A point source of 100,W emits light with 5 % efficiency. At a distance of 5,m from source, intensity

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8.Electromagnetic waves

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

A

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

B

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

C

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

D

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

(JEE MAIN-2023)

Solution

$I _{ EF }=\frac{1}{2} \times \frac{5}{4 \pi \times 5^2}$

$=\frac{1}{40 \pi}\,W / m ^2$

Standard 12

Physics

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