Question

Benda terletak didepan cermin cekung sejauh 30 cm.jika jarak fokus cermin 15 cm,perbesaran bayangan yang dihasilkan cermin adalah

Answer 1

1/f = 1/s + 1/s'
1/15 = 1/30 + 1/s'
1/s' = 1/15 - 1/30
1/s' = 2/30 - 1/30
1/s' = 1/30
s' = 30/1
s' = 30 cm

M = s'/s
    = 30/30
    = 1 

Answer 2

Diketahui.
So = 30 cm
f = 15 cm
ditanyakan. M =?
jawab.
1/f = 1/so + 1/si
1/15 = 1/30 + 1/si
-1/si = 1/30 - 1/15
-1/si = 1/30 - 2/30
-1/si = -1/30
1/si = 1/30
Si = 30 cm

M = |si/so|
M = 30/30
M = 1 kali