Question

A lens made of material of refractive index ${\mu _2}$ is surrounded by a medium of refractive index ${\mu _1}$. The focal length f is related as:
A. $f \propto (1 - {\mu _2} - {\mu _1})$
B. $f \propto \dfrac{1}{{({\mu _2} - {\mu _1})}}$
C. $f \propto \dfrac{1}{{({\mu _2} + {\mu _1})}}$
D. $f \propto \dfrac{1}{{(1 + {\mu _2} - {\mu _1})}}$

Answer 1

Hint: In order to find the answer, we can use the lens maker’s formula. According to the lens maker’s formula, the focal length depends on the relative refractive index of the lens and the radii of curvature of the two spheres which are used in making the lens by the following equation.
$\dfrac{1}{f} = \left( {{\mu _{rel}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where, ${R_1}$ is the radius of the first sphere and ${R_2}$ is the radius of the second sphere which is used to make the surface of the lens.

Complete step by step answer:
It is given that the refractive index of the lens is ${\mu _1}$
The Refractive index of the surrounding medium is given as ${\mu _2}$
We need to find the relation of focal length with the refractive index of the lens and the refractive index of the surrounding.
We know that the formula of the focal length is given by the lens maker’s formula. According to the lens maker’s formula, the focal length depends on the relative refractive index of the lens and the radii of curvature of the two spheres which are used in making the lens by the following equation.
$\dfrac{1}{f} = \left( {{\mu _{rel}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …………………...(1)
Where, ${R_1}$ is the radius of the first sphere and ${R_2}$ is the radius of the second sphere which is used to make the surface of the lens
${\mu _{rel}}$ is the relative refractive index given by the ratio of refractive index of lens to refractive index of medium.
${\mu _{rel}} = \dfrac{{{\text{refractive}}\,{\text{index}}\,{\text{of}}\,{\text{lens}}}}{{{\text{refractive}}\,{\text{index}}\,{\text{of}}\,{\text{medium}}}}$
Therefore,
${\mu _{rel}} = \dfrac{{{\mu _2}}}{{{\mu _1}}}$ …………...(2)
Now let us substitute the equation (2) in equation (1).
Then we get,
$\Rightarrow \dfrac{1}{f} = \left( {\dfrac{{{\mu _2}}}{{{\mu _1}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Therefore, on solving we get
$\Rightarrow \dfrac{1}{f} = \left( {\dfrac{{{\mu _2} - {\mu _1}}}{{{\mu _1}}}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
From this, we can see that,
$\Rightarrow \dfrac{1}{f} \propto \left( {\dfrac{{{\mu _2} - {\mu _1}}}{{{\mu _1}}}} \right)$
$\Rightarrow f \propto \dfrac{{{\mu _1}}}{{{\mu _2} - {\mu _1}}}$
$\Rightarrow f \propto \dfrac{1}{{({\mu _2} - {\mu _1})}}$

Therefore, option B is the correct answer.

Note:
From Lens maker’s formula
$\dfrac{1}{f} = \left( {\dfrac{{{\mu _2} - {\mu _1}}}{{{\mu _1}}}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
We can see that if net refractive index is decreased for example if we insert the lens in a denser medium than air then the value of ${\mu _2} - {\mu _1}$ will be less, then focal length will increase.