A) \[{{\operatorname{E}}_{r}}=-{{E}_{0}}\hat{i}\,cos\ \,(kz-\omega t)\]
B) \[{{\operatorname{E}}_{r}}={{E}_{0}}\hat{i}\ \,cos\ \,(kz+\omega t)\]
C) \[{{E}_{r}}=-{{E}_{0}}\hat{i}\ \,cos\ \,(kz+\omega t)\]
D) \[{{\operatorname{E}}_{r}}={{E}_{0}}\hat{i}\ \,\sin \ \,(kz+\omega t)\]
Correct Answer: B
Solution :
| Option [b] is correct. |
| Explanation: The phase of a wave changes by \[{{180}^{\operatorname{o}}}\] or n 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: |
| \[{{E}_{r}}={{E}_{0}}\,(-\hat{i})\ cos\,[k\ (-z)-\omega t+\pi ]\] |
| \[[\because \hat{z}\ =-\hat{z}\ and\ \hat{i}=-\hat{i}]\] |
| \[=-{{E}_{0}}\hat{i}\cos [\pi -(kz+\omega t)]\] |
| \[=-{{E}_{0}}\hat{i}[-\cos \{(kz+\omega t)\}]\] |
| \[={{\operatorname{E}}_{0}}\hat{i}\cos (kz+\omega t)\] |
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