Optics Question 312

Question: An electromagnetic wave going through vacuum is described by $ E=E _{0}\sin (kx-\omega t) $ ; $ B=B _{0}\sin (kx-\omega t) $ . Which of the following equation is true

Options:

A) $ E _{0}k=B _{0}\omega $

B) $ E _{0}\omega =B _{0}k $

C) $ E _{0}B _{0}=\omega k $

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

$ \frac{E _{0}}{B _{0}}=C. $ also $ k=\frac{2\pi }{\lambda } $

and $ \omega =2\pi \nu $ .

These relation gives $ E _{0}K=B _{0}\omega $



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