Question

A beam of light reflects and refracts at the surface between air and glass. The index of refraction of the glass is \(1.4\). If the refracted and reflected rays are perpendicular to each other, then the angle of incidence in the air is:

• A. \( \tan^{-1}(1.4) \)
• B. \( \sin^{-1} \left( \frac{1}{1.4} \right) \)
• C. \( \tan^{-1} \left( \frac{1}{1.4} \right) \)
• D. \( \sin^{-1} \left( \frac{1.4}{\pi} \right) \)

Hint:

When reflected and refracted rays are perpendicular, use \( i + r = 90^\circ \), and apply Snell’s law to relate sine and cosine.

Answer 1

The Correct Option is A

Given: Refracted and reflected rays are perpendicular, so angle between them = \(90^\circ\) Let \( i \) be angle of incidence and \( r \) be angle of refraction. Then: \[ i + r = 90^\circ \Rightarrow r = 90^\circ - i \] Using Snell’s Law: \[ n_1 \sin i = n_2 \sin r \quad \text{(air to glass)} \quad \text{with } n_1 = 1, n_2 = 1.4 \] \[ \sin i = 1.4 \sin(90^\circ - i) = 1.4 \cos i \Rightarrow \tan i = 1.4 \Rightarrow i = \tan^{-1}(1.4) \]