Question

A diverging lens with f = -31.5 cm is placed 14.0 cm behind a converging lens with f = 20.0 cm. Where will an object at infinity be focused?

Answer 1

Using the lens equation for first converging lens

1/f = 1/u + 1/v

u = object distance = +infinity

f = focal length = +20 cm

v = image distance = ?

1/v = 1/20 - 1/infinity

v = +20 cm

Now this image will be +20 cm right from the converging lens, Image's distacne from diverging lens will be

u1 = 14 - 20 = -6 cm = object distnace for diverging lens

f1 = focal length of diverging lens = -31.5 cm

v1 = image distance = ?

1/v1 = -1/31.5 + 1/6

v1 = 31.5*6/(31.5 - 6) = 7.41 cm

v1 = +7.41 cm (+ve sign means image to the right of diverging lens) OR (behind the diverging lens)

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