Question

Two lenses of power 3d and -5d are placed in contact to form a composite lens. An object is placed at a distance 50cm from this composite lense. find the position of image

Answer 1

Given: $P_1 = 3~d; P_2 = -5~d; s = 50~cm.$

Assuming both lenses are thin, the combined optical power will be the sum:

$P = P_1 + P_2 .$

We're going to make use of the thin lens equation:

$\frac{1}{s} + \frac{1}{s'} = \frac{1}{f} = P .$

Solving this for the image distance,

$s' = \frac{s}{Ps - 1} = \frac{s}{(P_1 + P_2)s - 1}.$

Entering the given values,

$s' = \frac{0.5~m}{(3~d - 5~d)*0.5~m - 1} = \frac{0.5~m}{-1 - 1} = -0.25~m = -25~cm.$

The minus sign (negative value) of the image distance indicates that the image is imaginary.