Question

A linearly polarized electromagnetic wave given as $\vec{E} = E_{0}\hat{i}\,cos \left(kz-\omega t\right)$ 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. $\vec{E}_{r} = -E_{0}\hat{i}\,cos \left(kz-\omega t\right)$
• B. $\vec{E}_{r} = E_{0}\hat{i}\,cos \left(kz+\omega t\right)$
• C. $\vec{E}_{r} = -E_{0}\hat{i}\,cos \left(kz+\omega t\right)$
• D. $\vec{E}_{r} = E_{0}\hat{i}\,cos \left(kz-\omega t\right)$

Answer 1

Answer: (B)

As the wall is perfectly reflecting, there is no. change in amplitude $E_0$.
Also the wall is optically inactive, so, there is no phase change.
After reflection, the wave travels along -ve z direction,
$\therefore \quad\vec{E}_{r} = E_{0}\hat{i}\,cos \left(-kz-\omega t\right)$
$= E_{0}\hat{i}\,cos \left(kz+\omega t\right)$
$\left(\because cos\left(-\theta\right) = cos\theta\right)$