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:
Gelombang Cahaya
5>?=5>165>?=5>1 C5>C1>7 3181G1 B5A9>7 >41 :D=@19 41<1= ;5894D@1> B581A981A9 %5C9;1 CDAD> 8D:1> @141 B91>7 81A9 ;141>7 ;141>7;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; *14112>41 C5<18 =5=@5<1:1A9 75<?=21>7 =5;1>9; =9B1<>G1 75<?=21>7 19A 41> 75<?=21>7C1<9 @1;18@5A25411>1>C1A175<?=21>75<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>B@?AC1B9 1BCA?>?=9 =9<9C5A 41> 75?7A169 ,18D;18 >41 21719=1>175:1<141>39A939A975<?=21>75<5;CA?=17>5C9;;8DBDB>G1 3181G1B5AC1@5>5A1@1>>G1$1F121>1C1B@5AC1>G11>@5AC1>G11>C5AB52DC 41@1C>41C5=D;1>@1412129>9 )<58;1A5>19CD@5<1:1A92129>9 45>71>219;
Cahaya merupakan gelombang elektromagnetik yang banyak digunakan untuk kepentingan teknologi komunikasi.
Bab
2
A B C gelombang
datang
S2
S1
terang
terang
terang
terang terang terang terang terang terang terang
terang
terang
terang gelap
gelap
gelap
gelap gelap gelap gelap gelap gelap gelap
gelap
gelap S0
Gambar 2.1
Interferensi pada celah ganda Young
A. Interferensi Cahaya
141<18 "!! !#
1AC9>G175<?=21>7=5=9<9;9!41> # +5@5AC9G1>7C5<18492181B@141@5=2181B1>75<?=21>7 218F175<?=21>73181G125AB961CB5@5AC981<>G175<?=21>72D>G9G19CD 41@1C25A9>C5A65A5>B9 )<58;1A5>19CDD>CD;=5>41@1C;1>9>C5A65A5>B9 3181G1@D>49@5A<D;1>BD=25A3181G1G1>7;?85A5>G19CDBD=25A3181G1 G1>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 Fresnell
a. Percobaan Celah Ganda Young
*5A3?211>9>949<1;D;1>?<58$160)45>71>=5>77D>1;1>4D1@5>7 81<1>7 *5>781<1>7@5AC1=1=5=9<9;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>1A B9>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?<58 4D1BD=25A3181G1+41>+G1>7;?85A5>41A981B9<@5=1>CD<1>4D1 Gambar 2.2
C1 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
!(&(.6//(/2(.%,%3+-104(2(.1/&%0)%*%8%-(3,%-%0.%*41%.41%.&(3+-65'%.%/&6-6.%5+*%0
Tes 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;
C B
P
r2
S2
S1 r1 d y D b r2 S2 r1 S1 Gambar 2.3
(a) Sinar gelombang dari celah S1 dan S2 berinterferensi di titik P. (b) Untuk Dd, r
1 dan r 2 dianggap
sejajar 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>7 3181G1 41A9 + 41> + 9>9 1;1> B1<9>7
25A9>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;9 1=@<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<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 <1G1A 45>71> =5=25A9;1> BD4DC 41A9 BD=2D C5>718
C5A8141@ 71A9B 75<1@ 1C1D 71A9B C5A1>7 C5AB52DC ->CD; =5>5>CD;1> 25B1A>G1 ;9C1 81ADB =5>78D2D>7;1>>G1 45>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=21A C5A<981C218F1B9>1A75<?=21>741>141<18B5:1:1A41>=5=25>CD; BD4DCC5A8141@BD=25A@DB1C *5A81C9;1>%/&%3 & ,5A>G1C1
BD4DCG1>74925>CD;?<58BD=2D@DB1C41>BD=2D 5>71>
45=9;91>;9C141@1C45>71>=D418=5>5>CD;1>218F1 B9> 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;C96 C5<18 49;5C18D9 218F1@1BC9 >?< 1C1D
