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3% of energy of 100 W bulb is converted ...

3% of energy of 100 W bulb is converted into visible light. Find the average intensity on a spherical surface 1 m away from the bulb consider the bulb is point source and medium is isotropic.

Text Solution

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Energy used per second in bulb,
U = 3 % of `P=100xx(3)/(100)`
`therefore U=3J`
Area of curcular surface considering it bulb at centre
`A=4pi r^(2)`
`A=4xx3.14xx(1)^(2)`
`therefore A = 12.56 m^(2)`
Average intensity of light on circular surface
`t=(U)/(A)=(3)/(12.56)=0.23885`
`therefore I ~~ 0.24 W//m^(2)`
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