Question

A point object O is placed at a distance of 20 cm in front of a equiconvex lens ${(}^a{\mu }_g=1.5)$ of focal length 10 cm. The lens is placed on a liquid of refractive index 2 as shown in the figure. An image will be formed at a distance h from a lens. The value of h is

• A. 5 cm
• B. 10 cm
• C. 20 cm
• D. 40 cm

Answer 1

The correct option is D 40 cm

As $\frac{1}{f}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}]$

$\frac{1}{10}=(1.5-1)(\frac{1}{R}+\frac{1}{R})=0.5\times \frac{2}{R}\Rightarrow R=10cm$

$\therefore \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

Refraction from first surface,

$\frac{1.5}{v_1}-\frac{1}{-20}=\frac{1.5-1}{+10}\Rightarrow v_1=\infty$

For the second surface,

$\frac{2}{v}-\frac{1.5}{\infty }=\frac{2-1.5}{-10}\Rightarrow v=-40cm\therefore h=40cm$