Fisika · b_Bab 2 Gelombang Cahaya

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24/08/2021 16:12:45

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a_Bab 1 Gejala Gelombang b_Bab 2 Gelombang Cahaya c_Bab 3 Gelombang Bunyi d_Bab 4 Medan Listrik e_Bab 5 Medan Magnetik f_Bab 6 Elektromagnetik g_bab 7 Radiasi Benda Hitam h_Bab 8 Fisika Atom i_Bab 9 Teori Relativitas Khusus j_Bab 10 Fisika Inti dan Radiaktivitas

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Gelombang Cahaya31A. InterferensiCahayaB. Difraksi CahayaC. 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:31Gelombang Cahaya 5>?=5>165>?=5>1C5>C1>73181G1B5A9>7>41:D=@1941<1=;5894D@1>B581A981A9 %5C9;1CDAD>8D:1>@141B91>781A9;141>7;141>7;9C141@1C=5>9;=1C99>418>G1@5<1>79G1>7=5>7891B935A18>G1<1>79C @1;18B525>1A>G1@5<1>799CD181G1=5AD@1;1>75<?=21>75<5;CA?=17>5C9; *14112 >41C5<18=5=@5<1:1A975<?=21>7=5;1>9;=9B1<>G175<?=21>719A41>75<?=21>7C1<9 @1;18@5A25411>1>C1A175<?=21>75<5;CA?=17>5C9;41>75<?=21>7=5;1>9;'1>DB91C5<18=5=1>611C;1>75<?=21>75<5;CA?=17>5C9;41<1=2941>7CA1>B@?AC1B91BCA?>?=9=9<9C5A41>75?7A169 ,18D;18>4121719=1>175:1<141>39A939A975<?=21>75<5;CA?=17>5C9;;8DBDB>G13181G1B5AC1@5>5A1@1>>G1$1F121>1C1B@5AC1>G11>@5AC1>G11>C5AB52DC41@1C>41C5=D;1>@1412129>9 )<58;1A5>19CD@5<1:1A92129>945>71>219; Cahaya merupakan gelombang elektromagnetik yang banyakdigunakan untuk kepentingan teknologi komunikasi.Bab2Sumber: Physics Today, 1995

Mudah dan Aktif Belajar Fisika untuk Kelas XII32ABCgelombangdatangS2S1terangterangterangterangterangterangterangterangterangterangterangterangteranggelapgelapgelapgelapgelapgelapgelapgelapgelapgelapgelapgelapS0Gambar 2.1Interferensi pada celahganda YoungA. Interferensi Cahaya 141<18 "!! !# 1AC9>G175<?=21>7=5=9<9;9! 41># +5@5AC9G1>7C5<18492181B@141@5=2181B1>75<?=21>7218F175<?=21>73181G125AB961CB5@5AC981<>G175<?=21>72D>G9G19CD41@1C25A9>C5A65A5>B9 )<58;1A5>19CDD>CD;=5>41@1C;1>9>C5A65A5>B93181G1@D>49@5A<D;1>BD=25A3181G1G1>7;?85A5>G19CDBD=25A3181G1G1>7=5=9<9;96A5;D5>B9B1=141>254161B5C5C1@ +D=25A3181G1G1>7;?85A5>41@1C491=1C9=5<1<D9@5A3?211>G1>749<1;D;1>?<58$160)41>3(40(..1.Percobaan Young dan Fresnella.Percobaan Celah Ganda Young*5A3?211>9>949<1;D;1>?<58$160)45>71>=5>77D>1;1>4D1@5>781<1>7 *5>781<1>7@5AC1=1=5=9<9;9B1CD<D21>7;539<41>@5>781<1>7;54D149<5>7;1@945>71>4D1<D21>7;539< *5A81C9;1>%/&%3 +9>1A=?>?;A?=1C9B49@5A?<5841A9<1=@DB521719BD=25A3181G1G1>7=5=1>31A=5<1<D935<18+ %5=D491>B9>1A41A935<18+49@1>31A;1>;5@5>781<1>7;54D1 D135<18@141@5>781<1>7;54D1G19CD+41>+G1>749@1B1>7B5:1:1A45>71>+1;1>25A6D>7B9B521719@5=1>31AB9>1AB9>1A;?85A5> %54D125A;1B41A935<1835<18+41>+9>925A9>C5A65A5>B9@141<1G1A "1B9<9>C5A65A5>B925AD@171A9BC5A1>741>71A9B75<1@ b.Percobaan Fresnell5>71>=5>77D>1;1>B52D18BD=25A3181G1+3(40(..=5=@5A?<584D1BD=25A3181G1+41>+G1>7;?85A5>41A981B9<@5=1>CD<1>4D135A=9> *5A81C9;1>%/&%3 Gambar 2.2Percobaan Fresnelluntukmenunjukkan interferensi cahaya.C1S1OPC2S2STo k o hAugustine Fresnell(1788–1827)Augustine Fresnell (1788–1827)adalah fisikawan Prancis yangmengembangkan teori gelombangtransversal cahaya berdasarkan hasilpenemuannya tentang lensa daninterferensi. Fresnellmemperlihatkan bahwa cahayamatahari terdiri atas bermacam-macam warna 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!(&(.6//(/2(.%,%3+-104(2(.1/&%0)%*%8%-(3,%-%0.%*41%.41%.&(3+-65'%.%/&6-6.%5+*%0Tes Kompetensi Awal @1;18G1>749=1;BD445>71>75<?=21>75<5;CA?=17>5C9; @1;183181G1C5A=1BD;75<?=21>75<5;CA?=17>5C9; 5A1@1;18;535@1C1>3181G141<1=AD1>781=@1 +52DC;1>B961CB961C75<?=21>73181G1

