I was asked the following question in a physics worksheet:
A $2.0 \,\mathrm{mW}$ laser is shone onto a metal plate. The cross-sectional area of the beam is $4.0\times 10^{-6} \, \mathrm m^2$. Assume all light is absorbed, and the laser is emits all light at $4.7\times 10^{14} \, \mathrm{Hz}$. Calculate the pressure due to the light.
My current approach was along the lines of: $$\begin{aligned} p &=\dfrac{hf}{c}\\ &=\dfrac{h\times4.7\times10^{14}}{3\times10^{8}}\\ &=1.0\times10^{-27}\mathrm{Ns}\\ \end{aligned}$$ $$\begin{aligned} f_{e^-}&=\tfrac{P}{E}=\tfrac{P}{hf}\\ &=\dfrac{2.0\times 10^{-3}}{hf}\\ &=6.45\times10^{15}\mathrm{Hz}\\ \end{aligned}$$ $$\begin{aligned} F&=\frac{\mathrm dp}{\mathrm dt}\\ &=p\times f_{e^-}\\ &=6.7\times10^{-12}\mathrm{N}\\ \end{aligned}$$ $$\begin{aligned} P&=\frac{F}{A}\\ &=\frac{6.7\times 10^{-12}}{4.0\times 10^{-6}}\\ &=1.7\times 10^{-6}\mathrm{Pa} \end{aligned}$$
But a classmate of mine used $P=Fv$ (i.e. power used to propel a point mass is equal to the force applied to that mass times the velocity of that mass) to solve for the force, and I don’t understand how that equation, which is used in linear kinematics, applies to a massless particle.
I understand the principle that light exerts pressure, but why does this formula which was derived for something entirely different apply in this context?
