Question

A plano-convex lens, when silvered at its plane surface it is equivalent to concave mirror of focal length $28\text{cm}$. When its curved surface is silvered, it is equivalent to a concave mirror of focal length $10\text{cm}$. The refractve index of the material of the lens is

• A. $\frac{9}{14}$
• B. $\frac{14}{9}$
• C. $\frac{17}{9}$
• D. None of these

Answer 1

The correct option is B $\frac{14}{9}$

Let $f_1$ be focal length of the plano convex lens.

Case-I : When it is a concave mirror of focal length $28\text{cm}$.

Focal length of plane silvered surface, $f_m=\infty$

Hence , $-\frac{1}{28}=\frac{2}{f_1}-\frac{1}{f_m}$

$\Rightarrow f_1=-56\text{cm}.....\left(1\right)$

Case-II : When it is a concave mirror of focal length $10\text{cm}$.
Here $f_m$ is focal length of silvered curved surface.

Hence, $-\frac{1}{10}=\frac{2}{f_1}-\frac{1}{f_m}$

Substituting the value of $f_1$ we get ,

$-\frac{1}{10}=\frac{2}{-56}-\frac{1}{f_m}$

$\Rightarrow \frac{1}{f_m}=\frac{1}{10}-\frac{1}{28}$

$\Rightarrow f_m=\frac{140}{9}\text{cm}$

Therefore, radius of curvature , $R=2f_m$

$\Rightarrow R=2\times \frac{140}{9}=\frac{280}{9}\text{cm}$

Now, using lens maker formula, focal length of the plano convex lens will be given by

$\frac{1}{f_1}=\frac{(\mu -1)}{R}$

Substituting the data obtained we get,

$\mu -1=\frac{280}{9}\times \frac{1}{56}$

$\Rightarrow \mu =1+\frac{5}{9}=\frac{14}{9}$

Hence , option (b) is the correct answer.