How Is the Formula for Radiation Pressure on a Reflecting Surface Derived?

[premed]

for a reflecting surface, the radiation pressure is equal to 2I/c where I is the intensity of light and c is the speed of light. I saw this in my physics homework manual but it was never mentioned in the book. thanks.

 

[olgranpappy]

First, recall that the intensity I is the average of the poynting (sp?) vector, so it is the energy flux. I.e., it's the number of incoming photons times their energy divided by time and divided by area (that's what "flux" means "whatever" flux is "whatever" per unit time per unit area)... that is
$$I=\frac{\hbar\omega N}{\rm time*Area}$$
where N is the number of photons and \hbar\omega is the energy of a single photon.

So

$${\rm Pressure}=\frac{\rm Force}{\rm Area}$$
and the Force is the total momentum transferred divided by the time. For reflection the photon bounces off so that it transfers an amount of momentum $\Delta p=2p$,
so the total force due to N reflecting photons is (2pN)/(time)
$${\rm Pressure}=\frac{2pN}{\rm time*Area}=\frac{2\hbar\omega N}{c{\rm * time*Area}}<br /> =\frac{2I}{c}$$

wheras, for absorption the momentum imparted by a single photon is $\Delta p=p$, which gives
$$P=\frac{I}{c}$$