07_bab6

A. Cahaya

B. PemantulanCahaya

C. PembiasanCahaya

D. Alat-Alat Optik

Cahaya dan Optika

,5C91@81A9=1>DB91C941;1;1>@5A>18C5A4125>41 49 B5;54141@1C C5A1 141>H1 3181H1181H1:D71 H1>7=5=2D1C>4141@1C=571>>4149 35A=9>

&D>7;9>>41 @5A>18=5=5A81C9;1> ;131 B@9?> @141=?29@141=?29B5@541=?C?A=5AD@1;1>B52D1835A=9> 35=2D>7 H1>7 497D>1;1> @5>75=D49 D>CD;=5

108 Mudah dan Aktif Belajar Fisika untuk Kelas X

A. Cahaya,19>B41>C5;>??=9 41@1C H1; C5AD>7;1@ ;1A5>1C5AB5491>H1C57H1>741>@5>5=D1>41

109Cahaya dan Optika

@5>5>H1(9621=5>H9=@D 218F1 @141 D=D=>H1 3181H1C5A49A9 1C1B B5;D=@D@1AC9;5749B52DC,.-1/ )"/b. Teori Gelombang

)1417:*(1525ADB181=5-5?A9)1AC9;571>=5>H1C1;1>218F13181H1=5AD@1;1>75


Aktivitas Fisika 6.1

karton

sinarpantul

sinar datang

pointerinframerah

2. Hukum Pemantulan#9;[email protected],8,6$5,5,.$@A?B5B@5=1>CD@14135A=9>41C1A

49 B531A1 219;>4141@1C=5>5=D;1>;5B9=@D 218F11 BD4DC 41C1>7 B1=145>71> BD4DC @1>CD1A41C1>771A9B>?A=1B9>1A@1>CDH1C11> 49;5>171>!D;D=)5=1>CD3. Pemantulan pada Cermin

5A=9>141H1>7=1=@D=5=1>CD7=5>75>19>H1 )5A=D;11> B5@5AC9 9CD 41@1C 25AD@1


A C

B

B'

D

s0 s'

benda bayangan

5A41B1A;1>8D;D=@5=1>CD49C9C9;25A1A@1>CDCD;BD4DCJC5A8141@71A9B>?A=17>41 41C1A;5C9;1>4125A49A94945@1>

35A=9>41C1AC5AB52DC)14135A=9>C5A71>>411719=1>1B961C21H1>71>H1>7 C5A25>CD;@14135A=9>41C1A

)5A81C9;1>$0%$4-9C9;,=5AD@1;1>21H1>71>41A9C9C9;,41@D>@A?B5B@5=25>CD;1>21H1>71>@14135A=9>41C1A1411A41C1>7,)B9>1A:1CD8@14135A=9>41C1A45>71>BD4DC41C1>71 ;5=D491> B9>1A 9>9 49@1>CD)5A81C9;1> :1 B9>1A ,9>1A 41C1>7,) B9>1A :1CD8@141 35A=9>45>71> BD4DC 41C1>72 ;5=D491> B9>1A 9>9 49@1>CD )5A81C9;1> :1 B9>1A

)5A@1>:1>71> B9>1A @1>CD B9>1A @1>CD4945>71> 71A9B C5A@DCDB@DCDB 41> 25A@?C?>71>49 C9C9;#149 71> C9C9;14174925>CD;41A9@5A@?C?>71>@5A@1>:1>71>4D1 B9>1A@1>CD71>31A1H1>7B1=121H1>71>25>414D149=5>B941>C97149=5>B9

