Question

Choose the correct answer from alternatives given.
A linearly polarized electromagnetic wave given as $\stackrel{-}{E}=E_0\stackrel{\wedge }{i}cos(kz-\omega t)$ 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. ${\stackrel{-}{E}}_r=-E_o\stackrel{\wedge }{i}cos(kz-\omega t)$
• B. ${\stackrel{-}{E}}_r=E_o\stackrel{\wedge }{i}cos(kz+\omega t)$
• C. ${\stackrel{-}{E}}_r=-E_o\stackrel{\wedge }{i}cos(kz+\omega t)$
• D. ${\stackrel{-}{E}}_r=-E_o\stackrel{\wedge }{i}sin(kz-\omega t)$

Answer 1

The correct option is B ${\stackrel{-}{E}}_r=E_o\stackrel{\wedge }{i}cos(kz+\omega t)$

Given: The equation of the linearly polarized electromagnetic wave is $\overrightarrow{E}=E_0\hat{i}cos(kz-\omega t)$.

To find: The equation of the reflected wave from $z=a$.

As the wave is reflected from the perfectly reflecting wall, it will only reverse the direction of motion of the wave along with a phase difference of $\pi$.

Therefore. Equation of reflected wave will be the same in magnitude but with negative directions.

Substitute $-\hat{i}$ and $-z$ for $\hat{i}$ and $z$ respectively in the equation of wave.

$E_r=E_0(-\hat{i})cos\left(k\right(-z)-\omega t+\pi )$ [$\because cos(\pi +\theta )=-cos\theta$]

$E_r=-E_{0}i\cos (kz+\omega t)$

i.e, opposite in direction.

Option $\left(B\right)$ is correct.