Question

A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is ${\theta _{iC}}$ ​ and the Brewster's angle of incidence is ${\theta _{iB}}$ ​, such that $sin{\theta _{iC}}/sin{\theta _{iB}} = n = 1.28$. The relative refractive index of the two media is:
A) $0.2$
B) $0.4$
C) $0.8$
D) $0.9$

Answer 1

Hint The critical angle for total internal reflection and the Brewster angle of incidence are both related to the ratio of the refractive index of the denser to the rarer medium. Find the Brewster angle in terms of $\sin {\theta _{iB}}$ and take its ratio with the critical angle.

Formula used:
-Critical angle of incidence: $\sin {\theta _{iC}} = \dfrac{{{n_2}}}{{{n_1}}}$
- Brewster’s angle of incidence: $\tan {\theta _{iC}} = \dfrac{{{n_2}}}{{{n_1}}}$

Complete step by step answer
The critical angle of incidence ${\theta _{ic}}$ is the highest incident angle possible after which the light ray will be deflected by the denser medium than getting refracted.
So we have
$\Rightarrow \sin {\theta _{iC}} = \dfrac{{{n_2}}}{{{n_1}}}$
If a ray of light is incident at an interface between two media in such a manner that the reflected and transmitted rays are at right angles to each other, the angle of incidence is called the Brewster angle and it is defined as:
$\Rightarrow \tan {\theta _{iC}} = \dfrac{{{n_2}}}{{{n_1}}}$
We can calculate $\sin {\theta _{iB}}$ from $\tan {\theta _{iB}}$ as
$\Rightarrow sin{\theta _{iB}} = \dfrac{{{n_2}}}{{\sqrt {\left( {n_1^2 + n_2^2} \right)} }}$
But since we want to find the ratio of
$\Rightarrow sin{\theta _{iC}}/sin{\theta _{iB}}$, we can write
$\Rightarrow sin{\theta _{iC}}/sin{\theta _{iB}} = \dfrac{{\dfrac{{{n_2}}}{{{n_1}}}}}{{\dfrac{{{n_2}}}{{\sqrt {\left( {n_1^2 + n_2^2} \right)} }}}} = 1.28$
Taking the square on both sides, we get
$\Rightarrow \dfrac{{\dfrac{{n_2^2}}{{n_1^2}}}}{{\dfrac{{n_2^2}}{{n_1^2 + n_2^2}}}} = 1.6384$
On simplifying, we get
$\Rightarrow \dfrac{{n_1^2 + n_2^2}}{{n_1^2}} = 1.6384$
Multiplying both sides $n_1^2$ and taking the terms containing $n_1^2$ on one side, we get
$\Rightarrow \dfrac{{n_2^2}}{{n_1^2}} = 0.6384$
Taking the square root on both sides, we get
$\dfrac{{{n_2}}}{{{n_1}}} = 0.8$ which corresponds to option (C).

Note
To solve such questions, we must be aware of the different phenomena about refraction of light in different mediums such as total internal reflection and Brewster’s angle of incidence. Also, since we’ve been asked about the relative refractive index, we only have to find the ratio of the refractive index of the two mediums and not individually.