Saved Bookmarks
| 1. |
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiationa.at a distance of 1m from the bulb b. at a distance of 10mAssume that the radiation is emitted isotropically and neglect reflection. |
|
Answer» SOLUTION :a.`A=4pi r^(2)=4xx3.14xx1^(2)m^(2)` `I=("POWER")/("Area")=(100xx 5//100)/(4xx3.14xx1^(2))=0.4 WM^(-2)` B.`I=(100xx 5//100)/(4xx3.14xx10^(2))=0.004 Wm^(-2)` |
|