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Q. A ray incident at a point at an angle of incidence of $60^{\circ}$ enters a glass sphere of refractive index $\sqrt{3}$ and is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is

Ray Optics and Optical Instruments

Solution:

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Refraction at $P$,
$\frac{sin\,60^{\circ}}{sin\,r_1} = \sqrt{3}$
$sin\,r_1 = \frac{1}{2}$ or $r_1 = 30^{\circ}$
Since, $r_2 = r_1$;
$\therefore r_2 = 30^{\circ}$
Refraction at $Q, \frac{sin\,r_2}{sin\,i_2} = \frac{1}{\sqrt{3}}$ or
$\frac{sin\,30^{\circ}}{sin\,i_2} = \frac{1}{\sqrt{3}}$ or
$i_2 = 60^{\circ}$
At point $Q, r'_2 = r_2 = 30^{\circ}$
$\therefore \alpha = 180^{\circ} - (r'_2 +i_2) $
$= 180^{\circ} -(30^{\circ} + 60^{\circ} ) = 90^{\circ}$