Answer
(a) The average power incident on a pupil is $1.6\times 10^{-16}~W$
(b) The average power emitted by the source is $3.1\times 10^{-9}~W$
Work Step by Step
(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 $3.1\times 10^{-9}~W$.
