Benda berada 5 cm di depan lensa cekung yang berjarak fokus 20 cm tentukan: a. Kekuatan lensa b. Letak bayangan c. Perbesaran bayangan d. Sifat bayangan
titania20
A. P = 100/f P = 100/20 P = 5 D b. 1/f = 1/s + 1/s' 1/20 = 1/5 + 1/s' 1/20 - 1/5 = 1/s' 1/20 - 4/20 = 1/s' -3/20 = 1/s' -20/3 = s' - 6,67 = s' c. M = s'/s = 6,67/5 = 1.334 kali d. maya, tegak, diperbesar
P = 100/20
P = 5 D
b. 1/f = 1/s + 1/s'
1/20 = 1/5 + 1/s'
1/20 - 1/5 = 1/s'
1/20 - 4/20 = 1/s'
-3/20 = 1/s'
-20/3 = s'
- 6,67 = s'
c. M = s'/s
= 6,67/5
= 1.334 kali
d. maya, tegak, diperbesar