Question

Let the refractive index of a denser medium with respect to a rarer medium be \(n_{12}\) and its critical angle be \(\theta_C\). At an angle of incidence \(A\) when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is \(90^\circ\). Angle \(A\) is given by :

• A. \(\frac{1}{\cos^{-1}(\sin\theta_C)}\)
• B. \(\frac{1}{\tan^{-1}(\sin\theta_C)}\)
• C. \(\cos^{-1}(\sin\theta_C)\)
• D. \(\tan^{-1}(\sin\theta_C)\)

Hint:

Reflected and refracted rays at \(90^\circ\) give a refraction angle \(r=90^\circ-A\) (Brewster's condition); use Snell's law with \(\sin\theta_C=\frac{1}{n_{12}}\).

Answer 1

The Correct Option is D

Angle of reflection \(=A\). If the angle between reflected and refracted rays is \(90^\circ\), the angle of refraction is \(r=90^\circ-A\).
Snell's law (denser to rarer): \(n_{12}\sin A=\sin r=\cos A\Rightarrow\tan A=\frac{1}{n_{12}}\).
Critical angle: \(\sin\theta_C=\frac{1}{n_{12}}\), so \(\tan A=\sin\theta_C\Rightarrow A=\tan^{-1}(\sin\theta_C)\).
Answer: (4) \(\tan^{-1}(\sin\theta_C)\)