41@1CC5A25>CD;?41C1A)A?B5B@5=25>CD;1>21H1>71>>H1B1=1B5@5AC9 @14125>41 C9C9;!17 C5A@5>C9>7 14121H1>71>8D;D=@5=1>CD B541H1>72D;1>25AD@1C9C9;1C1D71A9B>411;1>=5>41@1C;1>218F1D;DA1>21H1>71>25>41@5AB9B B1=145>71>D;DA1>25>41>H15>4141>21H1>71>81>H125A2541411>>H1)5A81C9;1>$0%$4 1791> ;9A9 25>41=5>:149 21791> ;1>1>21H1>71> 41> B521H1 )5A9BC9F1 9>9 49B52DC @5=21 B9B9 )0".)&+2"./&,+(1 141>H1 @5=21 B9B9 9>9 CD H1>785>41;492131=5D8DAD6>H181ADB492121H1>71>25>41@141 35A=9>41C1A 41@1C 49B9=@D218F1 B961C 21H1>71> 25>41 H1>7 4925>CD; ? 41C1A 141 B1=125B1A 45>71>D;DA1>25>41K C571;K :1A1;25>41 C5A8141@35A=9>B1=145>71> :1A1;21H1>71>C5A8141@

35A=9>

Gambar 6.5Ukuran benda persis sama denganukuran bayangan

Gambar 6.4Pembentukan bayangan padacermin datar

bayangan

S'S

P1

P2

benda

1

2

Gambar 6.6Ketika bercermin, posisi kiri bendamenjadi posisi kanan bayangan

matapengamat


(1B9>1A@1>CD71>B9>1A41C1>7=1;1 #149 D>CD;41@1C=571>B5:1>735A=9>41C1A=9>9=1779CD2D8B5B5?A1>7

$:$1*$11:$6$'$1%$:$1*$10$:$1H1>71>>H1C114171>H1>7C5A25>CD;41A9@5A@?C?>71>7

BD>7 B9>1AB9>1A 3181H1 B541>7;1>21H1>71>=1H114171>H1>74981B941A9@5A@?C?>71>@5A@1>:1>71>B9>1AB9>1A3181H1,5217193?>C?821H1>71>H1>74981B9@14135A=9>41C1A=5AD@1;1> 21H1>71>=1H1;1A5>1 21H1>71> C5AB52DC =5AD@1;1> @5A@?C?>71> 7BD>7 41A9@5A@1>:1>71>B9>1AB9>1A3181H141@D>3?>C?821H1>71>>H1C1C5A:149@14121H1>71>H1>74981B9?:1>7=9>9=D=BD1CD35A=9>H1>7492DCD8;1>171AB5?A1>7@5>1A9141C41@1C=57C9>779>H1%=5C5A$9$%)5A81C9;1>71=21A25A9;DC

70/$+%$:$1*$12/(+'7$%7$+&(40,1'$6$4)5A81C9;1>$0%$4 :9;1 B52D18 25>41 25A141 49 45@1> 4D1

2D1835A=9>H1>7=5=25>CD;BD4DC B1CDB1=11;1>C5A25>CD;B5:D=71> #D=71> 25A71>CD>7 @141 25A1@1 25B1ABD4DC 5A41B1A;1>81B9:D=71>H1>74925>CD;?H1>7=5=25>CD;BD4DC 14141C5A2D1835A=9>41C1AH1>7=5=25>CD;BD4DCJ!9CD>7:D=71>H1>7C5A25>CD;$9$%9;5C18D9 J+

360 1

+ J LJ+ L#149:D=71>H1>7C5A25>CD;2D18

J L +$5C5A1>71>

+ :D=71> BD4DC1>C1A14D12D1835A=9>

L

Contoh 6.3

G

F

H

E

A D C

B

cermin

Gambar 6.7Benda yang diletakkan di antara duabuah cermin datar dan membentuk

sudut 90 akan menghasilkan 3 buahbayangan.

cermin I

cerminII

bayangan

bayanganbayangan

benda

Contoh 6.2


Gambar 6.8Bagian-bagian cermin cekung

f = jarak fokusR = kelengkungan cermin

R

f

O

B

FP

a)

b)c)

