Question

A ray of light passes from a denser medium to a rarer medium at an angle of incidence \(i\). The reflected and refracted rays make an angle of \(90^\circ\) with each other. The angle of reflection and refraction are respectively \(r\) and \(r'\). The critical angle is given by

• A. \(\sin^{-1}(\tan r)\)
• B. \(\sin^{-1}(\cot r)\)
• C. \(\sin^{-1}(\tan r')\)
• D. \(\tan^{-1}(\sin i)\)

Hint:

Reflected and refracted rays perpendicular gives \(r + r' = 90^\circ\) with \(r = i\); then \(\sin C = \dfrac{\sin i}{\sin r'}\) reduces to \(\tan i\).

Answer 1

The Correct Option is A

By the law of reflection \(r = i\).
The reflected and refracted rays are perpendicular, so \(r + r' = 90^\circ\) and \(\sin r' = \cos r\).
Snell's law from the denser (\(n_1\)) to the rarer (\(n_2\)) medium: \(n_1\sin i = n_2\sin r'\), so \(\dfrac{n_2}{n_1} = \dfrac{\sin i}{\sin r'}\).
Critical angle: \(\sin C = \dfrac{n_2}{n_1} = \dfrac{\sin i}{\sin r'} = \dfrac{\sin r}{\cos r} = \tan r\).
Hence \(C = \sin^{-1}(\tan r)\).
Answer: (1) \(\sin^{-1}(\tan r)\)