Question

A candle (ho = 0.39 m) is placed to the left of a diverging lens (f = -0.054 m). The candle is do = 0.34 m to the left of the lens.

h_o = 0.39 m
f = -0.054 m
d_o = 0.34 m

Part (a) Write an expression for the image distance, d_i.

Part (b) Numerically, what is the image distance in meters?

Part (c) Is this real or virtual?

Part (d) Numerically, what is the image height, h_i?

Answer 1

a) 1/do + 1/di = 1/f

1/di = 1/f - 1/do

di = (do*f)/(do-f)

b) di = -0.34*0.054/(0.34+0.054) = -0.047 m

c)No. the image is virtual

d) m = -di/do = 0.047/0.34 = 0.137

hi = m*ho

= 0.137*0.39

= 0.053 m

Answer 2

Answer 3