P

F

O

Gambar 6.9Sinar istimewa pada cermincekung

Gambar 6.10Pembentukan bayangan di ruang IIIAB pada cermin cekung

O

B'

FP

A'

A

B II IIII

5A41B1A;1>$0%$4$0%$441>$0%$441@1C49B9=@D 218F11 :9;1 25>41 @141 AD1>7 "=1;1 21H1>71> H1>7 4981B9 14149@5A25B1A2 :9;125>4125A14149AD1>7""=1;121H1>71>H1>74981B9141H1C1 49 AD1>7 """ C5A2149@5A25B1A3 :9;1 25>41 25A141 @141 AD1>7 """=1;1 21H1>71> H1>7 4981B9

141H1C1 49 AD1>7 "" C5A2149@5A25B1A

Gambar 6.11Gambar bayangan jika letak bendanya diruang II

Gambar 6.12Gambar bayangan jika bendanya diruang I

OAPA'

B'

B

II IIII

F

OF AP A'

B'

B

II IIII

S'S

b. Pemantulan pada Cermin Cekung5A=9> 35;D>7 141 H1>7=5=9 35;D>7

5A=9> 35;D>7 25AB961C ;?>E5A75> H19CD 25AB961C=5>7D=@D B9>1A1791>21791> 35A=9> 35;D>7 C1=@1; B5@5AC9 @141$0%$4 (0%(167.$1%$:$1*$12/(+&(40,1&(.71*

%5C1;41>B961C21H1>71>H1>74925>CD;?35;D>725A71>CD>7@14141,52D1825>41H1>7494945@1>B52D1835A=9>35;D>7 1;1>=5=971> 45>71> B961CB961C C5AC5>CD )5A81C9;1>$0%$4 B961CB961C 35A=9> 35;D>749 1>C1A1>H11 ,9>1A 41C1>7 H1>7 B5:1:1A BD=2D DC1=1 49@1>CD=57 H1>7=571> BD=2DDC1=13 ,9>1A 41C1>7 H1>7=5 35A=9> 1;1>

49@1>CD;5=2171>B52D1825>41?35;D>741@1C49C5>CD;1>45>71>

31A1=5>771=21A;1>41A9 B9>1A 9BC9=5F1@141 35A=9> 35;D>71A1=571> 25>41 ? 35;D>7 141$0%$4 +$(% -".0* 35A=9> 35;D>7 4971=21A7;1@ 45>71> 21791>21791>>H1+$(% ("!1 71A9B 4971=21AD>CD;=57;1>C9>77925>4141@D>71A9BC941;@5A7=5AD@1;1>@5A@?C?>71> 1>C1A1 B9>1A @1>CD B9>1A @1>CD1=1-9C9; 9>9=5AD@1;1>21H1>71>41A9 C9C9;$5=D491> 71A9B C571; 9 4925A9>1=1H1>7=5AD@1;1>21H1>71>41A9C9C9;B589>771=5AD@1;1>21H1>71>41A9 71A9B


7%71*$1$16$4$-$4$.)2.75'$1-$4,-$4,.(/(1*.71*$1&(40,1!D2D>71> 1>C1A1 :1A1; 6?;DB 41> :1A9:1A9 ;57;D>71> 35A=9>

41@1C 4931A9 45>71>@5AC?71>$0%$4 )14171=21AC5AB52DCC1=@1;218F1B9>1AB5:1:1ABD=2DDC1=141C1>7

;5@5A=D;11>35A=9>35;D>7;5=D491>49@1>CD=51A 9>9=5>79;DC9 8D;D=@5=1>CD H19CD BD4DC 41C1>7 B1=145>71> BD4DC @1>CD B579C971B1=1;1;9B9B9B9B9#9;1B9>1A41C1>745;1CB5;171>BD=2DDC1=141@1C491>771@ B1=145>71> B589>7711A9 B9>949@5A?71># :1A1; 6?;DB :1A9:1A9 ;57;D>71> 35A=9>

