Question: Q. 1. A linearly polarized electromagnetic wave given as E=E0i^cos(kzωt) is incident normally on a perfectly reflecting infinite wall at z=a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as

(a) Er=E0i^cos(kzωt).

(b) Er=E0i^cos(kz+ωt).

(c) Er=E0i^cos(kz+ωt).

(d) Er=E0i^sin(kzωt).

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Solution:

Ans. Correct option : (b)

Explanation: The phase of a wave changes by 180 or π radian after got reflected from a denser medium. But the type of waves remains identical. Therefore, for the reflected wave, we have

z^=z^,i^=i^ and additional phase of π in the incident wave.

It is given that the incident electromagnetic wave is :

E=E0(i^)cos(kzωt)

Therefore, the reflelected electromagnetic wave is given by

$$ \begin{aligned} \mathrm{E}{r} & =E{0}(-\hat{i}) \cos [k(-z)-\omega t+\pi][\because \hat{z}=-\hat{z} \text { and } \hat{i}=-\hat{i}] \ & =-E_{0} \hat{i} \cos [\pi-(k z+\omega t)] \ & =-E_{0} \hat{i}[-\cos {(k z+\omega t)}] \ & =E_{0} \hat{i} \cos (k z+\omega t) \end{aligned} $$



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