NO 2 pakai cara (soal yang linda)

cermin cekung
f = R/2 = 3/2 m
s = 1 m ← di R1
s' = __? ← di R4

1/s + 1/s' = 1/f
1/s' = 1/f - 1/s
1/s' = 1/(3/2) - 1/1
1/s' = 2/3 - 3/3
1/s' = - 1/3
s' = - 3 m ← jwb

jarak bayangan terhadap cermin 3 m (maya(

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Diket :
Cermin Cekung (Positif)
Jari-jari, R = 3 m → f = R/2 = 3/2 m
jarak Linda, S = 1 m

Tanya :
Jarak bayangan Linda, S' = __?

Jawab :

1/s' = 1/f - 1/s
1/s' = 1/(3/2) - 1/1
1/s' = 2/3 - 1/1
1/s' = 2/3 - 3/3
1/s' = - 1/3
S' = - 3 m (tanda negatif artinya : bayangan maya, tegak dan jatuh dibelakang cermin cekung)


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
)|(
FZA