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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. |
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Answer» Solution :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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