Question

A ray of light from a denser medium strikes a rarer medium at an angle of incidence i If the angle of reflection is $\gamma$ and the angle of refraction is ${\gamma }^{\prime }$ and the reflected and refracted rays make an angle of ${90}^o$ with each other, then the critical angle will be:

• A. ${\sin }^{-1}\left(\tan \gamma \right)$
• B. ${\sin }^{-1}\left(\tan {\gamma }^{\prime }\right)$
• C. ${\tan }^{-1}\left(\sin \gamma \right)$
• D. ${\tan }^{-1}\left(\sin {\gamma }^{\prime }\right)$

Answer 1

Answer: (A)

${\sin }^{-1}\left(\tan \gamma \right)$

$\begin{array}{c}{\mu }_1\sin \gamma ={\mu }_2\sin {\gamma }^{\prime }-\left(1\right)\\ 90\gamma +90{\gamma }^{\prime }=90\text{(since given)}\\ \gamma +{\gamma }^{\prime }={90}^{\circ }-\left(2\right)\end{array}$

for critical angle $C$

${\mu }_1\sin i={\mu }_2\sin \gamma$

$\begin{array}{cc}{\mu }_1\sin c& ={\mu }_2\sin 90\\ c=& {\sin }^{-1}\left(\frac{{\mu }_2}{{\mu }_1}\right)\end{array}$

$\begin{array}{cc}& \text{equation}1{\mu }_1\sin \gamma ={\mu }_2\sin {\gamma }^1(\because i=\gamma )\end{array}$

$\begin{array}{c}{M}_1\sin \gamma ={\mu }_2\sin (90-{\gamma }^{\circ })\text{by -(2)}\\ {\mu }_1\sin \gamma ={\mu }_2\cos \gamma \\ \frac{M_2}{{\mu }_1}=\tan \gamma \\ C={\sin }^{-1}\left(\frac{{\mu }_2}{{\mu }_1}\right)={\sin }^{-1}\left(\tan \gamma \right)\\ \text{so option}A\text{is correct.}\end{array}$