A) \[\frac{3\times {{10}^{8}}}{|{{z}_{2}}-{{z}_{1}}|}\]
B) \[\frac{6\times {{10}^{8}}}{|{{z}_{2}}-{{z}_{1}}|}\]
C) \[\frac{1.5\times {{10}^{8}}}{|{{z}_{2}}-{{z}_{1}}|}\]
D) \[\frac{1}{{{t}_{1}}+\frac{|{{z}_{2}}-{{z}_{1}}|}{3\times {{10}^{8}}}}\]
Correct Answer: A
Solution :
| [a] \[\in ={{\in }_{o}}-{{e}^{i(kz-wt)}}\] |
| at \[t={{t}_{1}},z={{z}_{1}},E=o,\] |
| the next zero that occurs in its neighbourhood is at \[{{z}_{2}}\], the frequency of the electromagnetic wave at \[{{l}_{2}}\] |
| \[{{e}^{i(k{{z}_{1}}-w{{t}_{1}})}}={{e}^{i(k{{z}_{2}}-w{{t}_{2}})}}\] |
| \[k{{z}_{1}}-w{{t}_{1}}=k{{z}_{2}}-w{{t}_{2}}\] |
| \[({{t}_{2}}-{{t}_{1}})w=k(z-{{z}_{1}})\] |
| Where \[k=\frac{2\pi }{\lambda }=2\pi v\] |
| \[({{t}_{2}}-{{t}_{1}})=\frac{2\pi }{\lambda \times 2\pi v}({{z}_{2}}-{{z}_{1}})\] |
| \[({{t}_{2}}-{{t}_{1}})=\frac{1}{x\times v}({{z}_{2}}-{{z}_{1}})\] |
| \[\lambda \times v=\frac{({{z}_{2}}-{{z}_{1}}}{({{t}_{1}}-{{t}_{1}}}\] |
| \[C=\frac{({{z}_{2}}-{{z}_{1}})}{({{t}_{2}}-{{t}_{1}})}\] |
| \[({{t}_{2}}-{{t}_{1}})=\frac{({{z}_{2}}-{{z}_{1}})}{C}\] |
| Frequency is \[f\propto \frac{1}{t}\] then |
| \[\frac{1}{({{t}_{2}}-{{t}_{1}})}=\frac{C}{({{z}_{2}}-{{z}_{1}})}\] |
| Frequency\[=\frac{3\times {{10}^{8}}}{{{z}_{2}}-{{z}_{1}})}\] |
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