About 6% of the power of a 100 W light bulb is converted to visible radiation. The average intensity of visible radiation at a distance of 8 m is (Assume that the radiation is emitted isotropically and neglect reflection.)
Answer & explanation
Correct answer: option 3
Here, power of bulb = 100 W
As intensity, I = $\frac{\text { Power of visible light }}{\text { area }}=\frac{100 \times 6 / 100}{4 \pi(8)^2}=7.2 × 10^{-3} Wm^{-2}$