Question

A converging lens of focal length 7.90 cm is 17.2 cm to the left of a diverging lens of focal length -5.98 cm. A coin is placed 12.6 cm to the left of the converging lens. Calculate the location of the coin's final image. Calculate the magnification of the coin's final image.

Answer 1

Use the lens equation,

1/f = 1/v + 1/u

for the converging lens

1/7.9 = 1/v + 1/12.6

=> v = 21.178 cm

this will be to the right of the converging lens

so, the object distance for the diverging lens will be:

u' = 21.178 - 17.2 = 3.978 cm

using the lens equation for concave lens, we get:

- 1/5.98 = 1/3.978 + 1/v'

=> v' = - 2.388 cm

that is, the final image will be at a distance of 2.388 cm to the right of the diverging lens.

the magnification of the coin will be:

M = [vv'/uu'] = [2.388 x 21.178/(17.2 x 3.978)] = 0.7394.