Question

A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens. If the lens is removed, the point where the rays meet will move 5 cm closer to the mounting that holds. Find the focal length ( in cm) of the lens without sign.

Answer 1

Given u=10 cm, v=15 cm

$\frac{1}{f}$= $\frac{1}{v}$- $\frac{1}{u}$

$\frac{1}{f}$ =$\frac{1}{15}$- $\frac{1}{10}$

$\frac{1}{f}$ =$\frac{10-15}{10}$

$\frac{1}{f}$ =$\frac{-5}{150}$

f = - 30