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 1m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.
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P=100WP=100W
Power of visible radiation, P′=5×100100P′=5×100100=5W=5W
d=10md=10m
Hence, intensity of radiation,I=P′4πd2I=P′4πd2
=54π(10)2=54π(10)2=0.00398W/m2
P=100WP=100W
Power of visible radiation, P′=5×100100P′=5×100100=5W=5W
d=10md=10m
Hence, intensity of radiation,I=P′4πd2I=P′4πd2
=54π(10)2=54π(10)2=0.00398W/m2
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b answers is b 10 m is
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