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Soal dan Jawaban : Lensa tipis Bagian - 2

Written By onfisika on Monday, March 3, 2014 | 6:04 PM

Soal

Jarak fokus lensa gelas ( n = 1,5 ) di dalam alkohol ( n = 1,35) adalah 45 cm. Hitung jarak fokus dan kuat lensa tersebut di udara.

Jawaban

Diket : f = 45 cm ( Alkohol )

nalk = 1,35

ng = 1,5

nud = 1

Dit : f = …. ?

P = …. ?

Jawab : di alkohol

clip_image050[10] = clip_image114[1]clip_image077[2]

clip_image141 = clip_image143clip_image077[3]

clip_image141[1] = clip_image145clip_image077[4]

clip_image141[2] = clip_image147clip_image077[5]

clip_image141[3] = clip_image149 . clip_image077[6]

clip_image141[4] = clip_image129[1] . clip_image077[7]

clip_image077[8]= clip_image141[5] . clip_image132[1] = clip_image153

di udara

clip_image050[11] = clip_image114[2]clip_image077[9]

clip_image050[12] = clip_image156 . clip_image153[1]

clip_image050[13] = clip_image158 . clip_image153[2]

clip_image050[14] = clip_image160 . clip_image153[3]

clip_image050[15] = clip_image162

f = 10 cm = 0,1 m

P = clip_image050[16]= clip_image164 = 10 dioptri

 

Soal

Sebuah lensa plankonkaf mempunyai panjang fokus –25cm. Jari-jari kelengkungan salah satu permukaannya 12 cm. Hitung indeks bias lensa.

Jawaban

Diket : Plan Konkaf

f = - 25 cm

R1 = ~

R2 = - 12 cm ( berbentuk cekung )

n1 = 1

Dit : n2 = …. ?

Jawab :

clip_image050[17] = clip_image114[3]clip_image077[10]

clip_image167 = clip_image118[1]clip_image169

clip_image167[1] = clip_image122[3]clip_image171

clip_image167[2] = clip_image122[4] . clip_image173

n2 – 1 = clip_image167[3] . clip_image175

n2 = clip_image177+1

n2 = clip_image177[1] + clip_image179 = clip_image181 = 1,48

 

Soal

Sebuah lensa konkaf konveks mempunyai jari-jari kelengkungan 10 cm dan 12 cm terbuat dari kaca dengan indeks bias 1,6. Tentukan:

a. fokus lensa

b. kuat lensa

c. perbesaran bayangan jika sebuah benda diletakkan pada jarak 50 cm.

Jawaban

Diket : konkaf konveks

R1 = -10 cm

R2 = -12 cm

n1 = 1

n2 = 1,6

Dit : a. f = …. ?

b. P = …. ?

c. M = …. ? s = 50 cm

Jawab :

a) clip_image050[18] = clip_image114[4]clip_image077[11]

= clip_image184 . clip_image186

= clip_image188 . clip_image190

= clip_image188[1] . clip_image192

clip_image050[19] = clip_image194

f = -100 cm = -1 m

b) P = clip_image050[20]= clip_image196 = -1 dioptri

c) clip_image050[21] = clip_image098[1]+clip_image100[1] M = clip_image200

clip_image202= clip_image204 + clip_image100[2] M = clip_image207

clip_image100[3] = clip_image202[1]clip_image204[1] M = clip_image209X

clip_image100[4] = clip_image202[2]clip_image211

clip_image100[5] = clip_image213

S1 = clip_image215 cm

 

Soal

Sebuah lensa bikonveks mempunyai jari-jari kelengkungan 9 cm dan 18 cm. Pada jarak 24 cm ternyata bayangan yang terbentuk nyata pada jarak 24 cm dari lensa. Hitung :

a. Jarak fokus

b. Kekuatan lensa

c. Indeks bias lensa

Jawaban

Dik : Lensa bikonveks

clip_image216R1 = 9 cm

R2 = 18 cm

S = 24 cm

S1 = 24 cm ( nyata )

Dit : a. f = …. ?

b. P = …. ?

c. n2 = …. ?

Jawab :

a). clip_image050[22] = clip_image098[2]+clip_image100[6]
= clip_image218+clip_image218[1]
= clip_image220
clip_image050[23] = clip_image222
f = 12 cm
b). P = clip_image094[1]
P = clip_image225
P = clip_image227
P = 8clip_image229 dioptri

c) clip_image050[24] = clip_image114[5]clip_image077[12]

clip_image222[1] = clip_image118[2]clip_image232

clip_image222[2] = clip_image122[5]clip_image234

clip_image222[3] = clip_image122[6] . clip_image236

clip_image222[4] = clip_image122[7] . clip_image238

n2 – 1 = clip_image222[5] . clip_image241

n2 = clip_image160[1]+1

n2 = 1,5

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