berapa letak benda jika bayangan terletak 20 cm dibelakang cermin cekung yang mempunyai jarak fokus 40 cm

1/f = 1/so + 1/si
1/40 = 1/so + 1/20
1/so = 1/40 - 1/20
1/so = 2/80 - 4/80
1/so = -2/80
so = -80/2
so = -40

OPTIKA
• pemantulan cahaya

cermin cekung
f = 40 cm
s' = -20 cm (di belakang cermin)
s = ___? ← di ruang 1

1/s + 1/s' = 1/f
1/s = 1/f - 1/s'
1/s = 1/40 - 1/(-20)
1/s = 1/40 + 2/40
1/s = 3/40
s = 40/3 = 13⅓ cm ✔️

letak benda 13⅓ di depan cermin.