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 $