70753$'$&(40,1&(.71*

,5 B531A1 75?=5CA9 B961C 21H1>71> 25>41 41@1C49C5>CD;1> B531A1@5A89CD>71> 45>71> @5AB1=11> B521719 25A9;DC)5A81C9;1>$0%$4

)141$0%$4 C1=@1; 218F1 41> B521>7D>B589>771

(1

# =1;1 L

L

h s

h s L

)141$0%$4C1=@1;218F1 41> B521>7D>.>CD; 25A % / 6

.>CD; 25A % 6 /

B589>771 % % % 6 / / 6 6 / % / 6

(1 % /% / =1;1/ 6 // / 6

/ / 6 / 6 / // 6 / / 6 // // / /

// // // //

L

Gambar 6.14Letak dan sifat bayangan benda pada

cermin cekung

Gambar 6.13Hubungan antara fokus dan jari-

jari cermin cekung

P

R

F

R

f

A

O)

))

Keterangan:s = jarak benda ke cerminh = tinggi bendaf = jarak fokush' = tinggi bayangans' = jarak bayangan ke cermin

5>71>=5>77D>1;1>@5AB1=11>35A=9> C5AB52DC>4141@1C=5>5>CD;1> :1A1;21H1>71> B52D1825>4141@D>@5A25B1A1>21H1>71>

Konvensi tanda untuk pemantulan:K s bernilai + jika objek berada di

depan cermin (objek maya)bernilai jika objek di belakangcermin (objek nyata)

K s bernilai + jika objek berada didepan cermin (objek nyata)bernilai jika objek berada dibelakang cermin (objek maya)

K r, f bernilai + jika pusatkelengkungan berada di depancermin (cermin cekung)bernilai jika pusat kelengkunganberada di belakang cermin(cermin cembung).

Ingatlah

B

A P

E

D FO

f

s'R

s

h'

h


494569>9B9;1> B521719 @5A21>49>71> 1>C1A1 25B1A C9>779 21H1>71> 41>25B1A C9>779 25>41,531A1=1C5=1C9B41@1C49CD35=2D>7141H1>7=5=935=2D>7

5A=9> 35=2D>7 25AB961C 49E5A75> H19CD 25AB961C=5=1>31A;1> B9>1A1791>21791> 35A=9> 35=2D>7 B5@5AC9 @141$0%$4 (0%(167.$1%$:$1*$12/(+&(40,1&(0%71*

,52D1825>41H1>7494945@1> B52D1835A=9>35=2D>7=5=971>45>71>B961CB961CC5AC5>CD1H1>71>B52D1825>41?35=2D>741@1C49C5>CD;1>45>71>31A1=5>771=21A;1>41A9B9>1A9BC9=5F1@14135A=9>35=2D>7)5A81C9;1>$0%$4)141$0%$4C5A1A 41C1>7 H1>7 B5:1:1A BD=2D DC1=1 49@1>CD B51;1>1;1>

41C1>7 41A9 C9C9; 6?;DB2 B9>1A 41C1>7 H1>7=5>D:D C9C9; 6?;DB 49@1>CD B5:1:1A 45>71>

BD=2DDC1=13 B9>1A 41C1>7 H1>7=5>D:D C9C9; @DB1C ;57;D>71> 35A=9> 49@1>

CD;5=2125>41? 35=2D>7 C5A1A 9BC9=5F1 H1>7497D>1;1>1411A41> B9>1A5A41B1A;1>71=21A21H1>71> C5AB52DC41@1C49B9=@D218F125>41H1>7494945@1> B52D1835A=9>35=2D>7 B571>=1H1 C571;41>49@5A;539< 70753$'$&(40,1&(0%71*

