Question

A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to $60^\circ$. The refractive index of the prism material is $ \sqrt{2} $. The angle of incidence (in degrees) is ______ .

Hint:

For minimum deviation, the angle of incidence is equal to the angle of emergence. The relationship between the refractive index and the minimum deviation is given by the formula \( \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \).

Answer 1

Correct Answer: 45

Given that \( A = 60^\circ \), the formula for \( \mu \) simplifies to \[ \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \] This leads to \( \therefore \delta_m = 30^\circ \). Using the relation \( \delta_m = 2i - A \), where \( i = e \), we find \( \therefore i = 45^\circ \).