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Question
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.
Solution
Power rating of bulb, P = 100 W
It is given that about 5% of its power is converted into visible radiation.
∴ Power of visible radiation,
`"P'" = 5/100 xx 100` = 5 W
Hence, the power of visible radiation is 5 W.
(a) Distance of a point from the bulb, d = 1 m
Hence, intensity of radiation at that point is given as:
`"I" = ("P'")/(4pi"d"^2)`
= `5/(4pi(1)^2)`
= 0.398 W/m2
(b) Distance of a point from the bulb, d1 = 10 m
Hence, intensity of radiation at that point is given as:
`"I" = "P'"/(4pi("d"_1)^2)`
= `5/(4pi(10)^2)`
= 0.00398 W/m2
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