JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Wave Optics

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

Answer: (A)

Solution

At Brewster's incidence the reflected beam is completely polarized perpendicular to the plane of incidence. From the figure the incident ray travels in the

x

z

plane into the medium, so after reflection it moves in the

+ x

and

- z

sense and carries the phase

k x - k z - ω t

. Being polarized normal to the plane of incidence, its electric field lies in the interface and has no

z

-component. The reflected wave is therefore

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

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Wave Optics

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