## College Physics (4th Edition)

(a) The average power incident on a pupil is $1.6\times 10^{-16}~W$ (b) The average power emitted by the source is $3.1\times 10^{-9}~W$
(a) We can find the average power incident on a pupil: $I = \frac{P}{A}$ $P = I~A$ $P = I~\pi~r^2$ $P = (2.5\times 10^{-12}~W/m^2)~(\pi)~(4.5\times 10^{-3}~m)^2$ $P = 1.6\times 10^{-16}~W$ The average power incident on a pupil is $1.6\times 10^{-16}~W$ (b) We can find the average power emitted by the source: $I = \frac{P}{A}$ $P = I~A$ $P = I~(4\pi~r^2)$ $P = (2.5\times 10^{-12}~W/m^2)~(4\pi)~(10.0~m)^2$ $P = 3.1\times 10^{-9}~W$ The average power emitted by the source is $3.1\times 10^{-9}~W$.