91Ó°ÊÓ

The average force per unit area due to the wave is known as radiation pressure $p_{rad}$. Radiation pressure is the average divided by the absorbing area.

The formula for radiation pressure $p_{rad}$of the wave totally absorbed light is:

$p_{rad}=\frac{I}{C}$

The formula for radiation pressure of the wave totally reflected light is:

$p_{rad}=\frac{2I}{C}$

The average power momentum density $\left(p_{avg}\right)$is defined as the intensity $\left(1\right)$divided by the square of the speed of light $\left(I\right)$.

The formula for the average power momentum density $\left(p_{avg}\right)$is:

$p_{avg}=\frac{I}{c^2}$

Where, I is the intensity in $\mathrm{W}/{\mathrm{m}}^2$and c is the speed of light that is equal to .

Given that, I = 2500 $\mathrm{W}/{\mathrm{M}}^2$

Substitute the values in formula$p_{rad}=\frac{I}{c}$

Determine the radiation pressure in atmosphere:

$$\begin{gathered} {\mathrm{p}}_{\mathrm{rad}}=\left(8.33×{10}^{-6}\mathrm{Pa}\right)×\frac{1}{\left(1.103×{10}^5\right)} \\ =8.33×{10}^{-11}\mathrm{atm} \end{gathered}$$

Hence, the average radiation pressure on a totally absorbing section of the floor is $8.33×{10}^{-6}\mathrm{Pa}$or $8.33×{10}^{-11}\mathrm{atm}$

Determine the radiation pressure in pascals:

Substitute the values in formula $p_{rad}=\frac{2I}{c}$

$p_{rad}=\frac{2×2500}{3×{10}^{-6}\mathrm{Pa}}$

Determine the radiation pressure in atmosphere:

$p_{rad}=\left(16.66×{10}^{-6}Pa\right)×\frac{1}{1.103×{10}^5}$

$=16.66×{10}^{-10}\mathrm{atm}$

Hence, the average radiation pressure on a totally reflecting section of the floor is $=16.66×{10}^{-6}\mathrm{Pa}$or $16.66×{10}^{-10}\mathrm{atm}$.

Substitute the values in formula $P_{avg}=\frac{I}{c^2}$

$$\begin{gathered} P_{avg}=\frac{2500}{{\left(3×{10}^8\right)}^2} \\ =2.78×{10}^{-14}\mathrm{kg}/{\mathrm{m}}^2 \end{gathered}$$

Hence, average momentum density (momentum per unit volume) in the light at the floor is $2.78×{10}^{-14}\mathrm{kg}/{\mathrm{m}}^2$.