Question

A ray incident at a point at an angle of incidence of ${60}^o$ 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:

• A. ${50}^0$
• B. ${60}^0$
• C. ${90}^0$
• D. ${40}^0$

Answer 1

Answer: (C)

${90}^0$

Using the law of refraction for the first surface:

$\sin r=\frac{\sin 60}{\sqrt{3}}=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}$

$=\frac{1}{2}$

The angle of refraction will be $r={30}^0$

Now, applying the refraction at the second surface:

$\sqrt{3}\sin r=\sin e$

$\sqrt{3}\times \frac{1}{2}=\sin c=\frac{\sqrt{3}}{2}\Rightarrow e={60}^0$

Now, the sum of angles of a triangle is ${180}^{\circ }$. Use this property to find the angle.

$\alpha +r+e={180}^0$

$\Rightarrow \alpha ={180}^0-r-e$

$\alpha ={180}^0-{30}^0-{60}^0={90}^{\circ }$