Answer
a) $I_{avg}=0.12W/m^2$
b) the same
c) $I_{avg}=0.12W/m^2$
Work Step by Step
(a) We can find the required average intensity as follows:
$I_{avg}=\frac{P_{avg}}{A}$
We plug in the known values to obtain:
$I_{avg}=\frac{(0.050)(120W)}{(4)(3.14)(2.0m)^2}$
$I_{avg}=0.12W/m^2$
(b) We know that
$I_{avg}=\frac{P_{avg}}{A}$
We plug in the known values to obtain:
$I_{avg}=\frac{(0.050)(120W)}{4\times 3.14(2.0m)^2}=0.12W/m^2$
Thus, the intensity of the light is the same as that found in part(a).
(c) As $I_{avg}=\frac{P_{avg}}{4\pi r^2}$
We plug in the known values to obtain:
$I_{avg}=\frac{0.050(120)}{4(3.14)(2.0m)^2}$
$I_{avg}=0.12W/m^2$
