Saved Bookmarks
| 1. |
10 % of the total energy of a 100 W bulb is converted into visible light. Calculate the average intensity out a spherical surface which is at a distance of 1m from the bulb, consider the bulb to be a point source and let the medium be isotropic. |
|
Answer» Solution :Electrical energy used per second in bulb, U = P of `10%=100xx0.1` `THEREFORE U=10 J` Considering bulb as centre, area of circular SURFACE `A=4pi d^(2)=4xx3.14xx(1)^(2)=12.56m^(2)` Average INTENSITY on circular surface `I=(U)/(A)=(10)/(12.56)=0.796 Wm^(-2)` `therefore I=0.8 Wm^(-2)` (Approximate). |
|