Question

A light beam is incident at $\theta$ on an interface of air-glass such that the angle between reflected and the refracted beam is ${90}^0$. Then $\theta$ is

• A. ${\tan }^{-1}\frac{3}{2}$
• B. ${\tan }^{-1}2$
• C. ${\tan }^{-1}\frac{3}{4}$
• D. ${\tan }^{-1}\frac{4}{3}$

Answer 1

The correct option is A

${\tan }^{-1}\frac{3}{2}$

Step 1: Given data

Incident angle= $\theta$

Angle between the reflected and the refracted beam $={90}^0$

Step 2: Calculate the value of theta:

When angle between reflected and refracted beam is ${90}^0,\theta$ is the polarizing angle.

$$\begin{gathered} \therefore \tan i_p=\tan \theta \\ \Rightarrow \tan \theta =\mu \\ \Rightarrow \tan \theta =\frac{{\mu }_g}{{\mu }_a} \\ \Rightarrow \tan \theta =\frac{3}{2}\mathbf{}\mathbf{}\left[\because {\mu }_g=1.5,{\mu }_a=1\right] \\ \Rightarrow \theta ={\tan }^{-1}\frac{3}{2} \end{gathered}$$

Hence, option (A) is correct.