Question

Light with an energy flux of $18 \,W \,cm^{-2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20 \,cm^2$, 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)

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