91Ó°ÊÓ

The electromagnetic wave is normally incident on a flat surface.

Energy density is the intensity of e.m radiation divided by the speed of light and radiation pressure also has the unit of energy density. So both these physical quantities can be used to prove the required statement. Energy density is the amount of energy that can be stored in a given system, substance, or region of space. Energy density can be measured in energy per volume or per mass. The higher the energy density of a system or material, the greater the amount of energy it has stored.

Formulae are as follows:

Radiation pressure for perfect reflection, $P_r=\frac{2f{I}_0}{c}$

Radiation pressure for total absorption,$P_a=\frac{I_0}{c}$

The intensity of radiation incident onthesurface of surface area $A$

for time $t$,

The energy density of radiation with intensity $I$,

$u=\frac{I}{c}$

Where, P_r is the radiation pressure for perfect reflection, P_ais the radiation pressure for total absorption, c is the speed of light,Iis the intensity of radiation, A isthesurface area, and uis the energy density of radiation.

Let I₀ be the intensity of incident radiation.

Let fbe the fraction of radiation reflected from the surface, and (1-f) be the fraction of radiation absorbed. Then, the radiation pressure due to the reflected radiation beam will be,

$P_r=\frac{2I_0}{c}×f$

Similarly, the radiation pressure due to the absorbed radiation beam will be,

$P_a=\frac{I_0}{c}×(1−f)$

Therefore, the total pressure on the surface isthe sum of P_r and P_a.

$P_{total}=P_r+P_a=\frac{2I_0}{c}×f+\frac{I_0}{c}×\left(1-f\right)$

Taking $\frac{I_0}{c}$ common from expression,

$$\begin{gathered} P_{total}=P_r+P_a \\ =\frac{I_0}{c}·\left\{2f+\left(1-f\right)\right\} \end{gathered}$$

$P_{total}=\frac{(1+f)I_0}{c}$ …(1)

Now, imagine a cylinder of lengthl,and cross-sectional area A. Find outtheenergy of radiation, U, inside the cylinder.

Let’s assume, u is the energy density of radiation inside the cylinder, and then, by the definition of energy density, the following expression:

$u=\frac{U}{volumeofcylinder}=\frac{U}{Al}$

Hence, the energy of radiation inside the cylinder,

To find the intensity of radiation incident on the surface area A for time t, use the definition of intensity.

$\begin{array}{c}I=\frac{u}{At}\\ =\frac{uAl}{At}\\ I=\frac{ul}{t}\end{array}$

e.m radiation travels with speed c, so using this,

$c=\frac{I}{t}$

So the intensity of radiation becomes,

$$\begin{gathered} I=\frac{ul}{t} \\ I=uc \end{gathered}$$

As the e.m the radiation strikes the flat surface, part of radiation is reflected.

Therefore, the total radiation outside ofthecylinder is,

I_total = (incident radiation intensity+ reflected radiation intensity)

$\begin{array}{c}{I}_{total}=I_0+f{I}_0\\ I_{total}=(1+f)I_0\end{array}$

Hence,

$u=\frac{I_{total}}{c}$

$u=\frac{(1+f)I_0}{c}$ ----(2)

From equation (1)and (2), it isproved thatthe radiation pressure is equal to the energy density if the electromagnetic wave is incident normally on a flat surface.

Hence, it is proved thatthe radiation pressure is equal to the energy density if the electromagnetic wave is incident normally on a flat surface.

Definition of energy density and radiation pressure is used to prove the given statement.