Question: Q. 1. A linearly polarized electromagnetic wave given as $E=E_{0} \hat{i} \cos (k z-\omega t)$ 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) $E_{r}=-E_{0} \hat{i} \cos (k z-\omega t)$.
(b) $E_{r}=E_{0} \hat{i} \cos (k z+\omega t)$.
(c) $E_{r}=-E_{0} \hat{i} \cos (k z+\omega t)$.
(d) $E_{r}=E_{0} \hat{i} \sin (k z-\omega t)$.
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Solution:
Ans. Correct option : (b)
Explanation: The phase of a wave changes by $180^{\circ}$ or $\pi$ radian after got reflected from a denser medium. But the type of waves remains identical. Therefore, for the reflected wave, we have
$\hat{z}=-\hat{z}, \hat{i}=-\hat{i}$ and additional phase of $\pi$ in the incident wave.
It is given that the incident electromagnetic wave is :
$E=E_{0}(\hat{i}) \cos (k z-\omega t)$
Therefore, the reflelected electromagnetic wave is given by
$$ \begin{aligned} \mathrm{E}{r} & =E{0}(-\hat{i}) \cos [k(-z)-\omega t+\pi][\because \hat{z}=-\hat{z} \text { and } \hat{i}=-\hat{i}] \ & =-E_{0} \hat{i} \cos [\pi-(k z+\omega t)] \ & =-E_{0} \hat{i}[-\cos {(k z+\omega t)}] \ & =E_{0} \hat{i} \cos (k z+\omega t) \end{aligned} $$
