< PreviousGelombang 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>799CD 181G1=5AD@1;1>75<?=21>75<5;CA?=17>5C9; *141 12 >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, 1995Mudah 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 Fresnell 5>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 Fresnell untukmenunjukkan interferensi cahaya.C1S1OPC2S2STokohAugustine 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 D d, 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<18 41>35<18 G1>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;@141 B545=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>7 41> 141<18B5:1:1A41>=5=25>CD;BD4DCC5A8141@BD=25A@DB1C *5A81C9;1>%/&%3 & ,5A>G1C1BD4DC G1>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>1A 41> 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 pusatadalah337,210m3,610m2yDari soal diketahui:m = 1; d = 10–5 m; D = 0,2 msehingga12ydmD12ydDm353,610m10m10,2m2353,610m10m10,2m273,610m = 360 nmJawaban: cMudah 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>73181G1 1<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<18 K K K 45>71>141<189>45;B291B<1@9B1>C9@9B41> '9B1<;1>C521<<1@9B1>141<18=1;13?B B589>771 3?B 41> B9>45>71> C1> B589>771 3?B K C1> B9> 3?B K B9>B9>3?B 5>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>:149 3?B K B9>3?B 3?B KB9> 3?B 3?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;1 1;1>=5>:149 1C1D 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?B 41> 1C1D 3?B K 45>71>9>45;B291B41>=5=5>D89BG1A1C 41>> 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;B291B 45>71>@1>:1>775<?=21>7 I Contoh 2.3Gambar 2.5Interferensi oleh busa sabun.Sumber: www.designprodygzone.comGelombang 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?<58 3?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 XII38 96A1;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 Tunggal 96A1;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;1 B9>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; 1C1D C5A:149=1;B9=D=DC1=1@9C1C5A1>7C5>718B5@5AC949@5A<981C;1>@141%/&%3 Gambar 2.9Maksimum utama terjadiuntuk k = 0 atau = 0.maksimumutama 5>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.d543212d2dsin2dGelombang 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>:149 J D175<?=21>745>71>254161B5 1C1D2541BD4DC61B5 J49B52DC:D71B561B5 *5AB1=11>9>C5A65A5>B9=1;B9=D=41A9@?<1496A1;B9>G11;1>=5>:149 B9> B9> 1C1D B9> 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?<58 B9> >= >=>= >=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 = 1dpm330,410 m0,510 m10,4 m= 5,6 × 10–7 mJawaban: dContoh 2.62.Difraksi pada Kisi 96A1;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;BD4DC JC5A8141@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;1 B9> :D71=9>9=D=;54D1:9;1 B9> +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<18 G1>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.intensitasNext >