Gelombang Cahaya33CBPr2S2S1r1dyDbr2S2r1S1Gambar 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 C5AB52DC41@1C49<981C218F1+141<18BD=25AB9>1A=?>?;A?=1C9B + 41>+141<1821G1>71>41A9+?<5835A=9>41>5>71>45=9;91>B9>1AB9>1AG1>741C1>7@141<1G1AB5?<18?<1825A1B1<41A9+41>+ !5<?=21>73181G141A9+41>+9>91;1>B1<9>725A9>C5A65A5>B9@141<1G1A41>81B9<>G125A71>CD>7@141B5<9B9841A9<9>C1B1>;54D1B9>1A9CD *5A<D>41;5C18D9:9;1;54D1BD=25A3181G1=5=9<9;91=@<9CD4?G1>7B1=1@141C5=@1CC5A:149>G19>C565A5>B9=9>9=D=1;1>C5A25>CD;71A9B75<1@ +521<9;>G1:9;11=@<9CD4?C941;B1=19>C5A65A5>B9=9>9=D=>G1C941;75<1@B1=1B5;1<9 *5A81C9;1>%/&%3 75<?=21>73181G141C1>7=5>D:D35<1841>35<18G1>7C5A<5C1;@1412941>7 181G1C5AB52DCC5A496A1;B9?<58;54D135<18C5AB52DC41>=5>781B9<;1>@?<19>C5A65A5>B9@141<1G1A %9C141@1C=5>5>CD;1>49=1>1B5C91@@9C1C5A1>71C1D@9C175<1@C5A<5C1;@141<1G1A45>71>=5=25A9;1>BD4DC41A9BD=2DC5>718C5A8141@71A9B75<1@1C1D71A9BC5A1>7C5AB52DC ->CD;=5>5>CD;1>25B1A>G1;9C181ADB=5>78D2D>7;1>>G145>71> *5A81C9;1>%/&%3 & ,9C9;=5AD@1;1>B52D18C9C9;G1>7C5A<5C1;@141B545=9;91>AD@1B589>771:1A1;2;5*B1=145>71>;5* 45>71>45=9;91>B1=145>71>:1A1;41A9;5 "D2D>71>1>C1A1:1A1;41A9;541>9>9B1>71CAD=9C (1=D>;9C141@1C=5>G545A81>1;1>>G145>71>=5>71>771@218F1BDBD>1>:1A1;35<18C5A8141@<1G1A:1D8<529825B1A41A9@141:1A1;1>C1A1;54D135<18 1A971=21AC5A<981C218F1B9>1A75<?=21>741>141<18B5:1:1A41>=5=25>CD;BD4DCC5A8141@BD=25A@DB1C *5A81C9;1>%/&%3 & ,5A>G1C1BD4DCG1>74925>CD;?<58BD=2D@DB1C41>BD=2D 5>71>45=9;91>;9C141@1C45>71>=D418=5>5>CD;1>218F1B9>B9> K (34%/%%0 9=5AD@1;1>@5AB1=11>D>CD;=5>5>CD;1>:1A1;C5=@D81>C1A1B9>1A41>C5A8141@* ->CD;9>C5A65A5>B9=1;B9=D=9>C5A65A5>B9;?>BCAD;C96C5<1849;5C18D9218F1@1BC9>?<1C1D29<1>71>75>1@41A9@1>:1>775<?=21>7 (34%/%%0 941@1C49CD<9BB52171925A9;DC 12bLPembahasan 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 nmd.720 nmb.270 nme.1.800 nmc.360 nmSPMB 2003Pembahasan:Jarak pola gelap ke-1 ke pusatadalah337,2 10 m3, 6 10 m2yDari soal diketahui:m = 1; d = 10–5 m; D = 0,2 msehingga12ydmD12ydDm353, 6 10m 10m10,2 m2353, 6 10m 10m10,2 m273, 6 10m = 360 nmJawaban: c

Mudah dan Aktif Belajar Fisika untuk Kelas XII34B9> K 45>71> ->CD;C5A1>7@DB1C;9C1=5=25A9;1>@9C1C5A1>7@5AC1=1 41>B5C5ADB>G1 41@D>D>CD;9>C5A65A5>B9=9>9=D=9>C5A65A5>B945BCAD;C96@1BC929<1>71>71>:9<;1<9B5C5>71875<?=21>7(34%/%%0 941@1C49CD<9B;1>B52171925A9;DC B9> K45>71> ->CD;75<1@@5AC1=1;9C1=5=25A9;1>@9C175<1@;54D1 41>B5C5ADB>G1 2.Menentukan Jarak Pita Terang ke-m atau Pita Gelap ke-mdari Terang Pusat*141@5=2181B1>B525<D=>G1C5<1849B52DC;1>218F1@?<19>C5A65A5>B9@141<1G1A25AD@1@9C1C5A1>741>@9C175<1@ *5A81C9;1>%/&%3 %9C141@1C=5>5>CD;1>;54D4D;1>@9C1C5A1>7;51C1D@9C175<1@;5@141<1G1A *5A81C9;1>;5=21<9%/&%3 % )<58;1A5>1:1D8<529825B1A41A9@141BD4DC25A>9<19B1>71C;539< ->CD;BD4DCG1>7B1>71C;539<1;1>25A<1;DB9>C1> 1A9%/&%3 %;9C141@1C=5>5>CD;1>218F1B9>C1># K->CD;@9C1C5A1>7=1BD;;1>(34%/%%0 9;5(34%/%%0 9 B589>77149@5A?<58# K->CD;@9C175<1@=1BD;;1>(34%/%%0 9;5(34%/%%0 9 B589>77149@5A?<58 # K%5C5A1>71>:1A1;1>C1A35<18@141<1G1A#:1A1;C5A1>7 75<1@;541A9@DB1C:1A1;<1G1A;535<18@1>:1>775<?=21>73181G11<1=81<9>9 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.

Gelombang Cahaya35->CD;=5>5>CD;1>@1>:1>775<?=21>7B9>1AG1>749@1>31A;1>?<58<1=@D@9:1A>1CA9D=B9>1A9>949<5F1C;1>@1414D135<18G1>725A:1A1;== *141:1A1; =5C5A41A935<1849@1B1>7<1G1A $9;181B9<9>C5A65A5>B9@141<1G1A49@5A?<58:1A1;71A9BC5A1>7@DB1CB1=@1945>71>;5<9=1141<18==25A1@1;18@1>:1>775<?=21>7B9>1A>1CA9D=C5AB52DC%7%&9;5C18D9==L K=#==L K= = k# = =L K I$149@1>:1>775<?=21>7B9>1A>1CA9D=141<18 I Contoh 2.1*141B52D18@5A3?211>9>C5A65A5>B9497D>1;1>4D135<18B5=@9C $1A1;1>C1A1;54D135<189CD ==41>49<5C1;;1>@141:1A1; =;5<1G1A71A9B75<1@@5AC1=141A9@DB1CG1>7:1A1;>G1== "9CD>7<18@1>:1>775<?=21>73181G1>G1 %7%& == L K= =#==L K=9>C5A65A5>B975<1@1A9(34%/%%0 9 49@5A?<58# =K = = = K L K=L K= I$149@1>:1>775<?=21>73181G1>G1141<18 I Contoh 2.23.Interferensi oleh Lapisan Tipis*5=1>CD<1>3181G1=1C181A9?<58! #G1>74931=@DA45>71>19A1;1>=5=@5A<981C;1>71A9B71A9B25AF1A>1@141=9>G1; +@5;CAD=F1A>19>9=5=@5A<981C;1>141>G1 " ?<58<1@9B1>=9>G1;G1>7C9@9B9CD #>C5A65A5>B9C5AB52DC41@1C25AD@19>C5A65A5>B9=1;B9=D==1D@D>9>C5A65A5>B9=9>9=D= #>C5A65A5>B91>C1A175<?=21>7G1>749@1>CD<;1>?<58<1@9B1>C9@9B49CD>:D;;1>@141%/&%3 +525A;1BB9>1A41C1>7=5>75>19<1@9B1>C9@9B45>71>BD4DC41C1>71;1>49291B;1>41>B521791><17949@1>CD<;1>;5=21<9;5@5A=D;11> +9>1AG1>749@1>CD<;1>49<5F1C;1>@141B52D18<5>B1@?B9C9641>496?;DB;1>49C9C9;* 5A;1B3181G149C9C9;*=5AD@1;1>81B9<9>C5A65A5>B925A;1B3181G1 41> 45>71> 141<1825A;1B3181G1G1>749@1>CD<;1><1>7BD>741> 141<1825A;1B3181G1G1>7=5>71<1=9@5=291B1>C5A<5298418D<D;5=D491>49@1>CD<;1> Gambar 2.4Interferensi oleh lapisan tipis.PDABCidnrlensalapisantipis(1)(2)

