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

Power rating of bulb, P = 100 W

It is given that about 5% of its power is converted into visible radiation.

∴ Power of visible radiation,

`"P'" = 5/100 xx 100` = 5 W

Hence, the power of visible radiation is 5 W.

(a) Distance of a point from the bulb, d = 1 m

Hence, intensity of radiation at that point is given as:

`"I" = ("P'")/(4pi"d"^2)`

= `5/(4pi(1)^2)`

= 0.398 W/m²

(b) Distance of a point from the bulb, d₁ = 10 m

Hence, intensity of radiation at that point is given as:

`"I" = "P'"/(4pi("d"_1)^2)`

= `5/(4pi(10)^2)`

= 0.00398 W/m²