Question

A linearly polarized electromagnetic wave given as $\overleftarrow{E} = E_{0\backslash}\widehat{i}\cos(kz - \omega t)is$incident normally on a perfectly reflecting infinite well at z=a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as

• A. ${\overrightarrow{E}}_{r} = - E_{0}\widehat{i}\cos(kz - \omega t)$
• B. ${\overrightarrow{E}}_{r} = - E_{0}\widehat{i}\cos(kz + \omega t)$
• C. ${\overrightarrow{E}}_{r} = E_{0}\widehat{i}\cos(kz + \omega t)$
• D. $\overrightarrow{E}l_{r} = - E_{0}\widehat{i}\sin(kz - \omega t)$

Answer 1

Answer: (B)

${\overrightarrow{E}}_{r} = - E_{0}\widehat{i}\cos(kz + \omega t)$

Answer 2

: 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,

$\therefore{\overrightarrow{E}}_{r} = E_{0}\widehat{i}\cos( - kz - \omega t)$

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