Question

A plano-convex lens, when silvered at its plane surface is equivalent to a concave mirror of focal length 28 cm. When its curved surface is silvered and the plane surface is not silvered, it is equivalent to a concave mirror of focal length 10 cm, then the refractive 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 $\frac{14}{9}$

Given: A plano-convex lens, when silvered at its plane surface is equivalent to a concave mirror of focal length 28 cm. When its curved surface is silvered and the plane surface is not silvered, it is equivalent to a concave mirror of focal length 10 cm.

To find the refractive index of the material of the lens.

Solution:

Let $f_m$ be focal length of the curved silvered surface

Case (i): when it is a concave mirror of focal length 28cm

$f_m=\infty$

Hence $\frac{1}{28}=\frac{2}{f_1}+\frac{1}{f_m}⟹f_1=\mathrm{56....}\left(i\right)$

Case (ii): when it is a concave mirror of focal length 10 cm

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

Substituting the value of $f_1$, we get

$⟹\frac{1}{10}=\frac{2}{56}+\frac{1}{f_m}⟹\frac{1}{f_m}=\frac{1}{10}-\frac{1}{28}⟹\frac{1}{f_m}=\frac{28-10}{280}⟹f_m=\frac{280}{18}=\frac{140}{9}cm$

Therefore, the radius of curvature, $R=2f_m=2\times \frac{140}{9}=\frac{280}{9}cm$

Now,

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

Substituting the respective values, we get

$\frac{1}{56}=(\mu -1)\frac{1}{\frac{280}{9}}⟹\mu -1=\frac{280}{9}\times \frac{1}{56}⟹\mu =1+\frac{5}{9}=\frac{14}{9}$

is the refractive index of the material of the lens.