Mudah dan Aktif Belajar Fisika untuk Kelas XII36+5<9B98<9>C1B1>?@C9;G1>749C5=@D8?<58B9>1A41C1>789>771=5>:149B9>1A@1>CD<;5 41>B9>1A@1>CD<;5 141<18K K K 45>71>141<189>45;B291B<1@9B1>C9@9B41> '9B1<;1>C521<<1@9B1>141<18=1;13?BB589>771 3?B41>B9>45>71> C1>B589>771 3?BK C1>B9>3?BK B9>B9>3?B5>71>=5>77D>1;1>6-6/!0(..+64C5>C1>7@5=291B1>3181G1G1;>9B9>B9>49@5A?<58B5<9B98:1A1;C5=@D8;54D1B9>1A=5>:1493?BK B9>3?B3?B KB9>3?B3?B 3?B K+D@1G1C5A:1499>C5A65A5>B9=1;B9=D=49C9C9;*81ADB=5AD@1;1>;5<9@1C1>41A9@1>:1>775<?=21>7 ;1>C5C1@9B9>1A@1>CD<49=5>71<1=9@5AD2181>61B5=1;11;1>=5>:1491C1D K#>C5A65A5>B9=1;B9=D=B9>1A@1>CD<@141<1@9B1>C9@9B1;1>=5=5>D89@5AB1=11>3?B K45>71> (34%/%%0 925A<1;DD>CD;9>45;B291B<1@9B1>C9@9B<529825B1A41A9 1C1D 41@D>D>CD;=5=@5A?<589>C5A65A5>B9=9>9=D=;54D1B9>1A@1>CD<81ADB=5=9<9;9254161B5=1;1 #>C5A65A5>B9=9>9=D=41<1=1A18@1>CD<=5=5>D89@5AB1=11> 3?B41>1C1D 3?B K 45>71>9>45;B291B41>=5=5>D89BG1A1C41>> 1C1D 41>Gambar 2.7Interferensi oleh lapisan minyakyang tipis.Sumber:www.instckphoto.comGambar 2.6Interferensi oleh lapisanbusa sabun yang tipis.Sumber:www.funsci.com,5>CD;1>C521<<1@9B1>=9>9=D=G1>7492DCD8;1>BD@1G1C5A:1499>C5A65A5>B9@141B52D18<1@9B1>C9@9BG1>7=5=9<9;99>45;B291B45>71>@1>:1>775<?=21>7 I Contoh 2.3Gambar 2.5Interferensi oleh busa sabun.Sumber:www.designprodygzone.com

Gelombang Cahaya37Tugas Anda 2.1Coba Anda perhatikan kembaliGambar 2.5. Gelembung tersebutsebenarnya berwarna-warni.Mengapa demikian? Anda dapatmencari jawabannya dari bukureferensi atau internet.%7%&#>C5A65A5>B9=1;B9=D=@141<1@9B1>C9@9B=5=5>D89@5AB1=11>3?B 3?B+D@1G1C521<<1@9B1>=9>9=D=B5C9@9BC9@9B>G1=1;1 41>3?B B589>77149@5A?<58 I I I$149C521<<1@9B1>C9@9BG1>7492DCD8;1>141<18 I ,5>CD;1>@1>:1>775<?=21>7B9>1AG1>7497D>1;1>:9;1C5A:1499>C5A65A5>B9=9>9=D=?A45;54D141A9<1@9B1>C9@9B49D41A145>71>;5C521<1> >=BD4DC291BJ41>9>45;B291B<1@9B1> %7%&5>71>=5>77D>1;1>(34%/%%0 949@5A?<583?B >=3?BJ >=$149@1>:1>775<?=21>7G1>7497D>1;1>141<18 >= Contoh 2.4Kata Kunci• interferensi• sinar monokromatis• interferensi maksimum• interferensi minimumTes Kompetensi Subbab A(3,%-%0.%*'%.%/&6-6.%5+*%0 @1G1>749=1;BD49>C5A65>B93181G1 +52D1835<1871>4125A:1A1;==4925<1;1>735<1841>@141:1A1; =5C5A49<5C1;;1>B52D18<1G1A 5<1849B9>1A94D1B9>1A=?>?;A?=1C9B45>71>@1>:1>775<?=21>7>=41>>= 5A1@1;18:1A1;71A9BC5A1>7?A45;55=@1C;54D1B9>1A@141<1G1A +525A;1B3181G1=5<5F1C94D135<18B5=@9CG1>7B1CDB1=1<19>25A:1A1;== $1A1;35<18;5<1G1A =5C5A41>:1A1;1>C1A14D171A9BC5A1>7@141<1G1A141<18 L K 3= 5A1@1;18@1>:1>775<?=21>73181G1G1>7497D>1;1> ->CD;=5>7D;DA@1>:1>7B9>1A=5A1849<1;D;1>@5A3?211>B52171925A9;DC +9>1A29AD45>71>@1>:1>775<?=21>7>=49:1CD8;1>C571;<DADB@14135<1871>41 *?<19>C5A65A5>B9C5A:149@141<1G1AG1>725A:1A1; =41A935<18 !1A9BC5A1>7?A45@5AC1=125A:1A1;==41A971A9BC5A1>7@DB1C +5C5<189CDB9>1A=5A1849:1CD8;1>@14135<18 ,5A>G1C171A9BC5A1>7?A45@5AC1=125A:1A1;==41A971A9BC5A1>7@DB1C ,5>CD;1><18@1>:1>775<?=21>7B9>1A=5A189CD +52D18<1@9B1>C9@9B=5=9<9;99>45;B291B 497D>1;1>D>CD;=5<981C75:1<19>C5A65A5>B9 $9;1@1>:1>775<?=21>7>G1 IC5>CD;1>C521<=9>9=D=<1@9B1>C5AB52DCBD@1G1C5A:1499>C5A65A5>B9 B. Difraksi Cahaya*141@5<1:1A1>75C1A1>41>75<?=21>749%5<1B/C5<18492181B218F175<?=21>719AG1>7=5<5F1C9B52D18@5>781<1>745>71>B52D1835<18B5=@9C1;1>=5>71<1=9<5>CDA1> !5<?=21>7G1>741C1>741@1C25A25<?;B5C5<18=5<1<D935<18C5AB52DC *5=25<?;1>75<?=21>7G1>749B5212;1>?<58141>G1@5>781<1>725AD@135<1849B52DC496A1;B975<?=21>7 +1=181<>G145>71>75<?=21>73181G1G1>749<5F1C;1>@141B52D1835<18B5=@9C:D711;1>=5>71<1=9<5>CDA1>

