Question

Light with energy flux of 18$w/{cm}^2$ falls on a non reflecting surface of area 20${cm}^2$ at normal incidence the momentum delivered in $30$ minutes ise

• A. $1.2\times {10}^{-6}{kgms}^{-1}$
• B. $2.16\times {10}^{-3}{kgms}^{-1}$
• C. $1.8\times {10}^{-3}{kgms}^{-1}$
• D. $3.2\times {10}^{-3}{kgms}^{-1}$

Answer 1

The correct option is B $2.16\times {10}^{-3}{kgms}^{-1}$

$\begin{array}{c}ThetotalenergyfallingonthesurfaceisU=\left(18W/c{m}^2\right)\times \left(20c{m}^2\right)\times \left(30\times 60\right)=6.48\times {10}^{5}J\\ therefore,thetotalmomentumdelivered\left(forcompeleteabsorption\right)is\\ p=\frac{U}{c}=\frac{\left(6.48\times {10}^{5}J\right)}{3\times {10}^{8}m/s}=2.16\times {10}^{-3}kgm/s\\ theaverageforceexertedonthesurfaceis\\ F=\frac{p}{t}=\frac{2.16\times {10}^{-3}}{0.18\times {10}^4}=1.2\times {10}^{-6}kgm{s}^{-1}\end{array}$

Hence,

option $\left(A\right)$ is correct answer.