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A point source of 100 \mathrm{~W}emits light with 5 \% efficiency. At a distance of 5 \mathrm{~m} from the source, the intensity produced by the electric field component is: 

Option: 1

\frac{1}{2 \pi} \frac{W}{m^{2}}


Option: 2

\frac{1}{20 \pi} \frac{W}{m^{2}}


Option: 3

\frac{1}{10 \pi} \frac{W}{m^{2}}


Option: 4

\frac{1}{40 \pi} \frac{W}{m^{2}}


Answers (1)

best_answer

Light energy emitted per unit time =100 \times \frac{5}{100}=5 \mathrm{~W}

intensity at 5 m= \frac{\text { power }}{4 \pi(5)^2}=\frac{5}{4 \pi(5)^2}=\frac{1}{20 \pi}
However, \frac{1}{20 \pi} is due to Both E and B with equal Contributions

So, intensity due to \mathrm{E}=\frac{1}{40 \pi} \frac{\mathrm{W}}{\mathrm{m}^2}

Posted by

vishal kumar

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