Mudah dan Aktif Belajar Fisika untuk Kelas XII3896A1;B93181G1C5A:149:D71@14135<18B5=@9CG1>7C5A@9B18B5:1:1AB1CDB1=1<19>@141:1A1;G1>7B1=1 5<18B5=@9CG1>745=9;91>49B52DC %9B9141<18;5@9>71>;131G1>7497?A5B=5>DADC71A9BB5:1:1A41>21>G1;:D=<18>G1 $1A1;1>C1A14D135<1849B52DCC5C1@1>;9B9 1.Difraksi Celah Tunggal96A1;B9@14135<18CD>771<1;1>=5>781B9<;1>@?<171A9BC5A1>741>75<1@@141<1G1A 5<18CD>771<41@1C491>771@C5A49A91C1B2525A1@135<18B5=@9CG1>74921C1B9C9C9;C9C9;41>B5C91@35<189CD=5AD@1;1>BD=25A3181G1B589>771B1CDB1=1<19>>G141@1C25A9>C5A65A5>B9 *5A81C9;1>%/&%3 ->CD;=5>71>1<9B9B@?<1496A1;B935<18@141%/&%3 4921794D121791> *5A81C9;1>75<?=21>7 41> !5<?=21>7 =5>5=@D8<9>C1B1>G1>7<5298:1D8B525B1AB9>41A9@14175<?=21>7 +1=181<>G145>71>75<?=21>7 41>G1>7=5=9<9;92541<9>C1B1>B525B1AB9> #>C5A65A5>B9=9>9=D=G1>7=5>781B9<;1>71A9B75<1@C5A:149:9;1;54D175<?=21>725A254161B5 J1C1D2541<9>C1B1>>G1B1=145>71>@1>:1>775<?=21>7 B9>B9>$9;135<184921795=@1C21791>4941@1C71A9B75<1@;5C9;1B9>B9> "1<B5AD@145>71>9CD:9;1C5<184921795>1=21791>4941@1C71A9B75<1@;5C9;1B9>B9> +531A1D=D=41@1C49>G1C1;1>218F1@9C175<1@;5C5A:149:9;1B9> K %5C5A1>71><521A35<18BD4DCB9=@1>745E91B9 ->CD;1C1DC5A:149=1;B9=D=DC1=1@9C1C5A1>7C5>718B5@5AC949@5A<981C;1>@141%/&%3 Gambar 2.9Maksimum utama terjadiuntuk k = 0 atau = 0.maksimumutama5>71>=5>77D>1;1>@5>781<1>735<18CD>771<@141<1G1AC1=@1;@?<1496A1;B971A9BC5A1>7@DB1C41>71A9B75<1@;55=@1C=5=25>CD;BD4DCJC5A8141@71A9B>?A=1< $9;13181G1G1>7497D>1;1>=5=9<9;9@1>:1>775<?=21>7 IC5>CD;1><521A35<18G1>7497D>1;1> Contoh 2.5Gambar 2.8Difraksi cahaya pada celah tunggal.d543212d2dsin2d

Gelombang Cahaya39garis gelapterangdpusatQP->CD;=5>41@1C;1>@?<1496A1;B9=1;B9=D=2541<9>C1B1>41A99>C5A65A5>B9=9>9=D=81ADB49;DA1>7945>71> )<58;1A5>1;54D13181G1B561B5254161B5;54D1>G1=5>:149J D175<?=21>745>71>254161B5 1C1D2541BD4DC61B5J49B52DC:D71B561B5 *5AB1=11>9>C5A65A5>B9=1;B9=D=41A9@?<1496A1;B9>G11;1>=5>:149B9>B9>1C1DB9> K K 141<1829<1>71>71>:9< %7%&5>71>=5>77D>1;1>(34%/%%0 949@5A?<58B9>B9>J I=1;1 I$149<521A35<18>G1141<18 I ,5>CD;1><521A35<18=9>9=D=G1>7492DCD8;1>@141496A1;B935<18CD>771<:9;1499>79>;1>BD4DC496A1;B9>G1J41>@1>:1>775<?=21>7G1>7497D>1;1>>=D>CD;@?<1496A1;B9=1;B9=D= %7%&5>71>=5>77D>1;1>(34%/%%0 9 49@5A?<58B9> >=>=>= >=B9> $149<521A35<18=9>9=D=141<18 >= 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 × 10–7 mb.6,0 × 10–7 mc.5,2 × 10–7 md.5,6 × 10–7 me.0,4 × 10–7 mPPI 1983Pembahasan:Jarak titik api = jarak celah kelayar = = 40 cm.Gelap pertama m = 1dpm330,4 10 m 0,5 10 m10,4 m= 5,6 × 10–7 mJawaban: dContoh 2.62.Difraksi pada Kisi96A1;B93181G1C5A:149@D<1@1413181G1G1>7=5<1<D921>G1;35<18B5=@9C45>71>:1A1;35<18B1=1 5<18B5=@9CG1>745=9;91>49B52DC;9B9496A1;B91C1D49B9>7;1C;9B9 +5=1;9>21>G1;35<18@141B52D18;9B9B5=1;9>C1:1=@?<1496A1;B9G1>74981B9<;1>@141<1G1A '9B1<>G1@141415A18B5<521A 3=C5A41@1C 35<18 AC9>G1;9B9C5AB52DCC5A49A91C1B 35<18 3=1C1D 35<18 3= 5>71>45=9;91>:1A1;Gambar 2.10Difraksi pada kisiBB1AdCDGFETMPK1>C1A35<18141<18 3=L K3= *5A81C9;1>%/&%3