+D=DB@14135A=9>35=2D>7B1=1B5@5AC9AD=DBH1>725A 35;D>7 H19CD

L

#

% / % /

Gambar 6.15Bagian-bagian cermin cembungR = jari-jari kelengkungan cerminP = titik pusat kelengkungan cerminF = titik fokus cerminf = jarak titik fokus ke cermin

SU

O

f

F PR

O F P

Gambar 6.17Lukisan pembentukan bayangan olehcermin cembung

ab

c

O F P

Gambar 6.16Sinar-sinar istimewa pada cermincembung

!1 35=2D>7 B59 49B5212;1> 35=2D>7 C5A

2 D>CD; 25>41 >H1C1 49 45@1> 35A=9> 35=2D>7 B571>=1H1 #149>935=2D>725AC1>41>571C96

,52D1835A=9>35;D>7=5=971>=,52D1825>4149@141:1A1;=41A935A=9>C9>77925>413=!9CD>7@5A25B1A1>21H1>71>$9$%9;5C18D9

Contoh 6.4

2 D>CD; 25>41 >H1C1 49 45@1> 35A=9> 35=2D>7 B571>=1H1#149>935=2D>725AC1>41>571C96


p'

s s'

p

bendabayangan

P

/= = # =%3=5>71>=5>77D>1;1>(45$0$$1;49@5A?=4945@1>35A=9>C9>779>H13=41>@5A25B1A1>>H1;141493=4945@1>35A=9>35=2D>7H1>725A:1A9:1A93=1 9=1>171>2 5A1@1@5A25B1A1>>H13 %D;9B:1>H1B9>1A@141@5=25>CD;1>21H1>71>$9$%9;5C18D9/3= L3=# L L3=1 6 / /

L / L /

L / /L3=

2 L //

#149@5A25B1A1>>H1 ;171>C571;=1H141>49@5A;539


C. Pembiasan Cahaya71A>41=5=181=9 @5A9BC9F1 @5=291B1> 3181H1 25A9;DC

)5A81C9;1>>H17;D>71>H1>7=57;D@9 69 ;5 D41A1 ;18 3181H1>H1=5>H521A@1 @5A1>;131 @5=21C1B H1>7=5=25>CD; 2?7 4911=1;1> @5A9BC9F1 @5=291B1>

"*&/+ ."#.(/& 141


1. Indeks Bias">45;B291B41@1C49@1>41>7B521719BD1CD;5=1=@D1>=549D==5=

291B;1> =5=25


medium air(medium1)

medium kaca(medium2)

Gambar 6.20Sinar yang melalui dua mediumberbeda, yaitu air dan kaca.

,531A1 D=D= :9;1 3181H1 25A75A1; 41A9=549D= ;5=549D=9>45;B291B=549D=A5 B9>+ &++ .

)5AB1=11> C5AB52DC 41@1C :D71 49CD71>#6A5;D5>B941> @1>:1>775


1D>CD;=5>5>CD;1>;535@1C1>3181H149417B9BF1=5B525A;1B3181H1;54171>BD4DC41C1>7J$5=D491>B9BF1=5>31C1CBD4DC291BH1>7C5A:1494941H1C125B1A>H1

J#9;1;535@1C1>3181H149D41A1491>771@M=BC5>CD;1>;535@1C1>3181H14941B9>

2 & r

M B9>

JB9> J=B

M

=BM=B#149;535@1C1>3181H141n

2

i

r

n1

n2n

125A1414941B1A;?741H1= 19A+ B5@5AC9C1=@1;@14171=21A49B1=@9>7

)141;54125A1@1;1841B1A;?4971=1CB531A1C571;C1A1=1C141>71A9B>?A=171=21A

#9;1@5>71=1C=5B531A1C571;=5=5>D89@5AB1=11>C1>.B9>.41>C1>&B9>&,9>1A41C1>741A99;1>B589>771

C1> B9>C1> B9>

& & !. . ! 1C1D

1!

&.

