Question

A ray of light in air is incident at angle \(i\) on a face of an equilateral glass prism and is refracted through the prism. As \(i\) is varied, it is observed that the ray undergoes minimum deviation, when the \(i\) is three-fourth of the angle of the prism. Calculate the speed of light in the prism.

Hint:

At minimum deviation, prism behaves symmetrically: $i = e$.

Answer 1

Concept: For minimum deviation in prism: \[ i = e,\quad r = \frac{A}{2} \] Given: Equilateral prism: \[ A = 60^\circ \]

Step 1: Given relation \[ i = \frac{3}{4}A = \frac{3}{4} \times 60^\circ = 45^\circ \]

Step 2: Refraction angle At minimum deviation: \[ r = \frac{A}{2} = 30^\circ \]

Step 3: Snell’s law \[ \mu = \frac{\sin i}{\sin r} \] \[ \mu = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2} \]

Step 4: Speed of light \[ v = \frac{c}{\mu} = \frac{3 \times 10^8}{\sqrt{2}} \approx 2.12 \times 10^8 \ \text{m/s} \] \[ \boxed{v \approx 2.0 \times 10^8 \ \text{m/s}} \]