A linearly polarised electromagnetic wave given as `E=E_(0)haticos(kz-omegat)` is incident normally on a perfectly reflecting wall `z=a`. Assuming that the material of the optically inactive, the reflected wave will be give as

Options

A. `vecE_r=-E_0hati cos (kz-omegat)`
B. `vecE_r=E_0hati cos (kz+omegat)`
C. `vecE_r=-E_0hati cos (kz+omegat)`
D. `vecE_r=E_0hati sin (kz-omegat)`

Answer

b

Explanation

When a wave is reflected from denser medium, the reflected wave is without change in type of wave but
with a change in phase by `180^@` or `pi` radian. Therefore, for the reflected wave we use `z=-z, hati=-hati`
and additional phase of `pi` in the incident wave. The incident em wave is, `E=E_0haticos (kz-omegat)`
The reflected em wave is
`E_r=E_0(-hati)cos [k(-z)-omegat+pi]=-E_0hati cos [-(kz-omegat)+pi]`
`=E_0haticos [-(kz-omegat)] [:' cos ( theta+pi)=-cos theta]`
`=E_0haticos (kz+omegat) [:' cos (- theta)= cos theta]`