Mudah dan Aktif Belajar Fisika untuk Kelas XII40Gambar 2.12Difraksi minimum keduauntuk N banyak celah.Gambar 2.13Difraksi cahaya putih akanmenghasilkan pola berupapita-pita spektrum.cahayaputihkisidifraksimerahmerahmerahmerahbirubirubirubiruspektrumorde ke 2spektrumorde ke-1spektrumorde ke-0(putih)spektrumorde ke-1spektrumorde ke-2+52D18;9B945>71> 71A9B 3=49<5F1C;1>3181G1C571;<DADB45>71>@1>:1>775<?=21>7 !1A9BC5A1>7496A1;B9?A45@5AC1=1=5=25>CD;BD4DCJC5A8141@71A9B>?A=1<=1;B9=D=DC1=1 ,5>CD;1><18@1>:1>775<?=21>7 %7%&9;5C18D9 71A9B 3= L K3=B9>J 5>71>=5>77D>1;1>(34%/%0 9 49@5A?<58B9> L K3= L K3= I$149@1>:1>775<?=21>7G1>7497D>1;1>141<18 I %/&%3 =5=@5A<981C;1>B525A;1BB9>1A=?>?;A?=1C9BG1>749<5F1C;1>@141B52D18;9B941>=5>781B9<;1>@?<1496A1;B9@141<1G1A *?<1496A1;B925AD@171A9BC5A1>741>71A9B75<1@B531A125A71>C91> 96A1;B9=1;B9=D=C5A:149:9;1@141<1G1AC1=@1;71A9B71A9BC5A1>7 541<9>C1B1>G1>749<5F1C93181G1G1>741C1>741A94D135<1825A45;1C1>141<18 1C1D29<1>71>31318;1<9@1>:1>775<?=21>7>G1 *?<1496A1;B9=1;B9=D=DC1=1@141;9B9141<18B9>49=1>1141<18?A45496A1;B941>141<18:1A1;1>C1A35<181C1DC5C1@1>;9B9 96A1;B9=9>9=D=491>C1A1 =1;B9=D=C5A:149:9;1@141<1G1AC1=@1;71A9B71A9B75<1@41>=9>9=D=@5AC1=1B5BD418=1;B9=D=;5C5A:149:9;1B9>:D71=9>9=D=;54D1:9;1B9>+5217193?>C?8D>CD; 35<1849@5A?<58%/&%3 B541>7;1>D>CD;21>G1;35<1849@5A?<58%/&%3 %/&%3 =5=@5A<981C;1>3181G1@?<9;A?=1C9;@14135<18G1>7=5AD@1;1>3181G1@DC98 +9>1A@DC98@?<9;A?=1C9;C5A49A91C1B25A21719F1A>145>71>@1>:1>775<?=21>7C5A;539<F1A>1D>7D41>C5A25B1AF1A>1=5A18 5>71>45=9;1>F1A>1G1>7C5A45;1C45>71>141<18F1A>1D>7D41>G1>7C5A:1D8141<18F1A>1=5A18G1>7=5AD@1;1>B@5;CAD=F1A>1<5>7;1@G19CDD>7D29AD89:1D;D>9>7:9>77141>=5A18 +5C91@?A45496A1;B9=5>D>:D;;1>B@5;CAD=F1A>1 Contoh 2.7*5A>18;18>41=5<981C@5<1>79*5<1>79=5AD@1;1>B1<18B1CD75:1<11<1=B52171981B9<496A1;B9 ,D71B>41;D=@D<;1>9>6?A=1B9=5>75>19C5A:149>G1@5<1>79;5=D491>@A5B5>C1B9;1>CD<9B1>>414945@1>;5<1B Mari Mencari TahuGambar 2.11Difraksi minimum keduauntuk N = 2 celah.m = 2m = 12113213sindm = –1m = 0m = 1m = 20sind2Tantanganuntuk AndaTentukan daya urai sebuah celahdengan diameter 2 mm, jarak celahke layar 1 meter dengan panjanggelombang cahayanya 590 nm.intensitas

Gelombang Cahaya413.Daya Urai Optik# ! 141<18;5=1=@D1>B52D18<5>B1D>CD;=5=9B18;1>21G1>71>41A94D1C9C9;G1>7C5A@9B18B1CDB1=1<19>@141:1A1;=9>9=D= %5=1=@D1>@5A25B1A1>1<1C1<1C?@C9;=9B1<>G1<D@=9;A?B;?@41>C5A?@?>74921C1B9?<5841G1DA19<5>B141>:D714921C1B9?<58@?<1496A1;B9G1>7C5A25>CD;@14121G1>71>25>419CD %/&%3 =5=@5A<981C;1>218F121G1>71>G1>7C5A:149=5AD@1;1>@?<1496A1;B9G1>749B5212;1>?<58 !<D1BB9BC5=<5>B1?<581<1C1<1C?@C9;C5AB52DC *?<1496A1;B9G1>74925>CD;?<58B52D1835<182D<1CC5A49A91C1B25>CD;C5A1>7@DB1CG1>749;5<9<9>7939>39>C5A1>741>75<1@ +1<18B5?A1>79<=DF1>13' %8.(+)*=5>G9=@D<;1>218F14D12D18C9C9;BD=25AG1>7C5A1>7>G1B1=11;1>C5A<981CC5A@9B18:9;1=1;B9=D=B5>CA1<1C1D@DB1C@DB1C41A9@?<1496A1;B9G1>7B1CD25AC5@1C1><5C1;>G145>71>=9>9=D=@5AC1=141A9@?<1496A1;B9C9C9;G1>7<19> *?<1C5AB52DC41@1C49:5<1B;1>45>71>=5>77D>1;1>%/&%3 $1A9:1A9<9>7;1A1>C5A1>7G1>7C5A25>CD;41@1C491AC9;1>41G1@9B18@?<1496A1;B9G1>7C5A21C1B $9;13181G1=5<1<D9AD1>781=@11C1DD41A141G1DA1941A935<18<9>7;1A1>41@1C49C5>CD;1>45>71>@5AB1=11> K '5>DADC %8.(+)*41>(%04;A9C5A91:1A1;1>C1A1;54D1=1;B9=D=C5AB52DC@1<9>7;539<B1=145>71>:1A9:1A9<9>7;1A1>C5A1>7 '1;B9=D=G1>7;54D1:1CD8@141=9>9=D=G1>7@5AC1=11C1D:1A1;BD4DC1>C1A1;54D1@DB1C21G1>71>G19CDB9> K ->CD;BD4DCG1>7;539< K %5C5A1>71>41G1DA19=:1A1;25>4141A9<5>B1=@1>:1>775<?=21>73181G1=491=5C5A2D;11><5>B1=BD4DC45E91B9Gambar 2.14Bayangan dari optik fisis dua bendayang berdekatan karena (a) beririsandan (b) terpisah dengan baik.Gambar 2.15Lukisan sinar dari sumber cahayadari sebuah celah bulat.Dcelah bulatpola difraksircahaya sumber datang2s1s2LDContoh 2.8%5C9;1491=5C5A=1C149@5A25B1AB1=@19==25A1@1:1A1;=9>9=D=1>C1A14D1BD=25AC9C9;G1>7=1B9841@1C492541;1>?<58=1C1@141:1A1;3=41A9=1C1*1>:1>775<?=21>73181G149D41A1>=41>9>45;B291B=1C1*5A81C9;1>71=21A25A9;DC Merak 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 peacock’s tail are anexample of diffraction effect ofpeacock’s tail light wakes.Sumber: Biology Concepts & Connections,2006Informasiuntuk AndaInformation for You

