Physics · Wave Optics

JEE Main 2026 — 4 April, Morning Shift — Question 18

An unpolarized light is incident on the plane interface of air‑dielectric medium at Brewster angle. Identify the expression representing the reflected wave.

Option A:
$(E_x\hat{i}+E_y\hat{j})\sin(kx-kz-\omega t)$
Correct
Option B:
$(E_x\hat{i}+E_z\hat{k})\sin(kx+ky-\omega t)$
Option C:
$(E_x\hat{j}+E_y\hat{k})\sin(ky+kz-\omega t)$
Option D:
$(E_x\hat{i}+E_y\hat{j}+E_z\hat{k})\sin(kx+ky-kz-\omega t)$

Answer: A

Step-by-step solution

At Brewster angle, reflected light is plane polarised perpendicular to plane of incidence (y‑direction). Wave propagates along z, reflected wave has kx‑kz‑ωt phase

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Physics
Chapter: Wave Optics
Topic: Polarization of Light Waves