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) $\dfrac{9}{{14}}$
B) $\dfrac{{14}}{9}$
C) $\dfrac{{17}}{9}$
D) None of these.

Answer 1

In this problem, when the plane surface of a plano-convex lens which is silvered the lens is equivalent to a concave mirror with focal length $28$ $cm$. And when the curved surface is silvered, the plano-convex lens is silvered it is equivalent to a concave mirror of focal length $10$ $cm$. We need to find the refractive index of the material of the lens. First calculate the focal length of the plano-convex lens and the radius then using the lens maker formula, we can calculate the refractive index.

Complete step by step solution:
The refractive index of a lens, its focal length and the radius of the curvature are related using the lens maker formula. To apply the lens maker formula, first we need to find the focal length ${f_{net}}$ of the plano-convex lens, this is given using the net focal length formula:
$\dfrac{1}{{{f_{net}}}} = \dfrac{2}{{{f_1}}} + \dfrac{1}{{{f_m}}}$

Case 1 : We have a concave mirror with focal length ${f_{net}} = 28$ $cm$. As the focal length of plane mirror is ${f_m} = \infty$ , we have:
$\dfrac{1}{{{f_{net}}}} = \dfrac{2}{{{f_1}}} + \dfrac{1}{{{f_m}}}$
$\Rightarrow \dfrac{1}{{28}} = \dfrac{2}{{{f_1}}} + \dfrac{1}{\infty }$
$\Rightarrow \dfrac{1}{{28}} = \dfrac{2}{{{f_1}}} + 0$
$\Rightarrow {f_1} = 56$ $cm$.----equation $1$

Case 2 : When the focal length of the concave mirror is ${f_{net}} = 10$ $cm$.
In this case the convex surface is silvered and the plane surface is not silvered.
$\dfrac{1}{{{f_{net}}}} = \dfrac{2}{{{f_1}}} + \dfrac{1}{{{f_m}}}$
$\Rightarrow \dfrac{1}{{10}} = \dfrac{2}{{56}} + \dfrac{1}{{{f_m}}}$
$\Rightarrow \dfrac{1}{{{f_m}}} = \dfrac{1}{{10}} - \dfrac{1}{{28}}$
$\Rightarrow \dfrac{1}{{{f_m}}} = \dfrac{{18}}{{280}}$
$\Rightarrow {f_m} = \dfrac{{140}}{9}$ $cm$
The radius of curvature is calculated as:
$R = 2 \times {F_m}$
Substituting the value of ${f_m} = \dfrac{{140}}{9}$ , we get
$R = 2 \times \dfrac{{140}}{9}$
$\Rightarrow R = \dfrac{{280}}{9}$ $cm$.
Now, we have the radius of curvature and the focal length thus, we can find the refractive index using lens maker formula as:
$\dfrac{1}{{{f_1}}} = \left( {\mu - 1} \right)\dfrac{1}{R}$
Substituting the values, we have:
$\Rightarrow \dfrac{1}{{56}} = \left( {\mu - 1} \right)\dfrac{1}{{\dfrac{{280}}{9}}}$
$\Rightarrow \left( {\mu - 1} \right) = \dfrac{1}{{56}} \times \dfrac{{280}}{9}$
$\Rightarrow \left( {\mu - 1} \right) = \dfrac{5}{9}$
$\Rightarrow \mu = \dfrac{5}{9} + 1$
$\Rightarrow \mu = \dfrac{{14}}{9}$
The refractive index of the material of the lens is $\dfrac{{14}}{9}$ .

Thus, option B is the correct option.

Note: Follow the approach as taken in this problem. Lens maker formula relates the focal length, radius of curvature and the refractive index of the material of the lens. The value of $\dfrac{1}{\infty }$ is taken to be zero. The focal length for a plane mirror is considered to be at an infinite distance from the surface of the mirror.