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 esmitted isotropically and neglect reflection.

Answer 1

The average intensity of visible radiation at 1 m and 10 m are:

• About 5% of the power of a 100 W light bulb is converted to visible radiation.

       So, radiated power = P = 0.05 *  100 W = 5 W

• This power is radiated symmetrically through the sphere. Hence, equal distances from the source have equal intensity.

        So, area = S = $4\pi R^2$ , where R = distance at which intensity is calculated

• Intensity at a point r = I = $\frac{P}{S} = \frac{P}{4\pi R^2}$

• Average intensity at a distance of 1 m = $\frac{5}{4\pi *1^2 }$ = 0.4 W/m²

• Average intensity at a distance of 10 m = $\frac{5}{4\pi *10^2 }$ = 0.004 W/m²