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Rabu, 14 Agustus 2013

SOAL FISIKA

Dik: cermin cekung
M = 2x
S = 22 cm
Dit: S’ dan  f = ... ?
Jawban :
·         M  = I S’/S I
2   = S’/22
2x22  = S’
S’       =  44 cm
·         1/f = 1/S’ + 1/S
1/f = 1/5 + 1/44
1/f = 2+1/44
1/f = 2/44
F = 44/2 = 14,6 cm

11. Dik : cermin cekung
h =h
S= 4 cm
R = 14 cm
Dit : sifat bayangan yang dibentuk ?
Jawab :
 ‘’ gambar difoto ‘’
Maya, tegak, diperbesar

12. Dik: cermin cekung
S = 25 cm
M = 4x
Dit: f = ... ?
Jawab :
·         M = I S’/S I
4 = S’/25
4x25 = S’
S’ = 100 cm
·         1/f = 1/S + 1/S’
1/f = 1/25 + 1/100
1/f  = 1+4/100
1/f = 5/100
F = 100/5 = 20 cm

13. Dik : cermin cekung
S = 20 cm
R = 60 cm
Dit: letak bayangan dan sifat bayangan ?
Jawab :
‘’ gambar difoto’’
Dibelakang cermin, sifat : maya, diperbesar, tegak.

14. Dik: cermin cembung
F = -8 cm
S’ = 6 cm
Dit: S’ ..?
Jawab :
1/f = 1/S + 1/S’
-(1/8)= 1/6 + 1/S’
-(1/8) – 1/6 = 1/S’
-3-4/24 = 1/S’
-7/24 = 1/S’
-24/7 = S’
S’ = -3,42 cm

15. Dik : cermin cembung
S = 16 cm
S’ = -28,5 cm
Dit: f =..?
Jawab :
1/f = 1/S + 1/S’
1/f = 1/16 + (-1/28,5)
1/f = (28,5/456) – (16/456)
1/f = 12,5/456
1/f = 456/ 12,5
 F = 36, 48 cm
F = -36,48 cm (cermin c

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