29<1>71>75>1@41A9@1>:1>775<?=21>7 (34%/%%0 941@1C49CD<9B B52171925A9;DC
1
2 b
L
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
7,2 10 m3 3,6 10 m 3 2
y
Dari soal diketahui: m = 1; d = 10–5 m; D = 0,2 m
sehingga 1 2 yd m D 1 2 yd D m 3 5
3,6 10 m 10 m 1 0,2 m 2 3 5
3,6 10 m 10 m 1 0,2 m
2
B9>
K
45>71> ->CD;C5A1>7@DB1C;9C1=5=25A9;1> @9C1C5A1>7@5AC1=141>B5C5ADB>G1
41@D> D>CD; 9>C5A65A5>B9 =9>9=D= 9>C5A65A5>B9 45BCAD;C96
@1BC929<1>71>71>:9<;1<9B5C5>71875<?=21>7(34%/%%0 941@1C 49CD<9B;1>B52171925A9;DC
B9>
K
45>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-
m
dari Terang Pusat
*141 @5=2181B1> B525<D=>G1 C5<18 49B52DC;1> 218F1 @?<1 9>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>218F1
B9> 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>775<1@;541A9@DB1C
:1A1;<1G1A;535<18 @1>:1>775<?=21>73181G1 1<1=81<9>9
Tantangan
untuk Anda
->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;=5C5A 41A935<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 ==LK= # ==LK=
=
k #
= =
LK 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;54D1 35<189CD==41>49<5C1;;1>@141:1A1;=;5<1G1A71A9B75<1@@5AC1=141A9@DB1C G1>7:1A1;>G1== "9CD>7<18@1>:1>775<?=21>73181G1>G1
%7%&
==LK=
=
# ==LK=9>C5A65A5>B975<1@
1A9(34%/%%0 949@5A?<58 #
=K
= = =
K
LK
= LK= I
$149@1>:1>775<?=21>73181G1>G1141<18 I
Contoh
2.2
3. Interferensi oleh Lapisan Tipis
*5=1>CD<1>3181G1=1C181A9?<58!#G1>74931=@DA 45>71> 19A 1;1> =5=@5A<981C;1> 71A9B71A9B 25AF1A>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>7 G1>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>7
1;1>49291B;1>41>B521791><17949@1>CD<;1>;5=21<9;5@5A=D;11> +9>1A G1>7 49@1>CD<;1> 49<5F1C;1> @141 B52D18 <5>B1 @?B9C96 41> 496?;DB;1>49C9C9;* 5A;1B3181G149C9C9;*=5AD@1;1>81B9<9>C5A65A5>B9 25A;1B 3181G1 41> 45>71> 141<18 25A;1B 3181G1 G1>7 49@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 2.4
Interferensi oleh lapisan tipis. P
D
A
B C i
d
n r
lensa
lapisan tipis (1)
+5<9B98 <9>C1B1> ?@C9; G1>7 49C5=@D8 ?<58 B9>1A 41C1>7 89>771 =5>:149B9>1A@1>CD<;541>B9>1A@1>CD<;5141<18
KKK
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?B KC1>B9> 3?B K B9> B9>3?B
5>71>=5>77D>1;1>6-6/!0(..+64C5>C1>7@5=291B1>3181G1 G1;>9B9>B9>49@5A?<58B5<9B98:1A1;C5=@D8;54D1B9>1A=5>:149
3?B
K
B9> 3?B
3?BKB9> 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>: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 K
45>71>
(34%/%%0 925A<1;DD>CD;9>45;B291B<1@9B1>C9@9B<529825B1A 41A91C1D
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;B291B
41>=5=5>D89BG1A1C41>>1C1D41>
Gambar 2.7
Interferensi oleh lapisan minyak yang tipis.
Sumber:www.instckphoto.com
Gambar 2.6
Interferensi oleh lapisan busa 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
2.3
Gambar 2.5
Interferensi oleh busa sabun.
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.
