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(B) The total power of the bulb is $P_{total} = 100 \ W$. Since $20 \%$ of the power is converted to visible radiation,the power of visible radiation

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$5$

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(B) The total power of the bulb is $P_{total} = 100 \ W$. Since $20 \%$ of the power is converted to visible radiation,the power of visible radiation $P_{vis}$ is: $P_{vis} = 100 \ W 	imes \frac{20}{100} = 20 \ W$. The intensity $I$ of radiation emitted isotropically at a distance $r$ from a point source is given by the formula: $I = \frac{P_{vis}}{4 \pi r^2}$. Given $r = 5 \ m$,we substitute the values: $I = \frac{20}{4 \pi 	imes (5)^2} = \frac{20}{4 \pi 	imes 25} = \frac{5}{25 \pi} \ W/m^2$. Comparing this with the given expression $\frac{\alpha}{25 \pi} \ W/m^2$,we find that $\alpha = 5$.

About $20 \%$ of the power of a $100 \ W$ bulb is converted to visible radiation. Assuming that the radiation is emitted isotropically and neglecting reflection,the average intensity of visible radiation at a distance of $5 \ m$ is $\frac{\alpha}{25 \pi} \ W/m^2$. The value of $\alpha$ is

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