Question

A ray of light travelling in a medium of refractive index ri1 is incident on a medium of refractive index $n_2(<n_1)$ at an angle of incidence $Thaeta$. The reflected and refracted rays make an angle of ${120}^o$ with each other. Then the critical angle for the boundary separating the two media is

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

Answer 1

Answer: (C)

$si{n}^{-1}\left(\frac{tan\theta }{\sqrt{3-tan\theta }}\right)$

$\begin{array}{c}\sin i_c=\frac{n_2}{n_1}-\left(1\right)\\ 90-\gamma +90-\theta =120\\ \therefore \theta +\gamma ={60}^{\circ }\\ \therefore \frac{\sin \theta }{\sin \gamma }=\frac{n_2}{n_1}\\ \therefore \frac{\sin (60-\gamma )}{\sin \gamma }=\frac{n_2}{n_1}\end{array}$

$\begin{array}{c}\Rightarrow \frac{\sin 60\cos \gamma -\cos 60\sin \gamma }{\sin \gamma }=\frac{n_2}{n_1}\\ \Rightarrow \frac{\sqrt{3}}{2}\cot \gamma -\frac{1}{2}=\frac{n_2}{n_1}\\ \Rightarrow \frac{\sqrt{3}}{2}\cot (60-\theta )-\frac{1}{2}=\frac{n_2}{n_1}\\ \Rightarrow \frac{\sqrt{3}}{2}\left(\frac{1+\sqrt{3}\tan \theta }{\sqrt{3}-\tan \theta }\right)-\frac{1}{2}=\frac{n_2}{n_1}-\left(11\right)\end{array}$

from $\left(1\right)$ and $\left(11\right)$

$\therefore \sin i_e=\frac{1}{2}\left(\frac{\sqrt{3}+3\tan \theta \sqrt{3}\tan \theta }{\sqrt{3}-\tan \theta }\right)$

$\therefore \sin t_c=\frac{1}{2}\left(\frac{2\tan \theta }{\sqrt{3}\tan \theta }\right)$

$\Rightarrow i_c={\sin }^{-1}\left(\frac{\tan \theta }{\sqrt{3}-\tan \theta }\right)$

No option matches.