Question

A ray of light travels from a denser medium to a rarer medium. The reflected and the refracted rays are perpendicular to each other. If 'r' and '$r_1$' are the angle of reflection and refraction respectively and 'C' is the critical angle, then the angle of incidence is

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

Hint:

Whenever the reflected and refracted rays form a right angle, it meets the exact condition for Brewster's Law angle matching: $\mu = \tan i$. Since the denser medium refractive index is also $\frac{1}{\sin C}$, equating them gives $\cot i = \frac{1}{\sin C} \implies \tan i = \sin C$. This allows you to jump straight to the answer without a full geometric breakdown!

Answer 1

The Correct Option is C

Let $i$ be the angle of incidence. By the law of reflection, the angle of reflection is also equal to $i$ ($r = i$). Given that the reflected ray and refracted ray are completely perpendicular to each other, the geometry across the boundary flat line sets up the angle sum: $$r + 90^\circ + r_1 = 180^\circ \implies i + r_1 = 90^\circ \implies r_1 = 90^\circ - i$$ According to Snell's Law for light traveling from a denser medium into a rarer medium, the refractive index ($\mu$) of the denser host medium relative to the surrounding space is: $$\mu = \frac{\sin r_1}{\sin i} = \frac{\sin(90^\circ - i)}{\sin i} = \frac{\cos i}{\sin i} = \cot i$$ We also know that the refractive index is linked to the critical angle ($C$) by the standard reciprocal relation: $$\mu = \frac{1}{\sin C}$$ Equating both expressions for the refractive index $\mu$: $$\cot i = \frac{1}{\sin C} \implies \tan i = \sin C$$ $$i = \tan^{-1}(\sin C)$$
Final Answer:
The angle of incidence is $\tan^{-1}(\sin C)$, which corresponds to option (C).