The bulb as a point source, radiate light in all direction uniformly and it is given that only 5% of power is converted to visible radiations. (a) Let us assume a sphere of radius 1 m. Surface area A = 4πr2 = 4π (1)2 = 4 π m2 Intensity on the sphere I = $\frac{\text{Power}}{\text{Area}}$ = $\frac{100 \times 0.05}{4\pi}$ = 0.4 $\mathrm{W}\,\mathrm{m}^{-2}$ (b) Let us assume a sphere of 10 m radius. Surface area A = $4\pi r^2$ = 4 π $(10)^2$ = 400 π $\mathrm{m}^2$ Intensity on the sphere I = $\frac{\text{Power}}{\text{Area}}$ = $\frac{100 \times 0.05}{400\pi}$ = 0.004 W $\mathrm{m}^{-2}$