Question

Light with an energy flux of $18Wc{m}^{-2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20c{m}^2$, the average force exerted on the surface during a $30$ minute time span is

• A. $2.1\times 10^{-6}N$
• B. $1.2\times 10^{-6}N$
• C. $1.2\times 10^{6}N$
• D. $2.1\times 10^{6}N$

Answer 1

Answer: (B)

$\text{ Average force by light on surface is given by }$
$F_{av}=\frac{\Delta P}{\Delta t}$
$\Delta t=30min$
$=30\times 60s$
$=1800s$
$△P=\text{ Change in momentum }$
$\text{ Given that intensity of light }$
$I=\frac{18W}{c{m}^2}$
$\text{ Total power absorbed by surface }=I\times \text{ area of surface }$
$=\left(18W/c{m}^2\right)\left(20c{m}^2\right)$
$=360W$
$$\begin{gathered} {\text{ So }}^2,F_{av}=-\frac{E}{C.\Delta t}=-\frac{\text{ Power }}{C} \\ <br/>=-\frac{360}{3\times 10^{+8}} \end{gathered}$$
$=-120\times 10^{-8}N$
$F_{av}=-1.2\times 10^{-6}N$