## Radiation pressure of electromagnetic wave

When an electromagnetic wave strikes a surface then its momentum changes. the rate of change of momentum is equal to the applied force. this force acting on the unit area of the surface exerts a pressure called radiation pressure$(P_{rad})$.

Let us consider a plane electromagnetic wave incident normally on a perfectly absorbing surface of area $A$ for a time $t$. If energy $U$ is absorbed during this time then momentum $P$ delivered to the surface is given according to Maxwell's prediction by

$P=\frac{U}{C} \qquad(1)$

If $S$ is the energy flow per unit area per unit time i.e. Poynting vector then the energy density

$U=SAt \qquad(2)$

From equation $(1)$ and equation $(2)$

$P=\frac{SAt}{c}$

$P=UAt \qquad (\because U=\frac{S}{c})$

$\frac{P}{t}=UA \qquad (3)$

If average force $(F)$ acting on the surface, is equal to the average rate of change of momentum $(P)$, is delivered to the surface then

$F=\frac{P}{t} \qquad(4)$

Now from equation$(3)$ and equation$(4)$ we get

$F=UA \qquad(5)$

The radiation pressure $(P_{rad})$ exerted on the surface is

$P_{rad}=\frac{F}{A} \qquad(6)$

Now substitute the value of $F$ from equation$(5)$ in equation$(6)$ then we get

$P_{rad}=\frac{UA}{A}$

 $P_{rad}=U$

Hence, the radiation pressure exerted by a normally incident play electromagnetic wave on a perfect absorber is equal to the energy density of the wave.

For a perfect reflector or for a perfect reflecting surface, the radiation after reflection has momentum equal in magnitude but opposite in direction to the incident radiation. Then the momentum imparted to the surface will therefore be twice as on perfect absorber i.e.

 $P_{rad}=2U$