JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

An unpolarized light is incident on the plane interface of air-dielectric medium shown in figure. If the incident angle is equal to Brewster angle, identify the expression representing reflected wave.

A
$(E_{x} \hat{i} + E_{y} \hat{j}) sin (k x - k z - ω t)$
B
$(E_{x} \hat{i} + E_{z} \hat{k}) sin (k x + k y - ω t)$
C
$(E_{x} \hat{j} + E_{y} \hat{k}) sin (k y + k z - ω t)$
D
$(E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k}) sin (k x + k y - k z - ω t)$

Solution

The incident light travels along $+ z$ in the $x z$ -plane. After reflection at the interface, the $z$ -component of the wave vector flips sign, so the reflected wave propagates along the direction $(\hat{i} - \hat{k})$ giving a phase $k x - k z - ω t$ .

At Brewster's angle the reflected wave is completely polarised perpendicular to the plane of incidence. The plane of incidence is the $x z$ -plane, so the reflected $E$ must lie along $\hat{j}$ (perpendicular to that plane), with no $\hat{k}$ (longitudinal) component. The form matching both requirements is

$(E_{x} \hat{i} + E_{y} \hat{j}) sin (k x - k z - ω t),$

which is option (1). (Strictly, $s$ -polarised gives only the $\hat{j}$ piece, but among the four options only (1) carries the correct phase and excludes a $\hat{k}$ component.)

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