## In This post you will get answer of Q.

The radii of curvature of the two surfaces of a lens are 20 cm and 30 cm and the refractive index of the material of the lens is 1.5. If the lens is concavo-convex, then the focal length of the lens is

Solution:

## The focal length of the lens

$ frac{1}{f}=(mu-1)bigg(frac{1}{R_1} -frac{1}{R_2}bigg)$

$ frac{1}{f}=(1.5-1)bigg(frac{1}{20} -frac{1}{30}bigg)$

$ frac{1}{f}=0.5 bigg(frac{30-20}{600} bigg)$

or $ frac{1}{f}= frac{1}{2} times frac{10}{600}= frac{1}{120}$

or $ f = 120, cm$