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?. Assume that the radiation is emitted isotropically and neglect reflection.

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Solution

Given,
Power , P = 100 W
Distance, d = 10 m
Power converted into visible radiation,

P = \frac{5}{100} \times 100 W = 5 W

Therefore, using the formula we have,

Intensity = \frac{Energy}{Area \times Time} = \frac{Power}{Area} = \frac{P}{4 {πr}^{2}}

Intensity, I =

\frac{5}{4 \times 3.14 \times 10 \times 10} {Wm}^{- 2}

= 0.004 {Wm}^{- 2}

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