Mudah dan Aktif Belajar Fisika untuk Kelas XII42 5<18CD>771<B5<521A ==49B9>1A93181G1G1>7@1>:1>775<?=21>7>G1 I *?<1496A1;B949C1>7;1@@141<1G1AG1>7:1A1;>G13=41A935<18 ,5>CD;1>:1A1;1>C1A171A9B75<1@;5C97141>71A9BC5A1>7@DB1CD>CD;BD4DCG1>7;539<B9> C1> 5A1@1<521A35<18CD>771<G1>749@5A<D;1>BD@1G141@1CC5A:1499>C5A65A5>B9=1;B9=D=?A45;5C97145>71>BD4DC496A1;B9J41A9B525A;1BB9>1A=?>?;A?=1C9BG1>7=5=9<9;9@1>:1>775<?=21>7 I ,5>CD;1><1841G1DA19B52D1835<1845>71>491=5C5A ==41>:1A1;35<18;5<1G1A =5C5A45>71>@1>:1>775<?=21>73181G1>G1>= $1A1;4D1<1=@D=?29<141<18 =5C5A41>491=5C5A@D@9<=1C1B5?A1>71>1; == $9;1@1>:1>775<?=21>73181G1G1>749@1>31A;1>;54D1<1=@DC5AB52DCA1C1A1C1 I25A1@1;18:1A1;=?29<=1;B9=D=BD@1G1>G1<1<1=@D9CD=1B9841@1C49@9B18;1>?<58=1C1 +5?A1>71>1;=5=2D;1=1C1<521A<521AB589>771491=5C5A9A9B>G13= $9;1497D>1;1>B9>1A;D>9>745>71>@1>:1>775<?=21>7 I25A1@1;1841G1@9B18=1C11>1;C5AB52DC41<1=A1491>41>45A1:1CTes Kompetensi Subbab B(3,%-%0.%*'%.%/&6-6.%5+*%0C. Polarisasi Cahaya1.Polarisasi pada Kristal181G1G1>781>G1=5=9<9;91A1875C1A1>C5AC5>CD49B52DC3181G1C5A@?<1A9B1B9 941>7C1=@1;;54D4D;1>1A1875C1A1>C5AC5>CD41A93181G1C5A@?<1A9B1B949B52DC2941>7@?<1A9B1B9 $9;1B52D183181G11<1=91849<5F1C;1>@141B52D18;A9BC1<1A183181G1G1>7;5<D1A41A9;A9BC1<81>G141<1=B1CD1A18B1:1B589>77149B52DC3181G1C5A@?<1A9B1B9<9>51A $9;1;A9BC1<=5>G5A1@B521791>1A1875C1A>G1;A9BC1<9CD49B52DC*5A81C9;1>%/&%3 3181G141C1>7=9B1<>G1B9>1A1<1=9=1C181A9C941;C5A@?<1A9B1B949<5F1C;1>@141B52D18;A9BC1< %?=@?>5>E5AC9;1<49B5A1@?<58;A9BC1<41>3181G1G1>749C5ADB;1>C5A@?<1A9B1B9<9>51A +5<19>;A9BC1<@?<1A?94@D>41@1C=5=2D1C3181G1C5A@?<1A9B1B9 %/&%3 =5=@5A<981C;1>BDBD>1>4D1;5@9>7@?<1A?94B5:1:1A *?<1A?94@5AC1=149B52DC 41>;5@9>7G1>7;54D149B52DC 181G1G1>7;5<D1A41A9@?<1A?9481>G1=5=9<9;9B1CD1A1875C1A1>C5AC5>CD1C1D3181G1C5A@?<1A9B1B9;1A5>11A1875C1A1><19>>G149B5A1@ %/&%3 %=5=@5A<981C;1>@?<1A9B1C?A41>1>1<9B1C?AG1>749@1B1>7Kata Kunci• difraksi gelombang• kisi difraksi• sudut simpang (deviasi)• difraksi maksimum• difraksi celah tunggal• difraksi pada kisi• daya urai optik•apertur%7%&9;5C18D9491=5C5A<5>B1=1C1==L K=@1>:1>775<?=21>73181G1>=L K=9>45;B291B=1C1*1>:1>775<?=21>73181G1;5C9;1=5=1BD;9=1C149@5A?<5845>71> =1C1 = =!$1A1;C9C9;25>41;5<5>B13== 1G1DA19<5>B1=1C141@1C4989CD>7 = = =L K= Gambar 2.16Cahaya tak terpolarisasidilewatkan pada sebuah kristal.getaranhorizontaldiserapsempurnaoleh polaroidcahaya alamiyang datanggetaran vertikaldiserapsebagiancahaya diteruskanterpolarisasi linear

Gelombang Cahaya43Gambar 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.polarisatoranalisatorsumbercahayasumbercahayapolarisatoranalisatortidak adacahayacahaya yangditeruskanterpolarisasi12B5:1:1A1C1DB1=1;54D4D;1>>G13181G1G1>749C5ADB;1>C5A@?<1A9B1B9 *141%/&%3 &1A18CA1>B=9B91>1<9B1C?AC571;<DADB@1411A18CA1>B=9B9@?<1A9B1C?AC941;14175C1A1>G1>741@1C49C5ADB;1>1>1<9B1C?AB589>771>41C941;41@1C=5<981C3181G1 $9;125A;1B3181G11<1=91845>71>9>C5>B9C1B49<5F1C;1>@141B52D18@?<1A9B1C?A9451<9>C5>B9C1B3181G1G1>749<5F1C;1>141<181C1D ;1>C5C1@9:9;1;54D1>G149@1B1>725AB9<1>71>C941;1419>C5>B9C1B3181G1G1>741@1C=5<5F1C91>1<9B1C?A >1<9B1C?A25A6D>7B941<1==5>71>1<9B9BB9>1AG1>749<5F1C;1>?<58@?<1A9B1C?A $9;11>1<9B1C?A49@DC1A@141B11C9CD=1C1=5<981CB9>1AC5A1>7 +9>1A=5A54D@@141B11C@?<1A9B1C?A41>1>1<9B1C?AB1<9>7C571;<DADB41>C1;1413181G1G1>749C5ADB;1> '5>DADC5+(00(16+4%.64 K 9>C5>B9C1B3181G1G1>749<5F1C;1>@?<1A9B1C?A141<18 $9;1BD4DCBD=2D@?<1A9B1C?A41>1>1<9B1C?A141<18=1;19>C5>B9C1B3181G1B5C5<18=5<1<D91>1<9B1C?A141<18 3?BB589>7713?B K *5AB1=11>9>949B52DC:D716-6/%.6445>71>141<189>C5>B9C1B3181G1G1>7<5F1C1>1<9B1C?A41>141<18BD4DC1>C1A1@?<1A9B1C?A41>1>1<9B1C?A +1CD1>9>C5>B9C1B3181G1141<18F1CC =+525A;1B3181G11<1=91849<5F1C;1>@1414D1;5@9>7;131@?<1A?94G1>71A18@?<1A9B1B9B1CDB1=1<19>=5=25>CD;BD4DCJ $9;19>C5>B9C1B3181G11<1=918>G1141<18 . =C5>CD;1>9>C5>B9C1B3181G1G1>7C5<18=5<5F1C9;54D1;131@?<1A?94C5AB52DC %7%&5>71>=5>77D>1;1>(34%/%%0 9 49@5A?<583?B . =3?B. =. =$1499>C5>B9C1B3181G1G1>749<5F1C;1>141<18. =Contoh 2.92.Polarisasi pada Pemantulan dan Pembiasan*5A81C9;1>%/&%3 & +525A;1BB9>1A41C1>7G1>749<5F1C;1>@141@5A=D;11>2941>721C1B4D1=549D=G1>79>45;B291B>G125A2541=9B1<>G141>B521791>B9>1A49@1>CD<;1>41>B521791><17949291B;1> $9;1B9>1A@1>CD<41>B9>1A291BB1<9>7C571;<DADB=5=25>CD;BD4DCJB9>1A@1>CD<25AD@1B9>1AC5A@?<1A9B1B9<9>51A@?<1A9B1B9B5=@DA>1 *141B11C9CDBD4DC41C1>749B52DCBD4DC@?<1A9B1B9 +D4DC41C1>79>949B52DCBD4DC@?<1A9B1B91C1DBD4DC" %/&%3 &=5=@5A<981C;1>B9>1A41C1>7@1412941>721C1B +521791>49@1>CD<;1>41>B521791><17949291B;1> +5BD1945>71>"D;D=+>5<<9DBB9>B9>45>71>J1C1DK=1;141@1C49CD<9B;1>@5AB1=11>>G1G19CDB9>B9>KB9>3?BB9>3?BGambar 2.18Polarisasi pada (a) pemantulan dan(b) pembiasan.1terpolarisasicerminip2n1n2iprterpolarisasi

