Question

When unpolarised light from air incidents on the surface of a medium of refractive index $\sqrt{3}$, then the reflected light is totally polarised. The angle of refraction is

• A. 30$^\circ$
• B. 53$^\circ$
• C. 60$^\circ$
• D. 37$^\circ$

Hint:

Brewster's Law ($\tan\theta_B = n$) gives the polarizing angle. A key property at this angle is that the reflected and refracted rays are perpendicular to each other ($i+r=90^\circ$). This provides a quick way to find the angle of refraction once the angle of incidence is known.

Answer 1

The Correct Option is A

The problem states that when unpolarised light is incident, the reflected light is totally polarised. This occurs at a specific angle of incidence known as Brewster's angle, $\theta_B$.
Brewster's Law gives the relationship between Brewster's angle and the refractive index ($n$) of the medium:
$\tan(\theta_B) = n$.
We are given the refractive index of the medium as $n = \sqrt{3}$.
So, $\tan(\theta_B) = \sqrt{3}$.
This means the angle of incidence (Brewster's angle) is $\theta_B = 60^\circ$.
Let the angle of incidence be $i = \theta_B = 60^\circ$.
Let the angle of refraction be $r$.
We can find the angle of refraction using Snell's Law:
$n_1 \sin(i) = n_2 \sin(r)$.
Here, the light comes from air, so $n_1 = 1$. The medium has refractive index $n_2 = n = \sqrt{3}$.
$1 \cdot \sin(60^\circ) = \sqrt{3} \cdot \sin(r)$.
$\frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin(r)$.
$\sin(r) = \frac{1}{2}$.
Therefore, the angle of refraction is $r = 30^\circ$.
Alternatively, when light is incident at Brewster's angle, the reflected and refracted rays are perpendicular to each other. $i+r = 90^\circ$. $60^\circ + r = 90^\circ \implies r = 30^\circ$.