Question

About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) At a distance of 1 m from the bulb?
(b) At a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.

Answer 1

P = 100 W
n = 5%
Since the efficiency of the bulb is 5%, the effective power of the bulb.
P_eff = p x n = 100 x \(\frac{5}{100}\) = 5w
At a distance, ‘r’ the radiation coming from the bulb is distributed over the surface of a sphere of radius r i.e. over an area.
A = 4πr²
Therefore, intensity of a light at a distance of r,