Mudah dan Aktif Belajar Fisika untuk Kelas XII44C1> K 5>71>141<18BD4DC@?<1A9B1B91C1DBD4DC" 41>B5AC1141<189>45;B291B=549D=B1CD41>=549D=4D1 3.Polarisasi pada Pembiasan Ganda*141;A9BC1<;1<B9C1);D1AB1+9)=9;1C?@1B41>5B3181G141@1C=5>71<1=9@5=291B1>71>41;1A5>1=5=9<9;94D1>9<199>45;B291B *141%/&%3 C1=@1;1414D121791>B9>1AG1>749291B;1> +9>1A C941;=5>79;DC9@5=291B1>=5>DADC"D;D=+>5<<9DB1C1D49B52DCB9>1A9BC9=5F1B9>1A =5>79;DC98D;D=@5=291B1>+>5<<9DB1C1D49B52DCB9>1A291B 4.Polarisasi karena Hamburan181G1G1>741C1>7@141H1C71B1;1>=5>71<1=9@?<1A9B1B9B521791> <5;CA?>5<5;CA?>41<1=@1AC9;5<1;1>=5>G5A1@41>=5=1>31A;1>;5=21<9B521791>41A93181G1%/&%3 # # ! ! &1>79C@141B91>781A9C1=@1;25AF1A>129AD;1A5>1@5A9BC9F181=2DA1> *1AC9;5<@1AC9;5<D41A1=5>G5A1@B9>1A=1C181A941>=5=1>31A;1>>G1;5=21<9C5ADC1=1B9>1A29AD>G1 *141@17941>B?A581A9@1AC9;5<@1AC9;5<D41A11;1>=5>781=2DA;1><529821>G1;3181G129ADB589>771G1>7C5AB9B141A93181G1=1C181A9141<183181G1=5A18 D<1>C941;=5=9<9;91C=?B65AB589>771C941;41@1C=5>781=2DA;1>3181G1=1C181A9 )<58;1A5>19CD1C=?B65AD<1>C5A<981C75<1@ $9;13181G1C941;C5A@?<1A9B1B941C1>7@141BD1CD=549D=71B3181G1G1>74981=2DA;1>41@1CC5A@?<1A9B1B9B521791>1C1DB5<DAD8>G1 A18@?<1A9B1B9B545=9;91>AD@1B589>771C571;<DADBC5A8141@2941>7G1>74925>CD;?<5871A9BB9>1A41C1>741>71A9B@5>7<981C1> 5.Pemutaran Bidang Polarisasi%/&%3 =5=@5A<981C;1>B52D18 G1>7C5A49A91C1B4D12D18@?<1A?94G1>749@1B1>7@141B52D181<1CG1>749<5>7;1@945>71>B;1<145A1:1C41>;?C1;<1ADC1> *?<1A?94G1>745;1C45>71>BD=25A3181G149B52DC@?<1A9B1C?A41>G1>7<19>>G1141<181>1<9B1C?A 'D<1=D<1=1C14925<1;1>71>1<9B1C?AC941;=5<981C3181G1G1>741C1>775<1@ *5>D>:D;1>1<9B1C?A=5>D>:D;;1>BD4DC;5=D491>491>C1A1@?<1A9B1C?A41>1>1<9B1C?A49<5C1;;1>25:1>1;131G1>725A9B9<1ADC1>7D<1 181G1G1>7=5<1<D9@?<1A9B1C?A1;1>=5<5F1C9<1ADC1>9>9B525<D=B1=@19;51>1<9B1C?A +5C5<18491=1C9C5A>G1C1B5;1A1>7=1C1=5<981C141>G13181G1C5A1>7 &1ADC1>7D<141<1=81<9>925A6D>7B9B521719@5=DC1A2941>775C1A 71A=5>:149<529875<1@<1791>1<9B1C?A49@DC1AB589>771=5>D>:D;;1>BD4DC $14925B1A>G1BD4DC@DC1A1>2941>775C1A3181G1G1>749<1;D;1>?<58<1ADC1>7D<1141<18 K &1ADC1>7D<1C5AB52DC49B52DC<1ADC1> &1ADC1>C5AB52DC141G1>741@1C=5=DC1A2941>775C1A@?<1A9B1B9;5;9A941>141:D71G1>7;5;1>1> 5>71>1<1CB5=131=9>9?A1>741@1C=5>5>CD;1>Gambar 2.20Polarisasi karena hamburan.partikel-partikelgasgelombang datangtak terpolarisasigelombang hamburanterpolarisasiGambar 2.21Pemutaran bidang polarisasi.cahaya alamitak terpolarisasicahayaterpolarisasisumbuanalisatorsumbupolarisatorE00cosEnormaln1n2(1)(2)Gambar 2.19Polarisasi pembiasan ganda.

