Question

A light ray incident at a point on the surface of a glass sphere of $\mu = \sqrt{3}$ at an angle of incidence $60^\circ$. It is reflected and refracted at the farther surface of sphere. Find the angle between reflected and refracted ray.

• A. 90°
• B. 60°
• C. 30°
• D. 120°

Answer 1

Answer: (A)

90°

Answer 2

At the first surface (air to glass), the angle of incidence $i_1 = 60^\circ$. Using Snell's Law, $\mu_a \sin i_1 = \mu_g \sin r_1$, we get $1 \cdot \sin 60^\circ = \sqrt{3} \sin r_1$, which gives $\sin r_1 = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$, so $r_1 = 30^\circ$.

At the farther surface (glass to air), the angle of incidence $i_2$ is equal to the angle of refraction at the first surface, so $i_2 = r_1 = 30^\circ$. Using Snell's Law for refraction from glass to air, $\mu_g \sin i_2 = \mu_a \sin r_2$, we get $\sqrt{3} \sin 30^\circ = 1 \cdot \sin r_2$, which gives $\sin r_2 = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$, so $r_2 = 60^\circ$.

The angle between the reflected ray and the refracted ray at any surface is the sum of the angle of incidence and the angle of refraction at that surface. At the farther surface, this angle is $i_2 + r_2 = 30^\circ + 60^\circ = 90^\circ$.