Question

Sebuah lensa bikonkaf simetri berjari jari 20cm berindeks bias 3/2 berapa jarak fokus lensa ketika berada didalam sesuatu medium yang mempunyai indeksbias 1,4

Answer 1

Optik
• lensa

lensa bikonkaf
R1 = R2 = -20 cm
nl = 3/2
nm = 1,4 = 7/5
f = __?

Lens maker formula
1/f = (nl/nm - 1) (1/R1 + 1/R2)
1/f = {(3/2) / (7/5) - 1} (1/(-20) + 1/(-20))
1/f = (15/14 - 1) (-2/20)
1/f = (1/14) (-1/10)
1/f = -1/140
f = - 140 cm ← jwb