An unpolarized light is incident on the plane interface of air-dielectric medium shown in figure

Question

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. [2026]

Smart Achievers · Accepted Answer

(1)

The incident wave travels forward $(+ x)$ and down $(+ z)$ . Upon reflecting off the horizontal $x y$ -plane, it bounces back up. It now travels in the $+ x$ and $- z$ directions. This corresponds to the phase term sin $(k x - k z - ω t)$ , which is only present in Option (1).

Solution

Q.
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. [2026]

1 $(E_{x} \hat{i} + E_{y} \hat{j}) \sin (k x - k z - ω t)$

2 $(E_{x} \hat{i} + E_{z} \hat{k}) \sin (k x + k y - ω t)$

3 $(E_{x} \hat{j} + E_{y} \hat{k}) \sin (k y + k z - ω t)$

4 $(E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k}) \sin (k x + k y - k z - ω t)$

Ans.
(1)

The incident wave travels forward $(+ x)$ and down $(+ z)$ . Upon reflecting off the horizontal $x y$ -plane, it bounces back up. It now travels in the $+ x$ and $- z$ directions. This corresponds to the phase term sin $(k x - k z - ω t)$ , which is only present in Option (1).

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Solution

Q. 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. [2026]

1 (Exi^+Eyj^)sin(kx-kz-ωt)

2 (Exi^+Ezk^)sin(kx+ky-ωt)

3 (Exj^+Eyk^)sin(ky+kz-ωt)

4 (Exi^+Eyj^+Ezk^)sin(kx+ky-kz-ωt)

Ans. (1) The incident wave travels forward (+x) and down (+z). Upon reflecting off the horizontal xy-plane, it bounces back up. It now travels in the +x and -z directions. This corresponds to the phase term sin (kx-kz-ωt), which is only present in Option (1).