+ !+ ! =1;1

;5419;1>49@5A?H1>74971=1CB531A1C571;CD;BD4DC1>C1A1=1C1@5>71=1C41>71A9B>?A=1 B9> B9>

1!

&.

+& &. +

B9>& B9>J&

JB589>771 C1> C1>

J C1> C1> J

& ! !. !

=#149;541B5=D9;1>D>CD;BD4DC.J=5>:149=

Contoh 6.7

dasar kolam

4 m

nud

= 1

nair

= 4

3

r

d = 4 m

i r d '

i

,521719 ;?>B5;D5>B9 41A9!D;D= " ,>51A 41C1>7 41A9=549D=;DA1>7A1@1C;5=549D=


4. Sudut Kritis dan Pemantulan Sempurna&5>71@1 25A C1=@1; 25A;9 ;5C9;1 3181H1 :1CD8 @141

@5A=D;11>>H1&5>71@1 @141 B11C C5A9;=1C181A9 :1 A1H1 25A1B@11>7 19A)141;54D1 75:171CD B531A1 B5=@DA>1 1719=1>1 @5=1>CDB5=@DA>141@1C C5A:149 )5A81C9;1>$0%$4

1=21A C5AB52DC=5>D>:D;;1> 3181H1 25A75A1;41A9=549D= 7A1@1CD41A11C1D41A9=549D=25A9>45;B291B45;B291B1A291B>H1=5>:1D89 71A9B>?A=17>?H1 :D71>?:D;;1>?1A$5=D491>@141 B11C BD4DC41C1>749@5A25B1A B9>1A B9>1A41> B9>1A BD4DC291B@D>25AC1=21825B1A 1C1D B5=1;9>=5>:1D8971A9B>?A=17C5AC5>CDH1;>9B5@5AC9B9>1A3181H149291B;1>J C5A8141@71A9B>?A=1771 B9>1A291B>H1 B5:1:1A45>71> @5A=D;11> 2941>7 21C1B=549D= ;131D41A1 )141 ;51411>B5@5AC9 9>9 BD4DC41C1>749B52DC BD4DC;A9C9B

5>71>;1C1BD4DC;A9C9B1417;5C9;1B9>1A41C1>749291B;1>B525B1AJ#9;1BD4DC41C1>749@5A25B1A=5;1> 49@1>CD B531A1 B5=@DA>1AC9>H1 3181H1 C941;1;1>;5:D;;1>?1A)5A9BC9F19>9749B52DC@5=1>CDB5=@DA>1

5A41B1A;1> 81CDB5=@DA>181>H1 C5A:149 :9;1=5=5>D89 4D1 BH1A1C 25A9;DC1 181H141C1>741A9=549D= 7 A1@1C2 ,D4DC 41C1>7 n

2)

12 3

45

6i

udara

kacaiii

rrr

ri

r

Aktivitas Fisika 6.3

Sepotong plastik transparanterapung dipermukaan air. Agarsinar datang seperti pada gambardi atas dipantulkan sempurna olehpermukaan batas plastik air. Jikaindek bias plastik n

k dan indek bias

air na, berapa besar sudut ?

Tantanganuntuk Anda

udara

air

plastik

Pemantulan Sempurna

Tujuan PercobaanMengamati pemantulan sempurna.

Alat-Alat PercobaanGelas minum, lilin, dan air.

Langkah-Langkah Percobaan1. Isilah gelas minum dengan air sampai hampir penuh.2. Letakkan gelas tersebut di pinggir meja.3. Nyalakan lilin pendek dan letakkan dekat

dengan gelas.4. Perhatikan gambar berikut.

Amati dari arah yang berseberangandengan lilin dari arah agak dibawah permukaan.

5. Apa yang dapat Anda amati?6. Apa kesimpulan Anda dari kegiatan ini?

)A9>B9@@5=1>CD B5=@DA>149=1>611C;1>41?


