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About $20 \%$ of the power of a $100 \mathrm{~W}$ bulb is converted to visible radiation. Assuming that the radiations is emitted isotropically and neglecting reflection, the average intensity of visible radiation at a distance of $5 \mathrm{~m}$ is $\frac{\alpha}{25 \pi} \mathrm{W} / \mathrm{m}^2$. The value of $\alpha$ is
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5
Power, $\mathrm{P}=\frac{100 \times 20}{100}=20 \mathrm{~W}$
The average intensity of visible radiation at a distance
$$
\begin{aligned}
& 5(I)=\frac{P}{4 \pi r^2}=\frac{20}{4 \pi \times(5)^2} \\
& (I)=\frac{5}{25 \pi} \\
& \alpha=5
\end{aligned}
$$
The average intensity of visible radiation at a distance
$$
\begin{aligned}
& 5(I)=\frac{P}{4 \pi r^2}=\frac{20}{4 \pi \times(5)^2} \\
& (I)=\frac{5}{25 \pi} \\
& \alpha=5
\end{aligned}
$$
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