Question

A ray of light is incident on the face AB of a glass prism ABC having the vertex angle A equal to ${30}^{\circ }$. The face AC is silvered and a ray of light incident on the face AB retraces its path. If the refractive index of the material of prism is $\sqrt{3}$, the angle of incidence on the face AB is :

• A. 60${}^{\circ }$
• B. 30${}^{\circ }$
• C. 45${}^{\circ }$
• D. 90${}^{\circ }$

Answer 1

The correct option is A 60${}^{\circ }$

Now look at the diagram carefully.

Face AC is silvered means the light ray, falling on it will get reflected.

The ray re-traces its path means, it is moving backward along the same path through which it entered.

This is possible only if the ray is falling on AC perpendicularly.

$\underset{–}{Given:}$

$\bullet$ $\angle A={30}^{{}^O}$

$\bullet$ $\mu =\sqrt{3}$

$To$ $find:$ $i=$ $?$

$\underset{–}{Solution:}$

Look at the triangle $△APQ.$

In any triangle, the sum of all angles is always ${180}^{{}^O}$

Since ray PQ is perpendicular to AC, $\angle AQP={90}^{{}^O}$

In $△APQ,$ $\angle APQ$ $+$ $\angle A$ $+$ $\angle AQP$ $=180$

$\therefore \angle APQ$ $=$ $180-\angle A$ $-$ $\angle AQP$

$\therefore \angle APQ=180-30-90=60$

SInce line $PR$ is a normal, $\angle APR=90$

By angle addition property, $\angle APR=\angle APQ+\angle r$

$\therefore \angle r=\angle APR-\angle APQ=90-60=30$

By Snell's law, $\frac{\sin i}{\sin r}=\mu$

$\therefore \frac{\sin i}{\sin 30}=\sqrt{3}$

$\therefore \frac{\sin i}{\left(\frac{1}{2}\right)}=\frac{\sqrt{3}}{1}$ $\because$ $\sin 30=\frac{1}{2}$

$\therefore \left(\sin i\right)÷\left(\frac{1}{2}\right)=(2\times \sin i)=\frac{\sqrt{3}}{1}$

$\therefore \sin i=\frac{\sqrt{3}}{2}$

$\therefore i={\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=60$

$\underset{–}{Answer}:$ The angle of incidence on the face AB is ${60}^{{}^O}$