Question

The polarising angle of transparent medium is $\theta$. Let the speed of light in the medium be $v$. Then the relation between $\theta$ and $v$ is ($c=$ velocity of light in air)

• A. $\theta=\sin^{-1}(v/c)$
• B. $\theta=\tan^{-1}(v/c)$
• C. $\theta=\cot^{-1}(v/c)$
• D. $\theta=\cos^{-1}(v/c)$

Hint:

Brewster's angle depends on the refractive index of the interface.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question relates Brewster's law (polarizing angle) with the speed of light in a medium.

Step 2: Key Formula or Approach:
1. Brewster's Law: \( \tan \theta = \mu \)
2. Refractive index in terms of speed: \( \mu = \frac{c}{v} \)

Step 3: Detailed Explanation:
Substituting the expression for \( \mu \) into Brewster's Law:
\[ \tan \theta = \frac{c}{v} \] This can be rewritten using the cotangent function as:
\[ \frac{1}{\cot \theta} = \frac{c}{v} \implies \cot \theta = \frac{v}{c} \] Taking the inverse:
\[ \theta = \cot^{-1} \left( \frac{v}{c} \right) \]

Step 4: Final Answer:
The relation is \( \theta = \cot^{-1} \left( \frac{v}{c} \right) \).