Question

A point object is placed in air at a distance of $40$ cm from a concave refracting surface of refractive index $1.5$ If the radius of curvature of the surface is $20$ cm, then the position of the image is-

Answer 1

Given object distance u=40 cm, radius of curvature R=20 cm

refractive index $n_2=1.5$ and $n_1=1$

Now making use of the relation

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

$\frac{n_2}{v}=\frac{n_2-n_1}{R}+\frac{n_1}{u}$

$\frac{n_2}{v}=\frac{1.5-1}{-20}-\frac{1}{-40}$

$\frac{n_2}{v}=\frac{2}{-40}$

$v=1.5\times -20=-30$