Question

A plane electromagnetic wave is incident on an interface AB separating two media (refractive indices 𝑛₁ = 1.5 and 𝑛₂ = 2.0) at Brewster angle 𝜃𝐵, as schematically shown in the figure. The angle 𝛼 (in degrees) between the reflected wave and the refracted wave is:

• A. 120
• B. 116
• C. 90
• D. 74

Answer 1

The Correct Option is C

At Brewster’s angle \( \theta_B \), the reflected and refracted rays are perpendicular to each other. The Brewster angle can be calculated using the formula:

\[ \tan(\theta_B) = \frac{n_2}{n_1} \]

Substituting the given values:

\[ \tan(\theta_B) = 1.5 \quad \Rightarrow \quad \theta_B = \tan^{-1}(1.3333) \approx 53.1^\circ \]

Since the reflected and refracted rays are at \( 90^\circ \) to each other, the angle \( \alpha \) is:

\[ \alpha = 90^\circ \]

Thus, the correct answer is 90 degrees.