Question

If the speed of light in dense flint glass is $\dfrac{3}{5}$ times of its speed in the air, find the refractive index of dense flint glass.
A) 1.5
B) 1.75
C) 1.67
D) 1.1

Answer 1

Hint: The refractive index of a medium describes how light rays bend on entering the medium. This bending of light is called refraction. A denser medium has a higher value for its refractive index. Also, a decrease in the speed of light is observed when light travels from a rarer to a denser medium. Here we consider light to travel from air to dense flint glass.

Formula Used:
The ratio of the refractive indices of two mediums through which light passes through is given by, $\dfrac{{{n_2}}}{{{n_1}}} = \dfrac{{{v_1}}}{{{v_2}}}$ where ${n_1}$ , ${n_2}$ are the refractive indices of medium 1 and medium 2 respectively and ${v_1}$ , ${v_2}$ are the corresponding velocities of light in the two mediums.

Complete step by step answer:
Step 1: List the given parameters of the problem.
We consider the light to be travelling from air to dense flint glass. As dense flint glass is clearly the denser medium of the two, the velocity of light in it decreases and becomes $\dfrac{3}{5}$ times the velocity in air.
The refractive index of air is 1.
Step 2: Express the ratio of the refractive indices of the two mediums involved.
The ratio of the refractive indices of two mediums through which light passes through is given by, $\dfrac{{{n_2}}}{{{n_1}}} = \dfrac{{{v_1}}}{{{v_2}}}$ --------- (1),
where ${n_1}$ ${v_1}$ are the refractive index and velocity of medium 1 and ${n_2}$ , ${v_2}$ are the refractive index and velocity of light in medium 2.
Here, the air is the first medium and the dense flint glass is the second medium.
Let ${n_a}$ and ${n_g}$ be the refractive indices of air and glass respectively and let ${v_a}$ and ${v_g}$ be the corresponding velocities of light in the mediums.
Now using equation (1) we can express the ratio of the refractive indices of air and glass.
From (1), $\dfrac{{{n_g}}}{{{n_a}}} = \dfrac{{{v_a}}}{{{v_g}}}$ or, on rearranging we get, ${n_g} = {n_a}\dfrac{{{v_a}}}{{{v_g}}}$ ------ (2)
Step 3: Find the refractive index of dense flint glass using equation (2).
Equation (2) gives ${n_g} = {n_a}\dfrac{{{v_a}}}{{{v_g}}}$
Substituting the values for ${n_a} = 1$ and ${v_g} = \dfrac{3}{5}{v_a}$ in equation (2) we get, ${n_g} = 1 \times \dfrac{{{v_a}}}{{\left( {\dfrac{3}{5}{v_a}} \right)}}$
Simplifying we get, ${n_g} = \dfrac{5}{3} = 1.67$
∴ the refractive index or index of refraction of dense flint glass is ${n_g} = 1.67$
The correct option is C.

Note: Alternate method
The refractive index of dense flint glass can also be found out using the relation, $n = \dfrac{c}{v}$ where $n$ is the refractive index of the medium, $c$ is the speed of light in air and $v$ is the speed of light in the medium.
Here, the speed of light in a dense glass flint is given by $v = \dfrac{3}{5}c$ .
Then the index of refraction of dense flint glass will be $n = \dfrac{c}{{\left( {\dfrac{3}{5}c} \right)}} = \dfrac{5}{3}$
∴ the refractive index or index of refraction of dense flint glass is $n = \dfrac{5}{3} = 1.67$