Light with an energy flux of $9 \mathrm{Wcm}^{-2}$ falls on a nonreflecting surface at normal incidence. If…
Light with an energy flux of $9 \mathrm{Wcm}^{-2}$ falls on a nonreflecting surface at normal incidence. If the surface has an area of $20 \mathrm{~cm}^2$. The total momentum delivered for complete absorption in one hour is
$2.16 \times 10^{-4} \mathrm{kgms}^{-1}$
$1.16 \times 10^{-3} \mathrm{kgms}^{-1}$
$2.16 \times 10^{-3} \mathrm{kgms}^{-1}$
$3.16 \times 10^{-4} \mathrm{kgms}^{-1}$
Solution
We have total momentum as $\Delta p=\left(\frac{I A}{c}\right) t$ for non reflecting surface,
$\begin{aligned}
& \Delta p=\left(\frac{9 \times 10^4 \times 20 \times 10^{-4}}{3 \times 10^8} \times 3600\right) \\
& =216 \times 10^{-3} \mathrm{kgms}^{-1}
\end{aligned}$