GATE 2022 PH – Question 58

Correct answer: (A) Refraction angle $\pi/6$; (B) $\epsilon_2/\epsilon_1=3$; (C) The incident electric field may have yz-plane components

Explanation

The reflected field is $\mathbf E_R=A_R\cos\!\left[k_0\left(\tfrac{\sqrt3}{2}y-\tfrac z2\right)-\omega t\right]\hat x$.

**Reflected wave vector:** $\mathbf k_R=k_0\left(\tfrac{\sqrt3}2\hat y-\tfrac12\hat z\right)$. The normal to the interface $z=0$ is $\hat z$, so the angle of reflection $\theta_r$ satisfies $\cos\theta_r=\left|\dfrac{k_{Rz}}{k_0}\right|=\dfrac12$, that is $\theta_r=60^\circ=\dfrac\pi3$. The angle of incidence is also $\dfrac\pi3$. (So D, which says $\pi/6$, is false.)

**Brewster incidence:** the reflected and refracted rays are perpendicular, so the angle of refraction is
$$\theta_t=90^\circ-60^\circ=30^\circ=\frac\pi6\quad(\text{A is true}).$$

**Index ratio:** $\tan\theta_B=\dfrac{n_2}{n_1}=\tan60^\circ=\sqrt3$, so $\dfrac{\epsilon_2}{\epsilon_1}=\left(\dfrac{n_2}{n_1}\right)^2=3\quad(\text{B is true}).$

**Polarisation.** At Brewster's angle the reflected wave has its electric field perpendicular to the plane of incidence ($\hat x$ here). The component of the incident field in the plane of incidence ($yz$-plane) is completely transmitted, so the incident field can have components in the $yz$ plane (C is true).

Answer **A, B and C**.