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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Waves

About 5% of the power of a $100\: W$ light bulb is converted to visible radiation. What is the average intensity of visible radiation at a distance of $10\: m$ from the bulb?

$\begin {array} {1 1} (a)\;0.398\: W/m^2 & \quad (b)\;0.3098 \: W/m^2 \\ (c)\;0.0398\: W/m^2 & \quad (d)\;0.00398\: W/m^2 \end {array}$

1 Answer

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$P = 100\: W$
Power of visible radiation, $Pā€™ = \large\frac{5 \times 100}{100}$$ = 5W$
$d = 10m$
Hence, intensity of radiation,$ I = \large\frac{Pā€™}{4 \pi d^2}$
$= \large\frac{5}{4 \pi (10)^2}$$ = 0.00398\: W/m^2$
Ans : (d)
answered Feb 27, 2014 by thanvigandhi_1
 

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