Optics Question 180
Question: A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance 4.0 m from the source is
Options:
A) 64.7 V/m
B) 57.8 V/m
C) 56.72 V/m
D) 54.77 V/m
Show Answer
Answer:
Correct Answer: D
Solution:
Intensity of EM wave is given by $ I=\frac{P}{4\pi R^{2}}=v _{av}.c=\frac{1}{2}{\varepsilon _{0}}E _{0}^{2}\times c $
Therefore $ E _{0}=\sqrt{\frac{P}{2\pi R^{2}{\varepsilon _{0}}c}} $
$ =\sqrt{\frac{800}{2\times 3.14\times {{(4)}^{2}}\times 8.85\times {{10}^{-12}}\times 3\times 10^{8}}} $ = 54.77 $ \frac{V}{m} $
