About 10% 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 2m from the bulb?

(b) At a distance of 5 m?

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 2)

Answer 1

Concept:

The intensity of visible radiation is the quantity of visible light emitted per unit time, per unit area

\(I = \frac{P}{4\pi r^2}\)

Where P is the power of the source, P_i = Power of the visible radiation, and r is the distance from the source of radiation

Calculation:

10% of P = P_i

\(⇒ \frac{10}{100}\times 100 = 10 = P_i\)

⇒ P_i = 10 W

a) Given r = 2 m

\(I = \frac{P_i}{4\pi r^2}\)

\(⇒I = \frac{10}{4\pi 2^2}\)

⇒ I = 0.19 Watt/m²

b) r = 5m

\(⇒I = \frac{10}{4\pi 5^2}\)

\(⇒I = 0.0318\ W/m^2\)

Hence the answer is the option (1)