 1ADC1 *?1A?94 G17 45;1C 4571 BD=25A3181G1 49B52DC @?1A9B1C?A 41 G17 19G1 14118 119B1C?A'D1=D1 =1C1 49 251;17 119B1C?A C941; =5981C 3181G1 G1741C17 [email protected] *5D:D; 119B1C?A =5D:D;;1 BD4DC 155. ;5=D49149 1C1A1 @?1A9B1C?A 41 119B1C?A 495C1;;1 25:11 ;131 G17 25A9B91ADC1 7D1 181G1 G17 =51D9 @?1A9B1C?A 1;1 =55F1C9 1ADC1 99B525D= [email protected] ;5 119B1C?A +5C518 491=1C9 C5AG1C1 B5;1A17 =1C1=5981C 141G1 3181G1 C5A17 1ADC1 7D1 411= 81 99 25A6D7B9B521719 @5=DC1A 29417 75C1A 71A =5:149 5298 [email protected] 179 [email protected] B589771 =5D:D;;1 BD4DC $149 25B1AG1 BD4DC @DC1A129417 75C1A 3181G1 G17 491;D;1 ?58 1ADC1 7D1 14118 156. K 157. 1ADC1 7D1 C5AB52DC 49B52DC 1ADC1 1ADC1 C5AB52DC141 G17 [email protected] =5=DC1A 29417 75C1A @?1A9B1B9 ;5 ;9A9 41 141 :D71G17 ;5 ;11 571 11C B5=131= 99 ?A17 [email protected] =55CD;1Gambar 2.20Polarisasi karena hamburan.Gambar 2.21Pemutaran bidang polarisasi. 158. Gelombang Cahaya 45;?B5CA1B9 1ADC1 [email protected]; 1;C96 *?1A9=5C5A G17 ;8DBDB DCD; =55CD;1 ;?B5CA1B9 1ADC1 7D1 49B52DC1A9 25A21719 @5A3?211 [email protected];1 218F1 61;C?A61;C?A [email protected]:[email protected]:17 41 ;?B5CA1B9 1ADC1 +531A1 =1C5=1C9B [email protected] 49CD9B;1B521719 25A9;DC 159. K 160. %5C5A171 161. [email protected] 29417 75C1A;?B5CA1B9 [email protected]:17 1ADC1 C521BD4DC @DC1A1 :59B 1ADC1Contoh 2.10$9;1945;B291BD41A114118 162. [email protected][email protected]?;5B14118J,5CD;1945;B291B21?;5BC5AB52DC%7%9;5C18D9J 163. 571=577D1;1(34%/%%[email protected]?58C1 164. C1 165. C1J 166. $149945;B291B21?;5B14118 167. Contoh [email protected]:[email protected]@DC1A1:59B1ADC1G1J$9;1497D1;[email protected][email protected]?1A9B1B9G1 J89CD718;?B5CA1B91ADC19CD%7%9;5C18D9 3== J 168. J%[email protected]@5A?584571(34%/%%[email protected]?58 169. = 170. $149;?B5CA1B91ADC114118 171. L 172. 9;[email protected][email protected]?;5B14118J$9;1945;B291BD41A1 173. C5CD;118945;B291B21?;5BC5AB52DC @12919C5B9C1B3181G1G17;5D1A41A94D1;[email protected][email protected]=5=25CD;BD4DC 174. JB1CDB1=11941A93181G1=D1=D1,5CD;125B1AG1BD4DCG174925CD;?58;54D1;[email protected]?1A?94C5AB52DCKata Kunci bidang polarisasi dichroic polarisator analisator sudut polarisasi/sudut Brewster hamburan polarimeter larutan optik aktif sacharimeterTes Kompetensi Subbab C(3,%-%0.%*'%.%/6-6.%5+*%0 +52D18B1381A9=5C5A=5=99;[email protected]:17G1 3=25A9B91ADC17D14571;[email protected];1C1 175. 41=5=99;[email protected]:59BJ,5CD;[email protected][email protected]?1A9B1B9G1:9;[email protected];1B91A1CA9D= 176. #C5A65A5B9Refleksi (.1/%0).(-531/%)0(5+- 3?C?8G1 181G1*[email protected]*5=291B1!141Setelah mempelajari bab ini, tentu Anda menjaditahu bahwa cahaya merupakan gelombang elektro-magnetikyang dapat mengalami proses interferensi,difraksi, dan polarisasi. Dari semua materi pada bab ini,46 Mudah dan Aktif Belajar Fisika untuk Kelas XII 177. 