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. `E_(r) = -E_(0) hat(i) cos (kz - omega t)`
B. `E_(r) = E_(0) hat(i) cos (kz + omega t)`
C. `E_(r) = -E_(0) hat(i) cos (kz + omega t)`
D. `E_(r) = E_(0) hat(i) sin (kz - omega t)`

Answer

B

Explanation

The phase of a wave changes by `180^(@)` or `pi` radian after got reflected from a denser medium. But the type of waves remains identical.
Therefore, for the reflected wave, we have
`hat(z) - hat(z), hat(i) = -hat(i)` and additional phase of `pi` in the incident wave.
Incident electromagnetic wave. Then,
`E = E_(0)(-hat(i))cos (kz - omega t)`
Therefore, the reflected electromagnetic wave is given as :
`E_(r) = E_(0)(- hat(i))cos [k(-z)- omega t + pi] [because hat(z) = -hat(z) " and " hat(i) = -hat(i)]`
`= -E_(0)hat(i) cos [pi - (kz + omega t)]`
`= -E_(0)hat(i)0-cos {(kz + omega t)}]`
`= E_(0)hat(i) cos(kz + omega t)`