Ncert Exercises & Solutions

8.12 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.

(a)
Hint: $I=\frac{P}{4\pi d^{2}}$
Step 1: Find the power of visible radiation.
The power rating of the bulb, P = 100 W
As 5% of its power is converted into visible radiation, the power of visible radiation =P^'= 5 W
Step 2: Find the intensity of radiation at 1m.

$l = \frac{P'}{4 {πd}^{2}}$
$= \frac{5}{4 π {(1)}^{2}} = 0.398 W / m^{2}$

(b)
Hint: $I=\frac{P}{4\pi d^{2}}$
Step 1: Find the power of visible radiation.
The power rating of the bulb, P = 100 W
As 5% of its power is converted into visible radiation, the power of visible radiation =P^'= 5 W
Step 2: Find the intensity of radiation at 10m.

$l = \frac{P'}{4 {πd}^{2}}$
$= \frac{5}{4 π {(10)}^{2}} = 0.00398 W / m^{2}$

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