The electric field intensity produced by the radiations coming from a 200 W bulb at 6 m distance is E. What is the electric field intensity produced by the radiations coming from 100 W bulb at the same distance?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{E}{\sqrt{2}} \)
The intensity of the radiation from a point source at distance r is \( I = \frac{P}{4\pi r^{2}} \), and the intensity of an electromagnetic wave is proportional to the square of its electric field amplitude, \( I\propto E^{2} \). At the same distance r the intensity is proportional to the power, so \( E^{2}\propto P \), that is \( E\propto\sqrt{P} \).
\( \displaystyle \frac{E'}{E} = \sqrt{\frac{P'}{P}} = \sqrt{\frac{100}{200}} = \frac{1}{\sqrt{2}} \)
\( \displaystyle E' = \frac{E}{\sqrt{2}} \)