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 :
\[\in ={{\in }_{o}}-{{e}^{i(kz-wt)}}\] at \[t={{t}_{1}},z={{z}_{1}},E=o,\] the next zero that occurs in it?s 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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