Sebuah benda terletak 60 cm di depan cermin cekung yang memiliki jari-jari kelengkungan 30 cm . Perbesaran bayangan yang terjadi adalah Sama caranya yh ,makasihh:)
willyngham21
D1: s= 60 cm R= 30 f=1/2R (15) D2= M D3=1/f=1/s+1/s' 1/15=1/60+1/s' 1/s'= 4/60-1/60 =3/60 =1/20 s'=20 cm M=-s'/s =-{20/60} =1/3 kali
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awsahra
S = 60 R = 30 f = R/2 → f = 30/2 = 15 .. 1/f = 1/s + 1/s' 1/15 = 1/60 + 1/s' 1/15 - 1/60 = 1/s' 4-1/60 = 1/s' 3/60 = 1/s' 1/20 = 1/s' ( kali silang) s' = 20 cm .. perbesaran! M = s' / s = 20 / 60 = 1/3 kali
R= 30
f=1/2R (15)
D2= M
D3=1/f=1/s+1/s'
1/15=1/60+1/s'
1/s'= 4/60-1/60
=3/60
=1/20
s'=20 cm
M=-s'/s
=-{20/60}
=1/3 kali
R = 30
f = R/2 → f = 30/2 = 15
..
1/f = 1/s + 1/s'
1/15 = 1/60 + 1/s'
1/15 - 1/60 = 1/s'
4-1/60 = 1/s'
3/60 = 1/s'
1/20 = 1/s' ( kali silang)
s' = 20 cm
..
perbesaran!
M = s' / s = 20 / 60 = 1/3 kali