Question

A plano - convex lens becomes an optical system of $28\text{cm}$ focal length when its plane surface is silvered and illuminated from left to right as shown in fig.$\text{A}$. If the same lens is instead silvered on the curved surface and illuminated from other side as in fig.$\text{B}$, it acts like an optical system of focal length $10\text{cm}$. The refractive index of the material of lens is

• A. $1.50$
• B. $1.55$
• C. $1.75$
• D. $1.4$

Answer 1

The correct option is B $1.55$

We know that,

Power of silvered lens will be
$$P=2P_L+P_m$$
where , $P_L$ is power of lens and $P_m$ is power of mirror.

For plane mirror, $P_m=0$ $(\because f=\infty )$

Power of plano-convex lens, $P_L=\frac{\mu -1}{R}$

If plane surface is silvered then, power of silvered lens is given by,

$P=\frac{2(\mu -1)}{R}$

$\therefore$ Focal length of lens $f_1=\frac{R}{2(\mu -1)}........\left(1\right)$


If curved surface is silvered then power of silvered lens is given by

$P_2=2P_L+P_m$

$\Rightarrow P_2=\frac{2(\mu -1)}{R}+\frac{2}{R}=\frac{2\mu }{R}$

$\therefore$ Focal length , $f_2=\frac{R}{2\mu }......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{f_1}{f_2}=\frac{R}{2(\mu -1)}\times \frac{2\mu }{R}\Rightarrow \frac{f_1}{f_2}=\frac{\mu }{(\mu -1)}$

Given , $f_1=28\text{cm}$ and $f_2=10\text{cm}$

$\therefore \frac{28}{10}=\frac{\mu }{\mu -1}$

$\Rightarrow 28\mu -28=10\mu$

$\Rightarrow 18\mu =28$

$\Rightarrow \mu =1.55$

Hence, option (b) is the correct answer.