3?B 4571 BD4DC G17*5=DC1A19417*?1A9B1B9bagian mana yang menurut Anda sulit dipahami? CobaAnda diskusikan bersama teman atau guru Fisika Anda.Selain itu, coba Anda sebutkan manfaat yang Andaperoleh setelah mempelajari materi bab ini.Rangkuman 178. 181G1 C5A=1BD; 75?=217 55;CA?=175C9;[email protected]=21C1G1C941;=5=5AD;1=549D= #C5A65A5B9 14118 @5A9BC9F1 @57712D71 4D175?=2171C1D529841A975?=217G17;?85A5 [email protected]:9;1BD=25A3181G1G1;?85A51AC9G1=5=99;96A5;[email protected]+D=25A3181G1G17;[email protected][email protected]?211$160)413(40(.. #C5A65A5B93181G1=5781B9;[email protected][email protected]*[email protected];141A99C5A65A5B945BCAD;C96B197=55=18;11;[email protected]?=217G17=5=99;961B525A1F11*5AB1=11B59B98:1A1;[email protected]?=21714118B9 179. 4571 180. *?1C5A174981B9;141A99C5A65A5B9;?BCAD;C96B197 =57D1C;1 1;921C @57712D71 4D175?=217G17=5=99;961B5B1=1*5AB1=11B59B98:1A1;[email protected]?=21714118B94571 181. $1A1;[email protected];[email protected]?1C5A17;514118#$1A1;[email protected];[email protected][email protected];514118 182. # 96A1;B9 75?=217 14118 @A?B5B @5=25?;175?=217G1749B5212;[email protected]@1 3518 1C1D BD4DC @578117 G17=5781179 B521791 =D;1 75?=217 51845=9;9149B52DC;9B9496A1;B9$1A1;1C1A3518411=;[email protected];9B9 *141496A1;[email protected]@;5C5A:149:9;1B94571 183. B5417;[email protected];5C5A:149:9;1B9 184. 4571 K 185. 141182917171:9 186. 96A1;[email protected];9B9C5A:149:9;13181G1495F1C;[email protected]=5=99;9:1A1;G17B1=1 1G1 DA19 [email protected]; 14118 ;[email protected] B52D18 5B1DCD; =5=9B18;1 21G171 41A9 4D1 C9C9; [email protected][email protected]:1A1;=99=D= 187. *[email protected][email protected]?=217 1C DCD; =5G1A97 1A18 75C1A [email protected]?1A?94B118B1CD3?C?8G114118;A9BC1 188. #C5B9C1B3181G1G17495F1C;[email protected]@?1A9B1C?A;[email protected]@?1A?9414118 189. #C5B9C1B3181G1G17495F1C;[email protected]?A141184925CD; 1C1A1 BD=2D @?1A9B1C?A 41 BD=2D119B1C?A96A1;B9=5=2181B96A1;B9518,D77196A1;[email protected]%9B9*?1A9B1B9=5=2181B*[email protected]%A9BC1*[email protected]*5=1CD141*5=291B1C5A49A9 1C1B#C5A65A5B9'99=D=#C5A65A5B9'1;B9=D=*?1A9B1B9;1A511=2DA1Peta [email protected] =5711=9 190. Gelombang Cahaya 47Tes Kompetensi Bab 2 +.+*.%*4%.%*4%56,%7%%08%0)2%.+0)5(2%5'%0-(3,%-%0.%*2%'%6-6.%5+*%0 191. [email protected];B9?A45;54D1G174981B9;1?58;9B94571 3518 3=B525B1AJ'1;[email protected]:[email protected];1141181 I2 I3 I4 I5 I $9;1B91AG17:1CD8C571;[email protected]=D;11=9G1;[email protected]=9G1;945;B291BG1 41 @1:17 75?=217 41A9 3181G1 G17 =5781B9;19C5A65A5B9=1;B9=D==1;[email protected]=11G17=5=5D89141181 2 192. 34 193. 5 -CD;=55CD;[email protected]:1775?=217=??;A?=1C9B497D1;1 @5A3?211 0?D7 G17 41C1G1 B52171925A9;DC$1A1;1C1A1;54D13518G1==:1A1;3518;51G1A3=B5AC1:1A1;[email protected];5 [email protected];[email protected] 194. ==*1:17 75?=217 B91A =??;A?=1C9B C5AB52DC141181 = 4 =2 = 5 =3 = *[email protected]?2110?D74D1351825A:1A1; 195. ==495C1;;[email protected]:1A1; 196. =41A9B52D181G1A$9;1:1A1;C5A45;[email protected][email protected]=141 71A9B C5A17 ;5B5251B 14118== @1:1775?