Question

A thin bi-convex lens $\left( {n = 1.5} \right)$ has a focal length of $50\,{\text{cm}}$ in air. When immersed in a transparent liquid, the focal length is measured to be $250\,{\text{cm}}$. The refractive index of the liquid medium is:
(A) $1.33$
(B) $1.46$
(C) $1.30$
(D) $1.36$

Answer 1

First of all, we will find two expressions (one in air and the another in the medium) of focal length of using lens maker formula. We will reduce the two expressions and substitute the required values and manipulate accordingly.

Complete step by step answer:
In the given question,
The refractive index of the lens is $1.5$ .
The focal length of the lens in the air is measured to be $50\,{\text{cm}}$ .
When the lens is immersed in a transparent liquid, its focal length becomes $250\,{\text{cm}}$ .
We are asked to calculate the refractive index of the medium into which the lens is immersed.
To approach this problem, we must know that the focal length of any lens also depends on the surrounding medium and its refractive index. In some cases the focal length may decrease while in some cases it may increase. It is purely a relative phenomenon. This happens because this is the case of refraction. In refraction, the direction of rays depends on the refractive index of the two media.
We know,
$1\,{\text{cm}} = {10^{ - 2}}\,{\text{m}}$
So,
$50\,{\text{cm}} = 0.5\,{\text{m}}$ and,
$250\,{\text{cm}} = 2.5\,{\text{m}}$
We will apply the lens maker formula, which is given below:
In the air medium:
$\dfrac{1}{f} = \left( {\dfrac{{{n_2}}}{{{n_1}}} - 1} \right) \times \left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (1)
In the liquid medium:
$\dfrac{1}{{f'}} = \left( {\dfrac{{{n_2}}}{{{n_m}}} - 1} \right) \times \left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (2)
Where,
$f'$ indicates focal length of the lens inside the liquid.
$f$ indicates focal length of the lens in air.
${n_2}$ indicates the refractive index of the lens.
${n_1}$ indicates a refractive index of air.
${n_m}$ indicates a refractive index of the liquid medium.
${R_1}$ and ${R_2}$ indicates the radius of curvature of the two surfaces of the lens.
Now, dividing equation (1) by equation (2), we get:
$\dfrac{{\left( {\dfrac{1}{f}} \right)}}{{\left( {\dfrac{1}{{f'}}} \right)}} = \dfrac{{\left( {\dfrac{{{n_2}}}{{{n_1}}} - 1} \right)}}{{\left( {\dfrac{{{n_2}}}{{{n_m}}} - 1} \right)}} \times \dfrac{{\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}{{\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}$
$\dfrac{{f'}}{f} = \dfrac{{\left( {\dfrac{{{n_2}}}{{{n_1}}} - 1} \right)}}{{\left( {\dfrac{{{n_2}}}{{{n_m}}} - 1} \right)}}$ …… (3)
Substituting the required values in equation (3), we get:
\dfrac{{f'}}{f} = \dfrac{{\left( {\dfrac{{{n_2}}}{{{n_1}}} - 1} \right)}}{{\left( {\dfrac{{{n_2}}}{{{n_m}}} - 1} \right)}} \\\
\dfrac{{2.5}}{{0.5}} = \dfrac{{\left( {\dfrac{{1.5}}{1} - 1} \right)}}{{\left( {\dfrac{{1.5}}{{{n_m}}} - 1} \right)}} \\\
5 = \dfrac{{0.5}}{{\dfrac{{1.5}}{{{n_m}}} - 1}} \\\
\dfrac{{1.5}}{{{n_m}}} - 1 = \dfrac{{0.5}}{5} \\\
Manipulating further we get:
\dfrac{{1.5}}{{{n_m}}} = \dfrac{{0.5}}{5} + 1 \\\
\dfrac{{1.5}}{{{n_m}}} = 1.1 \\\
{n_m} = \dfrac{{1.5}}{{1.1}} \\\
{n_m} = 1.36 \\\
Hence, the refractive index of the liquid medium is $1.36$ .
The correct option is (D).

Note: Refractive index of a medium affects the focal length of lens, as these are based on the principle of refraction. Higher the refractive index of a medium, slower is the velocity of light in that medium.