Question

The angle of polarisation for a medium with respect to air is \( 60^\circ \). The critical angle of this medium with respect to air is:

• A. \( \sin^{-1} \sqrt{3} \)
• B. \( \tan^{-1} \sqrt{3} \)
• C. \( \cos^{-1} \sqrt{3} \)
• D. \( \sin^{-1} \frac{1}{\sqrt{3}} \)

Hint:

The critical angle is the angle of incidence at which total internal reflection occurs. It can be calculated using \( \sin(\theta_c) = \frac{1}{n} \), where \( n \) is the refractive index of the medium.

Answer 1

The Correct Option is D

We are given that the angle of polarisation for a medium with respect to air is \( 60^\circ \). The question asks for the critical angle of this medium with respect to air. The relationship between the angle of polarisation (\( \theta_p \)) and the critical angle (\( \theta_c \)) is given by Brewster's Law, which states that: \[ \tan(\theta_p) = \frac{n_2}{n_1} \] where \( n_1 \) is the refractive index of the medium and \( n_2 \) is the refractive index of air. Since \( \theta_p = 60^\circ \), we can find the refractive index \( n \) of the medium. 
Step 1: Applying Brewster's Law 
Using Brewster's Law for the angle of polarisation: \[ \tan(60^\circ) = \sqrt{3} = \frac{n_2}{n_1} \] \[ n_1 = \frac{1}{\sqrt{3}} \] where \( n_1 \) is the refractive index of the medium. 
Step 2: Finding the critical angle 
The critical angle \( \theta_c \) is related to the refractive index \( n_1 \) by: \[ \sin(\theta_c) = \frac{1}{n_1} \] Substituting the value of \( n_1 \): \[ \sin(\theta_c) = \frac{1}{\sqrt{3}} \] Thus, the critical angle is: \[ \theta_c = \sin^{-1} \left( \frac{1}{\sqrt{3}} \right) \] Thus, the critical angle of the medium with respect to air is \( \sin^{-1} \frac{1}{\sqrt{3}} \).