A) \[{{\operatorname{E}}_{0}}\omega \,\,=\,\,{{B}_{0}}k\]
B) \[{{\operatorname{E}}_{0}}k\,\,=\,\,{{B}_{0}}\omega \]
C) \[{{\operatorname{E}}_{0}}{{B}_{0}}\,\,=\,\,\omega k\]
D) None of these
Correct Answer: B
Solution :
As\[\frac{{{E}_{0}}}{{{B}_{0}}}=c\] \[\left( c =velocity of light \right)\] Also, \[k=\frac{2\pi }{\lambda },\,\,\,\omega =2\pi v\,\,and\,\,c=v\lambda \] So, combining all, we have \[{{E}_{0}}k={{B}_{0}}\omega \]
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