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

Hint: First you have to know the properties of lenses. A lens when silvered on its surface acts as a mirror. Its net focal length can be calculated by using the formula
$\dfrac { 1 }{ { f }_{ net } } \quad =\quad \dfrac { 2 }{ { f }_{ l } } \quad +\quad \dfrac { 1 }{ { f }_{ m } }$ where ${ { f }_{ l } }$ is the focal length of the lens and ${ { f }_{ m } }$ is the focal length of the silvered surface.

Complete step by step answer:
First we will find out the focal length of the plano-convex lens from the net focal length formula,
$\dfrac { 1 }{ { f }_{ net } } \quad =\quad \dfrac { 2 }{ { f }_{ l } } \quad +\quad \dfrac { 1 }{ { f }_{ m } }$. This formula comes from the fact that when a light ray falls on a lens silvered on one side, it undergoes refraction twice through the lens' non-silvered surface and reflection due to the silvered once.
Case 1: It is a concave mirror of focal length 28cm.
${ f }_{ m }\quad =\quad \infty$
Hence $\dfrac { 1 }{ 28 } \quad =\quad \dfrac { 2 }{ { f }_{ l } } \quad +\quad \dfrac { 1 }{ { f }_{ m } }$
$\Rightarrow \quad { f }_{ 1 }\quad =\quad 56$
Case 2: When it is a concave mirror of focal length 10cm.
Hence, $\dfrac { 1 }{ 10 } \quad =\quad \dfrac { 2 }{ { f }_{ l } } \quad +\quad \dfrac { 1 }{ { f }_{ m } }$
Substituting the value of ${ f }_{ l }$, we get
$\Rightarrow \quad \dfrac { 1 }{ 10 } \quad =\quad \dfrac { 2 }{ 56 } \quad +\quad \dfrac { 1 }{ { f }_{ m } }$
$\Rightarrow \quad \dfrac { 1 }{ { f }_{ m } } \quad =\quad \dfrac { 1 }{ 10 } \quad -\quad \dfrac { 1 }{ 28 }$
$\Rightarrow \quad \dfrac { 1 }{ { f }_{ m } } \quad =\quad \dfrac { 140 }{ 9 } cm$
Therefore, the radius of curvature,
$R\quad =\quad 2\times { f }_{ m }\quad =\quad \dfrac { 280 }{ 9 } cm$
Since, we know the radius of curvature for the curved surface of the lens and the net focal length of the lens, it is easy to find out the refractive index of the material of the lens by using the formula $\dfrac { 1 }{ f } \quad =\quad \left( \mu -1 \right) \dfrac { 1 }{ R }$. This comes from the lensmaker’s formula and denotes the refractive index of the material.
By substituting the values of focal length and radius of curvature in the equation
$\dfrac { 1 }{ 56 } \quad =\quad \left( \mu -1 \right) \dfrac { 1 }{ \dfrac { 280 }{ 9 } } \\\ \Rightarrow \quad \mu -1\quad =\quad \dfrac { 280 }{ 9 } \times \dfrac { 1 }{ 56 } \quad =\quad \dfrac { 5 }{ 9 } \\\ \Rightarrow \quad \mu \quad =\quad 1\quad +\quad \dfrac { 5 }{ 9 } \quad =\quad \dfrac { 14 }{ 9 }$

Therefore the correct answer is option B, $\dfrac { 14 }{ 9 }$

Note: While substituting the values of focal lengths and radius of curvature in the formulae, there is a huge chance for you to commit mistakes while considering the surfaces. You have to know at this point that the focal length of any plane mirror is $\infty$ because its radius of curvature is $\infty$ as its surface is straight and not curved.