Gelombang Cahaya45;?>B5>CA1B9<1ADC1>?@C9;1;C96 *?<1A9=5C5AG1>7;8DBDBD>CD;=5>5>CD;1>;?>B5>CA1B9<1ADC1>7D<149B52DC 1A925A21719@5A3?211>49B9=@D<;1>218F161;C?A61;C?AG1>7=5=5>71AD89BD4DC@DC1A2941>775C1A141<18:5>9B<1ADC1>C521<<1ADC1>@1>:1>741>;?>B5>CA1B9<1ADC1> +531A1=1C5=1C9B41@1C49CD<9B;1>B52171925A9;DC K %5C5A1>71>BD4DC@DC1A2941>775C1A;?>B5>CA1B9<1ADC1>@1>:1>7<1ADC1>C521<BD4DC@DC1A1>:5>9B<1ADC1> 9;5C18D9BD4DC@?<1A9B1B9BD1CD3181G1@14121<?;5B141<18J $9;19>45;B291BD41A1 C5>CD;1><189>45;B291B21<?;5BC5AB52DC @129<19>C5>B9C1B3181G1G1>7;5<D1A41A94D1;131@?<1A?94G1>749@1B1>7=5=25>CD;BD4DC JB1CDB1=1<19>41A93181G1=D<1=D<1 ,5>CD;1>25B1A>G1BD4DCG1>74925>CD;?<58;54D1;131@?<1A?94C5AB52DC $9;19>45;B291BD41A1141<18 BD4DC@?<1A9B1B93181G1@14121<?;5B141<18J ,5>CD;1>9>45;B291B21<?;5BC5AB52DC %7%&9;5C18D9J 5>71>=5>77D>1;1>(34%/%%0 949@5A?<58C1>C1> C1>J $1499>45;B291B21<?;5B141<18 +52D18B1381A9=5C5AG1>7@1>:1>7>G13=25A9B9<1ADC1>7D<1@1B9A45>71>@DC1A1>:5>9B<1ADC1>>G1J $9;1497D>1;1>B9>1A>1CA9D=@5=DC1A1>2941>7@?<1A9B1B9>G1 J89CD>7<18;?>B5>CA1B9<1ADC1>9CD %7%&9;5C18D93==J J %?>B5>CA1B9<1ADC1>41@1C49@5A?<5845>71>(34%/%%0 949@5A?<58 = $149;?>B5>CA1B9<1ADC1>141<18 L Contoh 2.10Contoh 2.11Kata 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=9<9;9C12D>7G1>7@1>:1>7>G1 3=25A9B9<1ADC1>7D<145>71>;5@5;1C1> 41>=5=9<9;9BD4DC@DC1A1>:5>9BJ ,5>CD;1>BD4DC@5=DC1A1>2941>7@?<1A9B1B9>G1:9;149@5A7D>1;1>B9>1A>1CA9D=

Mudah dan Aktif Belajar Fisika untuk Kelas XII46Setelah mempelajari bab ini, tentu Anda menjaditahu bahwa cahaya merupakan gelombang elektro-magnetik yang dapat mengalami proses interferensi,difraksi, dan polarisasi. Dari semua materi pada bab ini,Refleksibagian 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 181G1C5A=1BD;75<?=21>75<5;CA?=17>5C9;B589>771@5A1=21C1>>G1C941;=5=5A<D;1>=549D= #>C5A65A5>B9141<18@5A9BC9F1@5>7712D>71>4D175<?=21>71C1D<529841A975<?=21>7G1>7;?85A5> 181G141@1C25A9>C5A65A5>B9:9;1BD=25A3181G1>G1;?85A5>1AC9>G1=5=9<9;96A5;D5>B9B1=141>254161B5C5C1@ +D=25A3181G1G1>7;?85A5>41@1C491=1C9=5<1<D9@5A3?211>$160)41>3(40(.. #>C5A65A5>B93181G1=5>781B9<;1>@?<175<1@C5A1>7 *?<175<1@4981B9<;1>41A99>C5A65A5>B945BCAD;C96B1<9>7=5<5=18;1>1;921C@5>7712D>71>4D175<?=21>7G1>7=5=9<9;961B525A<1F1>1> *5AB1=11>B5<9B98:1A1;G1>749C5=@D875<?=21>7141<18B9> 45>71> *?<1C5A1>74981B9<;1>41A99>C5A65A5>B9;?>BCAD;C96B1<9>7=5>7D1C;1>1;921C@5>7712D>71>4D175<?=21>7G1>7=5=9<9;961B5B1=1 *5AB1=11>B5<9B98:1A1;G1>749C5=@D875<?=21>7141<18B9> 45>71> $1A1;41A9C5A1>7@DB1C;5@?<1C5A1>7;5141<18#$1A1;41A9C5A1>7@DB1C;5@?<175<1@;5141<18# 96A1;B975<?=21>7141<18@A?B5B@5=25<?;1>75<?=21>7G1>749B5212;1>?<58141>G1@5>781<1>725AD@1 35<18 1C1D BD4DC @5>781<1>7 G1>7=5>781<1>79B521791>=D;175<?=21>7 5<1845=9;91>49B52DC;9B9496A1;B9 $1A1;1>C1A35<1841<1=;9B949B52DCC5C1@1>;9B9 *141496A1;B935<18CD>771<@9C175<1@;5C5A:149:9;1B9>45>71> B541>7;1>@9C1C5A1>7;5C5A:149:9;1B9>45>71> K 141<1829<1>71>71>:9< 96A1;B9@141;9B9C5A:149:9;13181G149<5F1C;1>@14135<1835<18G1>7=5=9<9;9:1A1;G1>7B1=1 1G1DA19?@C9;141<18;5=1=@D1>B52D18<5>B1D>CD;=5=9B18;1>21G1>71>41A94D1C9C9;G1>7C5A@9B18B1CDB1=1<19>@141:1A1;=9>9=D= *?<1A9B1B9141<18@A?B5B@5>G1A9>71>1A1875C1ABD1CD75<?=21>7 <1CD>CD;=5>G1A9>71A1875C1A9>949B52DC@?<1A?94B1<18B1CD3?>C?8>G1141<18;A9BC1< #>C5>B9C1B3181G1G1>749<5F1C;1>@141@?<1A9B1C?A;5@9>7@?<1A?94141<18#>C5>B9C1B3181G1G1>749<5F1C;1>@1411>1<9B1C?A141<183?B45>71>BD4DCG1>74925>CD;1>C1A1BD=2D@?<1A9B1C?A41>BD=2D1>1<9B1C?A 96A1;B9=5=2181B96A1;B95<18,D>771<96A1;B9@141%9B9*?<1A9B1B9=5=2181B*?<1A9B1B9@141%A9BC1<*?<1A9B1B9@141*5=1>CD<1>41>*5=291B1>*?<1A9B1B9@141*5=291B1>!1>41C5A49A91C1B#>C5A65A5>B9'9>9=D=#>C5A65A5>B9'1;B9=D=#>C5A65A5>B9*?<1A9B1B9;1A5>1"1=2DA1>*5=DC1A1>941>7*?<1A9B1B9181G13?>C?8>G1(.1/&%0).(-531/%)0(5+-Peta Konsep41@1C=5>71<1=9

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