%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?BB589>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
2.4
Kata Kunci
• interferensi• sinar monokromatis • interferensi maksimum • interferensi minimum
Tes Kompetensi Subbab
A
(3,%-%0.%*'%.%/&6-6.%5+*%0
@1G1>749=1;BD49>C5A65>B93181G1
+52D1835<1871>4125A:1A1;==4925<1;1>735<18 41>@141:1A1;=5C5A49<5C1;;1>B52D18<1G1A 5<18 49B9>1A94D1B9>1A=?>?;A?=1C9B45>71>@1>:1>7 75<?=21>7>=41>>= 5A1@1;18:1A1;71A9B C5A1>7?A45;55=@1C;54D1B9>1A@141<1G1A
+525A;1B3181G1=5<5F1C94D135<18B5=@9CG1>7B1CD B1=1<19>25A:1A1;== $1A1;35<18;5<1G1A=5C5A 41>:1A1;1>C1A14D171A9BC5A1>7@141<1G1A141<18 LK3= 5A1@1;18@1>:1>775<?=21>73181G1
G1>7497D>1;1>
->CD;=5>7D;DA@1>:1>7B9>1A=5A1849<1;D;1> @5A3?211>B52171925A9;DC
+9>1A 29AD 45>71> @1>:1>7 75<?=21>7 >= 49:1CD8;1> C571; <DADB @141 35<18 71>41 *?<1 9>C5A65A5>B9C5A:149@141<1G1AG1>725A:1A1;=41A9 35<18 !1A9BC5A1>7?A45@5AC1=125A:1A1;==41A9 71A9BC5A1>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>7 75<?=21>7B9>1A=5A189CD
+52D18<1@9B1>C9@9B=5=9<9;99>45;B291B497D>1;1> D>CD; =5<981C 75:1<1 9>C5A65A5>B9 $9;1 @1>:1>7 75<?=21>7>G1 IC5>CD;1>C521<=9>9=D= <1@9B1>C5AB52DCBD@1G1C5A:1499>C5A65A5>B9
B. Difraksi Cahaya
96A1;B93181G1C5A:149:D71@14135<18B5=@9CG1>7C5A@9B18B5:1:1A B1CD B1=1 <19> @141 :1A1; G1>7 B1=1 5<18 B5=@9C G1>7 45=9;91> 49B52DC %9B9141<18;5@9>71>;131G1>7497?A5B=5>DADC 71A9B B5:1:1A 41> 21>G1; :D=<18>G1 $1A1; 1>C1A1 4D1 35<18 49B52DC C5C1@1> ;9B9
1. Difraksi Celah Tunggal
96A1;B9@14135<18CD>771<1;1>=5>781B9<;1>@?<171A9BC5A1>7 41>75<1@@141<1G1A 5<18CD>771<41@1C491>771@C5A49A91C1B2525A1@1 35<18B5=@9CG1>74921C1B9C9C9;C9C9;41>B5C91@35<189CD=5AD@1;1> BD=25A 3181G1 B589>771 B1CD B1=1 <19>>G1 41@1C 25A9>C5A65A5>B9 *5A81C9;1>%/&%3
->CD;=5>71>1<9B9B@?<1496A1;B935<18@141 %/&%3 492179 4D121791> *5A81C9;1>75<?=21>741> !5<?=21>7=5>5=@D8 <9>C1B1>G1>7<5298:1D8B525B1A B9> 41A9@14175<?=21>7 +1=1 81<>G145>71>75<?=21>741>G1>7=5=9<9;92541<9>C1B1>B525B1A
B9>
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B9> B9>
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B9> B9>
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B9> K
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Gambar 2.9
Maksimum utama terjadi untuk k = 0 atau = 0. maksimum
utama
5>71>=5>77D>1;1>@5>781<1>735<18CD>771<@141<1G1AC1=@1;@?<1496A1;B9 71A9BC5A1>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><521A 35<18G1>7497D>1;1>
Contoh
2.5
Gambar 2.8
Difraksi cahaya pada celah tunggal. d
5 4 3 2 1 2
d
2 d
sin 2 d
garis gelap
terang d
pusat Q P
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)<58;1A5>1;54D13181G1 B561B5254161B5;54D1>G1=5>:149J D175<?=21>745>71>2541 61B51C1D2541BD4DC61B5J49B52DC:D71B561B5 *5AB1=11>9>C5A65A5>B9 =1;B9=D=41A9@?<1496A1;B9>G11;1>=5>:149
B9>
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I
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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
dp
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
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Gambar 2.10
Difraksi pada kisi B
B1 A
d C D
G F E
T
M
P
K
Gambar 2.12
Difraksi minimum kedua untuk N banyak celah.
Gambar 2.13
Difraksi cahaya putih akan menghasilkan pola berupa pita-pita spektrum.
cahaya putih
kisi difraksi
me rah
merah merah
merah biru
biru
bir u
biru
spektrum orde ke 2
spektrum orde ke-1 spektrum orde ke-0 (putih) spektrum orde ke-1 spektrum orde ke-2
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>?A=1<=1;B9=D=DC1=1 ,5>CD;1><18@1>:1>775<?=21>7
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I
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+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; C5A49A9 1C1B 25A21719 F1A>1 45>71> @1>:1>7 75<?=21>7 C5A;539< F1A>1 D>7D 41> 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
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Mari Mencari Tahu
Gambar 2.11Difraksi minimum kedua untuk N = 2 celah. m = 2
m = 1
2
1 1
3
2 1
3
sin d
m = –1 m = 0 m = 1 m = 2
0
sin d
2
Tantangan
untuk Anda
3. Daya Urai Optik
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Gambar 2.14
Bayangan dari optik fisis dua benda yang berdekatan karena (a) beririsan dan (b) terpisah dengan baik.
