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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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Answer» <P> SOLUTION :Here, power of radiation = 5% of ELECTRICAL power`=100xx(5)/(100)` `therefore P=5W` Again, according to formula, `I=(P)/(4pi r^(2))` `=(5)/(4xx3.14xx(10)^(2))(because r =10 m)` `therefore I=3.981xx10^(-3)Wm^(-2)~~0.004 Wm^(-2)` |
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