Question

Two converging lenses having focal lengths of 11.0 cm and 21.0 cm are placed 50 cm apart, as shown in Figure P23.44. The final image is to be located between the lenses, at the position indicated. Figure P23.44 (a) How far to the left of the first lens should the object be positioned? cm (b) What is the overall magnification? (c) Is the final image upright or inverted, and is it real or virtual?

Answer 1

SOLUTION :

A)

By lens formula on second lens

v= 31-50 = -19 cm

f =21 cm

by lens formula

-1/u + 1/v = 1/f

-1/u - 1/19 = 1/21

-1/u = 1/21 + 1/19

= 0.10025

u = -9.975 cm

This is the image by first lens

v = 50-9.975 = 40.025 cm

f =11 cm

by lens formula again,

-1/u + 1/ v = 1/f

-1/u + 1/40.025 = 1/11

-1/u = 1/11 - 1/40.025

u = -15.168 cm

Hence the object should be placed at 15.168 cm left of first lens

b)

m = - (40.025 / 15.168) *(19 / 9.975) = -5.026

c)

inverted as it can be seen in figure