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About 5% of the power ofa 100 W lightbulb 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. |
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Answer» Solution :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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