Question

A ray of light travelling in a medium of refractive index $n_1$ is incident on a medium of refractive index $n_2(<n_1)$ at an angle of incidence $\theta$. The reflected and refracted rays make an angle of ${120}^0$ with each other. Then the critical angle for the boundary separating the two media is

• A. $si{n}^{-1}⟮\frac{1}{2}tan\theta ⟯$
• B. $si{n}^{-1}⟮\frac{1}{\sqrt{3}+tan\theta }⟯$
• C. $si{n}^{-1}⟮\frac{2tan\theta }{\sqrt{3}-tan\theta }⟯$
• D. $si{n}^{-1}\left(2cot\theta \right)$

Answer 1

The correct option is C $si{n}^{-1}⟮\frac{2tan\theta }{\sqrt{3}-tan\theta }⟯$

$\begin{array}{c}\text{Applying snell's}1a\omega \text{:}\\ n_1\sin {\gamma }_1=n_2\sin {\gamma }_2\\ \Rightarrow n_1\sin \theta =n_2\sin {\gamma }_2\\ \Rightarrow {\gamma }_2={\sin }^{-1}\left(\frac{n_1}{n_2}\sin \theta \right)\\ {\gamma }_2+\theta +{120}^{\circ }={180}^{\circ }\text{(according to question,)}\\ \end{array}$

$\begin{array}{c}\Rightarrow {\gamma }_2={60}^{\circ }-\theta \\ \Rightarrow {\sin }^{-1}\left(\frac{n_1}{n_2}\sin \theta \right)={60}^{\circ }-\theta \\ \Rightarrow \frac{n_1}{n_2}\sin \theta =\sin ({60}^{\circ }-\theta )\end{array}$

$\begin{array}{c}\Rightarrow \frac{n_1}{n_2}\sin \theta =\sin {60}^{\circ }\cos \theta -\cos {60}^{\circ }\sin \theta \\ \Rightarrow \frac{n_1}{n_2}=\frac{\sqrt{3}}{2}\times \frac{1}{\tan \theta }-\frac{1}{2}\\ \Rightarrow \frac{n_1}{n_2}=\frac{\sqrt{3}-\tan \theta }{2+\mathrm{an}\theta }\\ \Rightarrow \frac{n_2}{n_1}=\frac{2\tan \theta }{\sqrt{3}-\tan \theta }\end{array}$

$\begin{array}{c}\text{.}{\sin }^{-1}\left(\frac{n_2}{n_1}\right)={\sin }^{-1}\left(\frac{2\tan \theta }{\sqrt{3}-\tan \theta }\right)\\ \text{Lritical angle}={\sin }^{-1}\left(\frac{2\tan \theta }{\sqrt{3}-\tan \theta }\right)\\ \text{Hence, option}c\text{is correct.}\end{array}$