Question

The Brewster angle for air to glass transition of light is 
(Refractive index of glass = \( 1.5 \)

• A. \( \sin^{-1} \left(\frac{3}{2}\right) \)
• B. \( \cos^{-1} \left(\frac{3}{2}\right) \)
• C. \( \tan^{-1} \left(\frac{3}{2}\right) \)
• D. \( \cos^{-1} \left(\frac{2}{3}\right) \)

Hint:

Brewster's angle \( \theta_B \) is given by \( \tan \theta_B = n \). For air-to-glass transition, where \( n = 1.5 \), we use \( \theta_B = \tan^{-1} (1.5) \).

Answer 1

The Correct Option is C

Step 1: Understanding Brewster's Law 
Brewster's angle \( \theta_B \) is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface without any reflection. It is given by: \[ \tan \theta_B = n \] where \( n \) is the refractive index of the second medium (glass) concerning the first medium (air). 

Step 2: Apply the given values 
Given \( n = 1.5 \), the Brewster angle is: \[ \theta_B = \tan^{-1} (n) \] \[ \theta_B = \tan^{-1} \left(\frac{3}{2}\right) \] 

Step 3: Identify the correct option 
From the given answer choices, the correct expression is: \[ \tan^{-1} \left(\frac{3}{2}\right) \] Thus, the correct answer is Option (3).