Gambar 2.15
Lukisan sinar dari sumber cahaya dari sebuah celah bulat. D
celah bulat pola difraksi
r
cahaya sumber datang
2
s1
s2
D
Contoh
2.8
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*5A81C9;1>71=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
Informasi
untuk Anda
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Tes Kompetensi Subbab
B
(3,%-%0.%*'%.%/&6-6.%5+*%0
C. Polarisasi Cahaya
1. Polarisasi pada Kristal
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Kata Kunci
• difraksi gelombang • kisi difraksi
• sudut simpang (deviasi) • difraksi maksimum • difraksi celah tunggal • difraksi pada kisi • daya urai optik • apertur
%7%&
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Gambar 2.16
Cahaya tak terpolarisasi dilewatkan pada sebuah kristal.
getaran horizontal diserap sempurna oleh polaroid cahaya alami yang datang
getaran vertikal diserap sebagian
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.
polarisator analisator sumber
cahaya
sumber
cahaya polarisator analisator tidak ada cahaya cahaya yang
diteruskan terpolarisasi
1
2 B5:1:1A1C1DB1=1;54D4D;1>>G13181G1G1>749C5ADB;1>C5A@?<1A9B1B9
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K
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3?B
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Contoh
2.9
2. 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;BD4DC JB9>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>1A 41C1>7 @141 2941>7 21C1B +521791> 49@1>CD<;1>41>B521791><17949291B;1> +5BD1945>71>"D;D=+>5<<9DB
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Gambar 2.18
Polarisasi pada (a) pemantulan dan
1 terpolarisasi
cermin ip
2 n1
n2
ip
r
C1>
K
5>71>141<18BD4DC@?<1A9B1B91C1DBD4DC" 41>B5AC1
141<189>45;B291B=549D=B1CD41>=549D=4D1
3. Polarisasi pada Pembiasan Ganda
*141 ;A9BC1< ;1<B9C 1) ;D1AB1 +9) =9;1 C?@1B 41> 5B
3181G141@1C=5>71<1=9@5=291B1>71>41;1A5>1=5=9<9;94D1>9<19 9>45;B 291B *141%/&%3 C1=@1; 141 4D1 21791> B9>1A G1>7 49291B;1> +9>1AC941;=5>79;DC9@5=291B1>=5>DADC"D;D=+>5<<9DB 1C1D49B52DCB9>1A9BC9=5F1B9>1A=5>79;DC98D;D=@5=291B1>+>5<<9DB 1C1D49B52DCB9>1A291B
4. Polarisasi karena Hamburan
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5. Pemutaran Bidang Polarisasi
%/&%3 =5=@5A<981C;1>B52D18 G1>7C5A49A91C1B 4D12D18@?<1A?94G1>749@1B1>7@141B52D181<1CG1>749<5>7;1@945>71> B;1<1 45A1:1C 41> ;?C1; <1ADC1> *?<1A?94 G1>7 45;1C 45>71> BD=25A 3181G149B52DC@?<1A9B1C?A41>G1>7<19>>G1141<181>1<9B1C?A
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2941>775C1A3181G1G1>749<1;D;1>?<58<1ADC1>7D<1141<18
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Polarisasi karena hamburan. partikel-partikel
gas
gelombang datang tak terpolarisasi
gelombang hamburan terpolarisasi
Gambar 2.21
Pemutaran bidang polarisasi. cahaya alami
tak terpolarisasi cahaya terpolarisasi
sumbu analisator sumbu
polarisator E0
0cos
E
normal
n1 n2
(1)
(2)
Gambar 2.19
;?>B5>CA1B9<1ADC1>?@C9;1;C96 *?<1A9=5C5AG1>7;8DBDBD>CD;=5>5> CD;1>;?>B5>CA1B9<1ADC1>7D<149B52DC
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Contoh
2.10
Contoh
2.11
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
Setelah mempelajari bab ini, tentu Anda menjadi tahu bahwa cahaya merupakan gelombang elektro-magnetik yang dapat mengalami proses interferensi,
Refleksi
bagian mana yang menurut Anda sulit dipahami? Coba Anda diskusikan bersama teman atau guru Fisika Anda. Selain itu, coba Anda sebutkan manfaat yang Anda
Rangkuman
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