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

To solve this problem, let's start by using Brewster's Law, which tells us that the angle of polarization occurs when the refractive index of the medium is the tangent of the angle of polarization. Mathematically, this can be represented as:

\[ n = \tan 60^\circ = \sqrt{3} \]

Here, \( n \) is the refractive index of the medium.

Next, we need to find the critical angle (\( C \)). The critical angle is the angle of incidence that provides an angle of refraction of \( 90^\circ \). It is given by the equation:

\[ \sin C = \frac{1}{n} \]

Substituting the value of \( n = \sqrt{3} \) gives:

\[ \sin C = \frac{1}{\sqrt{3}} \]

Therefore, the critical angle \( C \) is:

\[ C = \sin^{-1} \frac{1}{\sqrt{3}} \]

The correct answer is \( \sin^{-1} \frac{1}{\sqrt{3}} \).

Answer 2

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 \( \sin^{-1} \frac{1}{\sqrt{3}} \).