=2173181G1G17=5G91A9141181 197. I 4 I2I 5 I3 I *[email protected]=25A3181G1;?85A5141181 ;54D1G1B171C25A45;1C12 [email protected]?G1B1=13 [email protected]=14 254161B5;[email protected] ;54D1G1=5=131A;[email protected]@1B1 939(5FC?C5A:149;1A5175:111 496A1;B92 @?1A9B1B93 [email protected] 9C5A65A5B95 A56A1;B9 *[email protected]?2110?D735187141:9;1:1A1;1C1A14D13518G149:149;14D1;19B5=D1=1;1:1A1;[email protected] ;19B5=D12;19B5=D13 M;19B5=D14 N;19B5=D15 [email protected];25AD218 @12919C5B9C1B3181G1=D1=D1G17=55F1C94D1 @1CA5C1A41B9 141181C1A1 ;54D1 @1C=5=25CD; BD4DC J 9C5B9C1B 3181G1 G17495F1C;1?58;[email protected] 198. 4 199. 2 200. 5 201. 3 [email protected]?1A?94=5=99;[email protected][email protected]=5=25CD;BD4DCJ9C5B9C1B3181G1G1749C5ADB;11;1B52149745711 202. 4 203. 2 5 204. 3 205. [email protected]@[email protected]:1A1; ==49B91A9C571;DADB!1A9BC5A17;5C971C5A5C1;==41A971A9BC5A17;[email protected]:1A1;G1 206. =41A9 3518 *1:17 75?=217 B91A G17 [email protected];19141181L 207. K==2 L 208. K==3 209. L 210. K==4L 211. K==5 L 212. K== 213. +525A;1B B91A =??;A?=1C9B 4571 @1:1775?=217L 214. K=41C17C571;[email protected];9B9$9;[email protected];CAD=?A45;[email protected];9B9=1;1:[email protected]=;9B9141181L 215. 4L 216. 2 L 217. 5 L 218. 3 L 219. 1?;;1311;1=5781B9;[email protected]@?1A9B1B9951A:9;[email protected]=5=25CD;BD4DC1 J 4 J2 J 5 220. J3 J 221. %7%.%*2(35%08%%0(3+-65+0+'(0)%05(2%5 222. D135184571:1A1; ==49B91A9C571;DADB!1A9BC5A17;5C971C5A5C1;==41A971A9BC5A17;[email protected]:1A1; 223. [email protected]@1:[email protected];19 [email protected]@9B18 224. ==$9;1:1A1;[email protected]@ 1G1A 225. 3= 41 :1A1; 1C1A1 71A9B [email protected]@[email protected]=114118 [email protected];18 @1:17 75?=217 3181G1 G17497D1;148 Mudah dan Aktif Belajar Fisika untuk Kelas XII $9;[email protected][email protected];1;5411=19A21719=11;[email protected]@1419C5A65A5B9G17C5A:149 +52D18B1381A9=5C5A=5=99;9C12D7G17499B91ADC17D1 @1:17G1 3= %?B5CA1B9 7D1 G17497D1;1 226. [email protected]:59B1ADC114118 @5A 227. 3=$9;1497D1;1B91A1CA9D=C5CD;1 @5=DC1A1 29417 @?1A9B1B9 3181G1 ?581ADC1 228. +525A;1B 3181G1 =??;A?=1C9B 49:1CD8;1 @[email protected];125A45;1C14571:1A1; 229. ==*?19C5A65A5B9G17C5A:14949C17;[email protected]@141:1A1; 3=41A935189;5C18D9218F1:1A1;[email protected]@5AC1=149B52518;9A9;[email protected]@5AC1=1 49 B52518 ;11 14118 == *1:1775?=21725A;1B3181G1141181 230. =2=3 =4=5 231. =!()+10%. 232. +D1CD3181G1=55A17935187141G17=5=99;9:1A1;1C1A3518 233. 3=B545=9;9189771C5A25CD;@[email protected]@1411G1AG1725A:1A1;3=%5C9;[email protected][email protected]?1C5A17141183==1;[email protected]:1775?=2173181G1G17497D1;1C5AB52DC141181= 4 =2 = 5 =3 =# %810 234. [email protected];CAD=?A45;[email protected]=5A18?A45;[email protected];[email protected];1;[email protected]@1:1775?=217B91AD7D41B91A=5A1814118123 4 5 B5=D1:1F121B118