A point source of 100,W emits light with 5% e | vedclass

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(B) The power of the source is $P = 100\,W$. The efficiency is $5\%$,so the power of the emitted light is $P_{light} = 100 	imes 0.05 = 5\,W$.At a di

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

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(B) The power of the source is $P = 100\,W$. The efficiency is $5\%$,so the power of the emitted light is $P_{light} = 100 	imes 0.05 = 5\,W$.At a distance $r = 5\,m$,the total intensity $I$ is given by $I = \frac{P_{light}}{4 \pi r^2} = \frac{5}{4 \pi 	imes 5^2} = \frac{5}{100 \pi} = \frac{1}{20 \pi} \, W/m^2$.The intensity produced by the electric field component $(I_{EF})$ is half of the total intensity because the energy is shared equally between the electric and magnetic field components in an electromagnetic wave.Therefore,$I_{EF} = \frac{1}{2} I = \frac{1}{2} 	imes \frac{1}{20 \pi} = \frac{1}{40 \pi} \, W/m^2$.

$A$ 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:

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