Gambar 6.24(a) Pemantulan sempurna

pada serat optik.(b) serat optik.

(a) (b)

Contoh 6.8!9CD>7BD4DC;A9C9B25AH1>7=5=945;B291B=DC (

+& +

&; J#149BD4DC;A9C9B25A141B9>

(& +5 + 1C1D B9>

(

+& + L$5C5A1>71>&;BD4DC;A9C9B

( 1 1sint

i r

5 Pembiasan pada Kaca Plan Paralel$131@@1A1:1>7>H1

492D1C B5:1:1A $131 @ @1A1 D>CD;=5>71=1C9:1>H1B9>1AH1>7=5>7141>D>CD;=5>5>CD;1>9>45;B291B ;131 C5AB52DC

,9>1A 41C1>7 41A9 D41A1=51A=5>7141A9 1A18B9>1AB5=DH1@5A75B5A1>B9>1AC5AB52DC41@1C4989CD>7B52171925A9;DC

)5A81C9;1> ( B9> ( (

0& 6 .

L

">45;B291B H1>7 C9>779 1;1>=5>71;921C;1> BD4DC;A9C9B ;539771B9>1A41C1>745>71>BD4DC41C1>7H1>7C941;C5A71CD B5=@DA>1

1791> 7=5AD@1;1>@5=2D>7;DB C5A2D1C41A9 @7B9=54D>7921791> 9>C9(13181H11C1D 75


(1cos

d

rL

5A41B1A;1> ( 49@5A?(45$0$$1 ; 41>(45$0$$1 ; 49@5A?! 0. & 6 . B589>771 25B1A @5A75B5A1> B9>1A

0

B9> L3?B

! & ..

L

$5C5A1>71>& BD4DC 41C1>7 !C5211A;131@@1A17B525B1AJ#9;1C52125B1A@5A75B5A1>H1>7491 J L J 3?BJ #149;5C9;1=57C5A2D1C41A975B

@1A1>H1>7;54D1B9B9>H14921C1B92941>7@5A=D;1>H1>7A>5=25>CD;BD4DCC5AC5>CD B1CD B1=1 ,D4DC C5AB52DC 49>1=1;1> BD4DC @5=291BB9=2?7@5A=D;11>@A9B=125A6D>7B9 B521719 2941>7@5=291B)5A81C9;1>:1>H1B9>1AH1>7=57 41A9 D41A1=5>D:D 2941>7 @5A=D;11> @A9B=1 1;1>49291B;1>=5>45;1C971A9B>?A=1;5C9;13181H1=5>9>771@A9B=1=5>D:DD41A13181H1C5AB52DC1;1>49291B;1>=5>:1D8971A9B>?A=1 & 6 .

J 6 6

)5A81C9;1> J . 1J . 2

Gambar 6.27Prisma segitiga

Gambar 6.28Jalannya sinar pada prisma

NN

sinardatang

sinarbias

)


J 6 6

J J J 6 6 . 6 6 &

1 2r i L,D4DC 45E91B9>H1 J J J6 6 6 6

& L..&&.L.&

1 2i r L$5C5A1>71> BD4DC@5=291B@A9B=1 . BD4DC291B41A9 B9>1A=1BD;& BD4DC41C1>7 B9>1A=1BD; . BD4DC291B41A9 B9>1A;57 B9>1A ;571>BD4DC41C1>7J25A1@1;18BD4DC45E91B9H1>749177D>1;1>@5AB1=11>

1 2i r(1&41> C59=5>717

=5>781B9 BD4DC 45E91B9 C5AC5>CD ,11A 41C1>7 C5AC5>CD@1BC91;1>=5>781B9>99=D=$1@1>;?>49B9;8DBDB

B9> B9>& +. + 45>71>++D41A141>+;131

B9> B9> B9> .. .'9CD;1>B52171925A9;DC

2 1 1i r r

2 60 19,47 40,53i $5=D491>>9CD;1>B52171925A9;DC

B9> B9> B9> B9> J+. & . .+

B9> 1A3B9> J. . . #149BD4DC45E91B9H1>7491


7


8. Lensa$3


F2

F1A

B

O

ruangbenda

I I I

ruangbenda

I I

ruangbenda

I

ruangbenda

IV

ruangbayangan

(IV)

ruangbayangan

(I)

ruangbayangan

(III)

ruangbayangan

(II)

