Physics - Modern Physics Question with Solution | TestHub

Momentum of incident light per second

${\mathrm{p}}_1=\frac{\mathrm{E}}{\mathrm{c}}=\frac{60}{3\times {10}^8}=2\times {10}^{-7}$

Momentum of reflected light per second

${\mathrm{p}}_2=\frac{60}{100}\times \frac{\mathrm{E}}{\mathrm{c}}=\frac{60}{3\times {10}^8}=1.2\times {10}^{-7}$

Force on the surface = change in momentum per second

$={\mathrm{p}}_2-\left(-{\mathrm{p}}_1\right)={\mathrm{p}}_2+{\mathrm{p}}_1=\left(2+1.2\right)\times {10}^{-7}=3.2\times {10}^{-7} \mathrm{N}$