A plane polarized electromagnetic wave propagating in y-z plane is incident at the
interface of two media at Brewster's angle. Taking z = 0 as the boundary
between the two media, the electric field of the reflected wave is given by
E\vecโ = A cos k_0 โ3
2 y - 1
2 z - ๐t x เท
then which among the following statements are correct?
Options:
(A) The angle of refraction is $\frac{\pi}{6}$
(B) Ratio of permittivity of the medium of refraction ($\epsilon_2$) with respect to incident medium ($\epsilon_1$) is $3$
(C) The incident wave can have components of its electric field in the $y$-$z$ plane
(D) The angle of reflection is $\frac{\pi}{6}$
Official Answer Key:
A,B,C
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.58)**
**1. Electromagnetic Wave Optics (Brewster's Angle)**:
- Brewster's law states that at the polarizing angle $\theta_B$:
$\tan\theta_B = \frac{n_2}{n_1} = \sqrt{\frac{\epsilon_2}{\epsilon_1}}$
- At Brewster's incidence, reflected and refracted rays are perpendicular:
$\theta_B + \theta_r = \frac{\pi}{2} \implies \theta_r = \frac{\pi}{2} - \theta_B$
**2. Detailed Step-by-Step Evaluation**:
- For Brewster angle $\theta_B = \pi/3 = 60^\circ$:
$\theta_r = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$
**Statement (A) is TRUE**.
- Ratio of dielectric permittivities:
$\frac{\epsilon_2}{\epsilon_1} = \tan^2(\theta_B) = \tan^2(60^\circ) = (\sqrt{3})^2 = 3$
**Statement (B) is TRUE**.
- Brewster reflection eliminates the component with electric field in the plane of incidence (p-polarized wave). For non-zero reflection, the incident wave can have electric field components in the plane of incidence. **Statement (C) is TRUE**.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, B, C)**.
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