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Question

(D) The incident wave is traveling in the $+z$ direction,given by the form $\cos(kz - \omega t)$.When an electromagnetic wave is incident normally on

Vedclass · Accepted Answer

The reflected wave will be $3.1 \cos \left[(1.8)z + (5.4 	imes 10^6)tight] \hat{i} 	ext{ N/C}$.

Vedclass · Answer

(D) The incident wave is traveling in the $+z$ direction,given by the form $\cos(kz - \omega t)$.When an electromagnetic wave is incident normally on a perfectly reflecting wall,the reflected wave travels in the opposite direction ($-z$ direction).The reflected wave will have the form $\cos(kz + \omega t)$.Given the incident wave $E_i = 3.1 \cos \left[(1.8)z - (5.4 	imes 10^6)tight] \hat{i}$,the reflected wave $E_r$ must have the same amplitude and frequency but a reversed propagation direction.Thus,$E_r = 3.1 \cos \left[(1.8)z + (5.4 	imes 10^6)tight] \hat{i} 	ext{ N/C}$.Therefore,option $D$ is correct.

$A$ linearly polarized electromagnetic wave in vacuum is given by $E = 3.1 \cos \left[(1.8)z - (5.4 \times 10^6)t\right] \hat{i} \text{ N/C}$. It is incident normally on a perfectly reflecting wall at $z = a$. Choose the correct option.

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