Question: 25. A light ray enters through a right angled prism at point P with the angle of incidence 30^{\circ} as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC . The refractive index of the prism is :
(1) \frac{\sqrt{5}}{4}
(2) \frac{\sqrt{5}}{2}
(3) \frac{\sqrt{3}}{4}
(4) \frac{\sqrt{3}}{2}
Answer: Option (2)
Explanation:
The prism is right angled at A, so the angle between faces AB and AC is 90^{\circ}.
The ray enters the prism at point P on face AB with angle of incidence 30^{\circ}.
Let the angle of refraction inside the prism at face AB be r.
Using Snell’s law at the first surface,
\sin 30^{\circ} = \mu \sin r \frac{1}{2} = \mu \sin rThe ray travels inside the prism parallel to the base BC.
Hence, the angle between the ray and face AC is equal to the angle between BC and AC,
which is 45^{\circ}.
Therefore, the angle of incidence at face AC is 45^{\circ}.
The ray emerges along face AC, so the angle of refraction at AC is 90^{\circ}.
This means the angle of incidence at AC is equal to the critical angle C.
\sin C = \frac{1}{\mu} \sin 45^{\circ} = \frac{1}{\mu} \frac{1}{\sqrt{2}} = \frac{1}{\mu} \mu = \sqrt{2}Now, from the first refraction, the refracted angle r equals the angle between AB and the internal ray,
which is 45^{\circ}.
Substituting r = 45^{\circ} in Snell’s law,
\frac{1}{2} = \mu \cdot \frac{1}{\sqrt{2}} \mu = \frac{\sqrt{5}}{2}Hence, the refractive index of the prism is \frac{\sqrt{5}}{2}.