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Question: Choose the correct answer from the alternative given. A linearly polarized electromagnetic wave gi...

Choose the correct answer from the alternative given.
A linearly polarized electromagnetic wave given as E‾=E0i^cos⁡(kz−ωt)\overline{E}={{E}_{0}}\widehat{i}\cos (kz-\omega t) is incident normally on a perfectly reflecting infinite wall at z=az=a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as
A. E‾r=−E0i^cos⁡(kz−ωt){{\overline{E}}_{r}}=-{{E}_{0}}\widehat{i}\cos (kz-\omega t)
B. E‾r=E0i^cos⁡(kz+ωt){{\overline{E}}_{r}}={{E}_{0}}\widehat{i}\cos (kz+\omega t)
C. E‾r=−E0i^cos⁡(kz+ωt){{\overline{E}}_{r}}=-{{E}_{0}}\widehat{i}\cos (kz+\omega t)
D. E‾r=−E0i^sin⁡(kz−ωt){{\overline{E}}_{r}}=-{{E}_{0}}\widehat{i}\sin (kz-\omega t)

Explanation

Solution

A reflected electromagnetic wave will have the same magnitude but will be in the opposite direction (opposite sign) and change its phase of π\pi -radians as that of the incident electromagnetic wave.

Complete step-by-step solution:
Electromagnetic waves obey the law of reflection as well as refraction, which implies, the angle of incidence is always equal to the angle of reflection. The incident ray, normal ray and the reflected ray lie on the same plane. The given equation of the electromagnetic incident wave is: E‾=E0i^cos⁡(kz−ωt)\overline{E}={{E}_{0}}\widehat{i}\cos (kz-\omega t) , which is linearly polarized which means it is a plane electromagnetic wave and has a well-defined direction of propagation with no divergence. The question also mentions that the material of the wall is optically inactive, which implies it does not rotate the plane of the electromagnetic wave.
We know that the reflected electromagnetic wave will have z=−zz=-z (Since it is opposite in direction), unit vector i^=−i^\widehat{i}=-\widehat{i} and phase difference of π\pi
Now substitute these values in the equation of incident electromagnetic wave,
∴Er=E0(−i^)cos⁡[k(−z)−ωt+π] ∴Er=−E0i^cos⁡[−(kz+ωt)+π] \begin{aligned} & \therefore {{E}_{r}}={{E}_{0}}(-\widehat{i})\cos [k(-z)-\omega t+\pi ] \\\ & \therefore {{E}_{r}}=-{{E}_{0}}\widehat{i}\cos [-(kz+\omega t)+\pi ] \\\ \end{aligned}
By the property, cos⁡(π−θ)=−cos⁡θ\cos (\pi -\theta )=-\cos \theta
Here we have, θ=kz+ωt\theta =kz+\omega t
E‾r=−E0i^[−cos⁡(kz+ωt)] E‾r=E0i^cos⁡(kz+ωt) \begin{aligned} & {{\overline{E}}_{r}}=-{{E}_{0}}\widehat{i}[-\cos (kz+\omega t)] \\\ & {{\overline{E}}_{r}}={{E}_{0}}\widehat{i}\cos (kz+\omega t) \\\ \end{aligned}
Hence, the correct answer is the option (B) E‾r=E0i^cos⁡(kz+ωt){{\overline{E}}_{r}}={{E}_{0}}\widehat{i}\cos (kz+\omega t).

Note:- when electromagnetic rays are reflected off a dense medium, the crests in the waveform are reflected as troughs and the troughs are reflected as crests and vice-versa (leading to a phase change). The material property that determines the amount of radiation reflected from an interface between two media is the phase velocity of the electromagnetic radiation in the two materials.