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) $${75}^{\circ }$$
B) $${30}^{\circ }$$
C) $${45}^{\circ }$$
D) $${60}^{\circ }$$

Answer 1

Hint: When a ray of light within a medium like (glass or water) is completely reflected from the surrounding surfaces back to the medium, is called total internal reflection. This phenomenon occurs when the angle of incidence is greater than the critical angle.

Complete step by step solution:
Step I:
Since the given ray of light is travelling in two mediums ( glass sphere and air), it follows Snell’s Law. According to Snell’s Law, the ratio of sines of angle of incidence and refraction are constant for a wave if it is made to pass between two given mediums.

Step II:
The total internal reflection occurs at point B, so point B is symmetric about the normal. But the incident ray is parallel to the final outgoing ray, so the incident ray is parallel to the normal at point B. The refraction at point A occurs in accordance with Snell's Law.

Let $$\theta$$be the angle of incidence and $$r$$ be the angle of refraction. The refractive index of glass is $$\sqrt{3}$$.

Step III:
So according to Snell’s Law:
$${\displaystyle \frac{\sin \theta }{\sin r}}=\sqrt{3}$$---(i)
As OA and OB is the radius of the sphere, so the angles A and B will be equal.
$\Rightarrow$ $$\mathrm{\angle }A=\mathrm{\angle }B=r_1$$
Where$$r_1$$ is the angle of refraction.
OB is the normal and is parallel to the incident ray. So it can be taken as transversal.
$\Rightarrow$ $$\mathrm{\angle }AOB=180-\theta$$.

Step IV:
Sum of angles of a triangle is $${180}^{\circ }$$
$\Rightarrow$ $$\mathrm{\angle }AOB+\mathrm{\angle }A+\mathrm{\angle }B={180}^{\circ }$$
Substituting the values of all the angles of triangle,
$\Rightarrow$ $$180-\theta +r_1+r_1={180}^{\circ }$$
$\Rightarrow$ $$180-\theta +2r_1=180$$
$\Rightarrow$ $$-\theta =-2r_1$$
$\Rightarrow$ $$\theta =2r_1$$---(ii)

Step V:
Substituting value of $$\theta$$ in equation (i),
$\Rightarrow$ $${\displaystyle \frac{\sin 2r_1}{\sin r_1}}=\sqrt{3}$$
Using identity $$\sin 2\theta =2\sin \theta \cos \theta$$ in the numerator,
$\Rightarrow$ $${\displaystyle \frac{2\sin r_1\cos r_1}{\sin r_1}}=\sqrt{3}$$
$\Rightarrow$ $$2\cos r_1=\sqrt{3}$$
$\Rightarrow$ $$\cos r_1={\displaystyle \frac{\sqrt{3}}{2}}$$
The value of cos is $${\displaystyle \frac{\sqrt{3}}{2}}$$ when $$r_1={30}^{\circ }$$.

Step VI:
Substituting value of$$r_1$$ in equation (ii),
$\Rightarrow$ $$\theta =2r_1$$
$\Rightarrow$ $$\theta =2\times 30$$
$\Rightarrow$ $$\theta ={60}^{\circ }$$

Step VII:
The angle of incidence is $${60}^{\circ }$$.

Option (D) is the right answer.

Note: It is to be noted that even if light is internally reflected at the interface, but at the same interface refraction of light also takes place. This means that whenever a light ray gets reflected from the surface of a sphere, some part of the light escapes and undergo refraction of light. This can decrease the intensity of the light ray everytime it undergoes reflection. After some time it completely gets diminished in the form of refracted rays from the surface of the glass sphere.