Question 169: EM waves

Question

Nearly 10% of the power of a $$110 \mathrm{~W}$$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $$1 \mathrm{~m}$$ from the bulb to a distance of $$5 \mathrm{~m}$$ is $$a 	imes 10^{-2} \mathrm{~W} / \mathrm{m}^{2}$$. The value of 'a' will be _________.

Accepted Answer

$\mathbf{P}^{\prime}=10 \%$ of $110 \mathbf{W}$
$=\frac{10}{100} 	imes 110 \mathrm{~W}$
$=11 \mathrm{~W}$
$\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_1^2}-\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_2^2}$
$=\frac{11}{4 \pi}\left[\frac{1}{1}-\frac{1}{25}ight]$
$=\frac{11}{4 \pi} 	imes \frac{24}{25}$
$=\frac{264}{\pi} 	imes 10^{-2}=84 	imes 10^{-2} \mathrm{~W} / \mathrm{m}^2$

EM waves

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