Cermin cekung berjari jari 8cm tinggi benda 2cm terletak 1cm di depan cermin. Tentukan: A. Jarak bayangan B. Perbesaran C. Tinggi bayangan

Cermin Cekung (positif)
R = 8 cm → f = R/2 = 8/2 = 4 cm

tinggi benda, h = 2 cm
jarak benda, s = 1 cm

A) Jarak bayangan, S' = ___?

1/s + 1/s' = 1/f

1/1 + 1/s' = 1/4

1/s' = 1/4 - 1/1

1/s' = 1/4 - 4/4

1/s' = - 3/4

s' = - 4/3 cm (Tanda minus : Maya, tegak, dibelakang cermin cekung)

B) Perbesaran, M = ___?

M = - s'/s
M = - (- 4/3) / 1

M = + 4/3 kali (Tanda + artinya Tegak, Diperbesar)

C) Tinggi bayangan, h' = ___?

h' = M x h

h' = 4/3 x 2
 
h' = 8/3 cm