Question

Give pictorial representation of polarized and unpolarised light. Polarizing angle for a transparent medium is 60°. Find the value of angle of refraction and refractive index of the medium.

Hint:

Remember Brewster's law relates the polarizing angle and the refractive index: \( \tan \theta_p = n \). This is useful for finding the refractive index of a medium when the polarizing angle is known. For angles of refraction, use Snell's Law: \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \).

Answer 1

Unpolarized light is light that oscillates in all directions perpendicular to the direction of propagation. Polarized light, on the other hand, only oscillates in one direction. The process of polarizing light involves passing it through a polarizing filter, which only allows light oscillating in a specific direction to pass through.
The diagram below shows the difference between polarized and unpolarized light:
In the case of the polarized light, the oscillations are restricted to one plane, while in unpolarized light, the oscillations are in multiple planes.
Now, the problem gives the polarizing angle \( \theta_p \) for a transparent medium as \( 60^\circ \). The polarizing angle is related to the refractive index \( n \) of the medium using Brewster's Law, which is given by: \[ \tan \theta_p = n, \] where \( \theta_p \) is the polarizing angle. Substituting the given value of \( \theta_p = 60^\circ \): \[ \tan 60^\circ = n. \] Since \( \tan 60^\circ = \sqrt{3} \), we have: \[ n = \sqrt{3}. \] Thus, the refractive index of the medium is \( n = \sqrt{3} \approx 1.732 \).
Next, we can find the angle of refraction using Snell's Law, which is: \[ n_1 \sin \theta_1 = n_2 \sin \theta_2, \] where \( n_1 \) and \( n_2 \) are the refractive indices of the two media, \( \theta_1 \) is the angle of incidence, and \( \theta_2 \) is the angle of refraction. Since the angle of incidence \( \theta_1 \) is equal to the polarizing angle \( \theta_p = 60^\circ \), and we know the refractive index of air is approximately 1, we can use: \[ 1 \cdot \sin 60^\circ = \sqrt{3} \cdot \sin \theta_2. \] Solving for \( \sin \theta_2 \): \[ \sin \theta_2 = \frac{\sin 60^\circ}{\sqrt{3}} = \frac{\frac{\sqrt{3}}{2}}{\sqrt{3}} = \frac{1}{2}. \] Thus, \( \theta_2 = 30^\circ \). Therefore, the angle of refraction is \( 30^\circ \).