Question

A right-angled prism made from a material of refractive index $\mu$ is kept in air. A ray PQ is incident normally on the side AB of the prism. Find (in terms of $\mu$ ) the maximum value of $\theta$ up to which this incident ray necessarily undergoes total internal reflection at the face AC of the prism.

Answer 1

Hint:In this question, we know that if the angle between two lines is the same as the angle between their perpendiculars and one angle of right angles prism is always ${90}^{\circ }$. The angle of incidence should be minimum for total internal reflection.

Complete step by step solution:
In this question we have given that a right-angled prism has the refractive index $\mu$. We need to calculate the maximum value of $\theta$ up to which this incident ray necessarily undergoes total internal reflection at the face AC of the prism.

In this question let us assume that the angle of incidence is $i$ and the angle of incidence is incident on $AC$.
The angle between two lines is same as the angle between their perpendiculars so we can write,
$i=A$
Since the prism is right angled, we can write,
$$\Rightarrow \theta ={\displaystyle \frac{\pi }{2}}-A$$
Since $i=A$ we can write,
$\Rightarrow \theta ={\displaystyle \frac{\pi }{2}}-i......\left(1\right)$
We know that refractive index is expressed as,
$\mu ={\displaystyle \frac{1}{\sin i}}$
As we know that the angle of incidence should be minimum for the total internal reflection,
$i_c={\sin }^{-1}{\displaystyle \frac{1}{\mu }}......\left(2\right)$
Here, $i_c$ is a critical angle.

Now the maximum value of $\theta$ for total internal expression can be calculated by substituting the expression of $i_c$ to the equation (1) as,
$\Rightarrow \theta ={\displaystyle \frac{\pi }{2}}-i_C$

Now we substitute the value of critical angle as expressed in equation (2) in above equation to get the maximum value of $\theta$.
$\therefore {\theta }_{max}={\displaystyle \frac{\pi }{2}}-{\sin }^{-1}{\displaystyle \frac{1}{\mu }}$

Therefore, the maximum value will be ${\theta }_{max}={\displaystyle \frac{\pi }{2}}-{\sin }^{-1}{\displaystyle \frac{1}{\mu }}$.

Note:As we know that the minimum angle for total internal reflection is the critical angle. The angle of the right-angled prism is ${90}^{\circ }$. The maximum value of $\theta$ is calculated corresponding to the minimum value of $i$.