no 2 dan 3 pake cara :)

2.
R = 400 cm, f = 1/2.R = 200 cm = 2 m
s = 18 m
s' = jarak bayangan
M = perbesaran bayangan
a.
1/2 = 1/s + 1/s'
1/2 = 1/18 + 1/s'
1/s' = 1/2 - 1/18
1/s' = (9-1)/18
1/s' = 8/18
s' = 18/8 = 2,25 m
b.
M = s'/s
M = 2,25/18
M = 1/8 kali

4.
s = 24 cm
f = 8 cm
h = 4 cm
h' = tinggi bayangan
h = tinggi benda
M = perbesaran bayangan
s' = jarak bayangan
a.
1/f = 1/s + 1/s'
1/8 = 1/24 + 1/s'
1/s' = 1/8 - 1/24
1/s' = (3-1)/24
1/s' = 2/24
s' = 12 cm
b.
M = s'/s
M = 12/24
M = 1/2 kali
c.
M = h'/h
1/2 = h'/4
h' = 4/2
h' = 2 cm
d.
Bayangan besifat nyata, terbalik, dan diperkecil. Benda ada di ruang 3 sedangkan bayangan di ruang 2.