Answer
(a) $0.28KW/m^2$
(b) $4.6cm$
Work Step by Step
(a) We know that
$I_{avg}=\frac{P_{avg}}{A}$
We plug in the known values to obtain:
$I_{avg}=\frac{0.50\times 10^{-3}}{\pi(0.750\times 10^{-3})^2}=0.28KW/m^2$
(b) The required distance can be determined as
$I_{laser}=\frac{P_{bulb}}{4\pi r^2}$
This can be rearranged as
$r=\sqrt{\frac{P_{bulb}}{4\pi I_{laser}}}$
We plug in the known values to obtain:
$r=\sqrt{\frac{(0.050)(150)}{4\pi(280)}}=4.6cm$