Gambar 6.34Penamaan ruang pada lensacembung

!.'+!(0%-"%.34+!."!6!.'!.0!$!,%.2!#%-"4.',3?;B2/6>3;033?>=A=3?>/


K

*30B/6:37@;3;7:79773?;B9//3?;B9//


!5!"793A/6B7 1;2723>/


e. Susunan Dua Lensa:/A/:/A=>A79@3>3?A7;79?=@9=>2/=


Gambar 6.40Bagian-bagian mata

fovea

pangkalsaraf optik

retinasklera

iris

kornea

pupil

cairandi sekitar mata

otot siliar

lensa

1. Mata(3?


c. Cacat Mata dan Penanggulangannya(/2/93


2/


1%2")/0) !3! 4!%/A/ AB/;3;7:797 =A=A @7:7/? E/3?A7 >/2/;/A/

/2/8/?/9


/


matapengamat

bayangan

L

okuler

benda

sob s'ob sok

FokF 'ob

Fob

objektif

(a)

Lokuler

benda

sob s'ob sok

Fok

F 'ob

Fob

objektif

(b)

matapengamat

=0 =9 K#3A3?/3?03@/?/


Apabila jarak bayangan yangdihasilkan lup adalah (s = x),artinya mata berakomodasipada jarak x sehinggaperbesaran sudutnya adalah

anguler

PP PPM = +

f x

Untuk mata berakomodasimaksimum, bayangan yang di-bentuk oleh lup terletak dititik dekat mata (s = PP)s = x = PPx = PP

anguler

PP PPM = +

f PP

PP= +

f1

Untuk mata tidakberakomodasi, bayangan yangdibentuk oleh lup terletak dititik jauh mata (s = PR).Untuk mata normal,PR = maka s = . Agarjarak s = maka bendaharus diletakkan di titikfokus F.s = x = x =

anguler

PP PPM = +

f x

PP= +

f0

PP=

f

Ingatlah(3?03@/?/


(3?03@/?/3?03@/?/>/2/@//A;/A/A72/903?/9=;=2/@7/2/:/6=0L=9

Tugas Anda 6.2Anda mungkin pernah melihathasil suatu pencitraan luar angkasayang diperoleh dari teleskopHubble. Bagaimanakah prinsipFisika yang diterapkan padateleskop luar angkasa tersebut?

4. Teropong atau Teleskop+3?=>=;3?B>/9/A79 E//A275B


objektif(+)

okuler()

L

s'ob s

oks

ob = ~

s'ok

= ~

matapengamat

Gambar 6.50Pembentukan bayangan olehteleskop bintang pada mata yangberakomodasi maksimum.

=0 =9 K(/


# %,%2+/01)2-!+3:3@9=>>?7@;/27@30BA8B5/A3:3@9=>07 A3:3@9=>>?7@;/07/@/


bayangan

sob

= ~

okuler()

objektif(+)

s'ob

sok

Fok

FobF

ok

matapengamat

Gambar 6.53Pembentukan bayangan olehteleskop panggung oleh mata yangberakomodasi.

*30B/6A3:3@9=>/:7:37E/3?03@/?/


*30B/69/;3?/;3;7:797A7A79/>7;;/D/:B/3?7@;/275B/9/;3?/?34:39@ :3


3?=:36B?/7/


@3>3?A7

@74/A@74/A/A7>//A707/@9//A//2/>3?;B9//

07_bab6

Documents

Transcript of 07_bab6