Question

A ray of light travelling in a transparent medium of refractive index $\mu $, falls on a surface separating the medium from air at an angle of incidence of ${45^ \circ }$. For which of the following values of $\mu $ the ray can undergo total internal reflection?
(A) $\mu = 1.33$
(B) $\mu = 1.40$
(C) $\mu = 1.50$
(D) $\mu = 1.25$

Answer 1

Hint
In this problem, the ray of light travels from one medium to another medium. The main condition for total internal reflection is, the light must travel from one medium to another medium. So, this problem is solved by using the total internal reflection formula.

Complete step by step solution
Given that, Angle of incidence, $i = {45^ \circ }$
For total internal reflection, the angle of incidence is greater than critical angle,
$\sin i > \sin C\,..................\left( 1 \right)$
Where, $i$ is the angle of incidence, $C$ is the critical angle.
The sine of critical angle is equal to the reciprocal of the refractive index of the medium.
So, $\sin C = \dfrac{1}{\mu }$
Here, we have to find the refractive index so keep the refractive index on one side and other terms in other side, then the above equation is written as,
$\mu = \dfrac{1}{{\sin C}}\,.......................\left( 2 \right)$
Substitute the $\sin C$ value from equation (1) in the equation (2), then the above equation is written as,
$\mu > \dfrac{1}{{\sin i}}$
Substitute the angle of incidence in the above equation, then the above equation is written as,
$\mu > \dfrac{1}{{\sin {{45}^ \circ }}}$
From trigonometry, the value of $\sin {45^ \circ }$ is $\dfrac{1}{{\sqrt 2 }}$, then substitute this value in above equation,
$\mu > \dfrac{1}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}}$
On further simplification,
$\mu > \sqrt 2 $
By taking square root,
$\mu > 1.414$
Therefore, from the given option, the option (C) is greater than $1.414$.
Hence, the option (C) is the correct answer.

Note
This problem is also solved by another method by using Snell’s law. But when we are using Snell’s law to solve this problem, we have to assume that the angle of incidence and critical angle are equal. And the refractive index of air value can also be used to solve this problem when we are using Snell’s law.