Question

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

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

Answer 1

Answer: (D)

As $\frac{1}{f}=\left(\mu -1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$\frac{1}{10}=\left(1.5-1\right)\left(\frac{1}{R}+\frac{1}{R}\right)$
$=0.5\times \frac{2}{R}$
$\Rightarrow R=10cm$
$\therefore \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
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,
$\frac{2}{v}-\frac{1.5}{\infty }=\frac{2-1.5}{-10}$
$\Rightarrow v=-40cm$
$\therefore h=40cm$