Question

The polarising angle for a medium is found to be \( 60^\circ \). The critical angle of the medium is

• A. \( \sin^{-1} \left( \frac{1}{2} \right) \)
• B. \( \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \)
• C. \( \sin^{-1} \left( \frac{1}{\sqrt{3}} \right) \)
• D. \( \sin^{-1} \left( \frac{1}{4} \right) \)
• E. \( \sin^{-1} \left( \frac{2}{\sqrt{3}} \right) \)

Hint:

Remember: \(\tan i_p = \mu\) and \(\sin C = 1/\mu\) . Therefore, a very direct shortcut is \(\sin C = 1 / \tan i_p = \cot i_p\) . For \( i_p = 60^\circ \), \( \cot 60^\circ = 1/\sqrt{3} \)

Answer 1

The Correct Option is C

Concept: This problem combines Brewster's Law and the concept of the Critical Angle for total internal reflection.
• Brewster's Law: Relates the refractive index (\( \mu \)) to the polarizing angle (\( i_p \)): \( \mu = \tan i_p \).
• Critical Angle (\( C \)): Relates the refractive index (\( \mu \)) to the angle at which total internal reflection starts: \( \mu = \frac{1}{\sin C} \implies \sin C = \frac{1}{\mu} \).

Step 1: Find the refractive index \( \mu \).
Given the polarizing angle \( i_p = 60^\circ \): \[ \mu = \tan 60^\circ = \sqrt{3} \]

Step 2: Solve for the critical angle \( C \).
Using the relation for the critical angle: \[ \sin C = \frac{1}{\mu} \] \[ \sin C = \frac{1}{\sqrt{3}} \] \[ C = \sin^{-1} \left( \frac{1}{\sqrt{3}} \right) \]