Question

A ray of light on a glass sphere (refractive index $\sqrt{3}$ suffers total internal reflection before emerging out exactly parallel to the incident ray. The angle of incidence was:

• A. ${75}^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${60}^o$

Answer 1

The correct option is D ${60}^o$

The entire path of the light is symmetric about the normal at B (Total internal reflection point). But, as Incident ray is parallel to the final outgoing ray, we have incident ray parallel to the normal at B.

At the point A, the refraction is governed by the Snell's law.

$\frac{sin\theta }{sin{r}_1}=\sqrt{3}$ -------------(A)

As $OA=OB$ (radius of sphere), we have angles at $A$ and $B$ equal. i.e., $\angle A=\angle B=r_1$

As $OB$ is parallel to the incident ray, and the normal acts as transversal $\angle AOB=180-\theta$

Thus, $(180-\theta )+2\times r_1=180$ (Sum of angles of triangle)

i.e., $\theta =2\times r_1$

Using this in equation (A),

$\frac{sin\left(2r_1\right)}{sin\left(r_1\right)}=\sqrt{3}\Rightarrow \frac{2sin\left(r_1\right)cos\left(r_1\right)}{sin\left(r_1\right)}=\sqrt{3}\Rightarrow cos\left(r_1\right)=\frac{\sqrt{3}}{2}$

Thus, $r_1={30}^o\Rightarrow \theta ={60}^o$

Incidence angle of the incident ray = ${60}^o$