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About 5% of the power of a 100 W light ...

About `5%` of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.

Text Solution

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The energy crossing per unit area per unit time perpendicular to the direction of propagation is called intensity.
(a) Average intensity of visible radiation at `1m=(5%" of 100W")/(4pi(1)^(2))=0.4W//m^(2)`.
(b) Average intensity of visible radiation at `10m=(5%" of 100 W")/(4pi(10)^(2))=0.04W//m^(2)`.
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