Question

Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a × 10-2 W.m2. The value of 'a' will be _____.

Answer 1

Pradiation =0.1×110=11 W
ΔIradiation1=Iradiation1–Iradiation2
=$11(\frac{1}{4π}–\frac{1}{4π×25})$
$=\frac{11×24}{4π×25}$
=84×10–2 W/m2
So, the answer is 84.