Question

A linearly polarized electromagnetic wave given as $E=E_0\hat{i}\cos (kz-\omega t)$ 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. ${\overrightarrow{E}}_r=-E_0\hat{i}\cos (kz-\omega t)$
• B. ${\overrightarrow{E}}_r=E_0\hat{i}\cos (kz+\omega t)$
• C. ${\overrightarrow{E}}_r=-E_0\hat{i}\sin (kz+\omega t)$
• D. ${\overrightarrow{E}}_r=E_0\hat{i}\sin (kz-\omega t)$

Answer 1

The correct option is B ${\overrightarrow{E}}_r=E_0\hat{i}\cos (kz+\omega t)$

When a wave is reflected from the denser medium, the reflected wave is without a change in the type of wave but with a change in phase by ${180}^o$ or $\pi$ radian.

Therefore, the reflacted wave is with $z=-z,i=-i$ and additional phase of $\pi$ in the incident wave.

Here the incident EM wave is, $E=E_{0}i\cos (kz-wt)$

Hence, the reflected EM wave is

$E_r=E_0(-i)\cos \left[k\right(-z)-\omega t+\pi ]=-E_{0}i\cos [-(kz+\omega t)+\pi ]$

$=E_{0}i\cos (kz+\omega t)$
Hence, the correct option is $\left(B\right)$