Chapter 25 - Electromagnetic Waves - Problems and Conceptual Exercises - Page 902: 51

(a) $83 mW/m^2$ (b) $0.83 mW/m^2$

(a) We know that $I_{avg}=\frac{P_{avg}}{4\pi r^2}$ We plug in the known values to obtain: $I_{avg}=\frac{65\times 10^3}{4\pi (250)^2}=83mW/m^2$ (b) $I_{avg}=\frac{P_{avg}}{4\pi r^2}$ We plug in the known values to obtain: $I_{avg}=\frac{65\times 10^3}{4\pi (2500)^2}=0.83mW/m^2$