Dual Nature of Matter

JEE Main level, single correct, medium

135 of 200

Question 135 of 200

A point source of 100W100W emits light with 55% efficiency. At a distance of 5m5m from the source, the intensity produced by the electric field component is:

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Answer: (B) 140πWm2\dfrac{1}{40\pi }\dfrac{W}{{m}^{2}}

Total power emitted =100×5100=5W=100\times \dfrac{5}{100}=5W
Now intensity due to electric field will be half of the total intensity. Therefore,
IE=12×powerarea=12×54π×52{I}_{E}=\dfrac{1}{2}\times \dfrac{\text{power}}{\text{area}} =\dfrac{1}{2}\times \dfrac{5}{4\pi \times {5}^{2}}
=140πWm2=\dfrac{1}{40\pi }\dfrac{W}{{m}^{2}}

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