An intense light source radiates uniformly in all directions. At a distance of5.0 m from the source, the radiation pressure on a perfectly absorbing surface is $9.0\times {10}^{-6}Pa$. What is the total average power output of the source?

The power transported per unit area is known as the intensity (l).

The formula used to calculate the intensityis:

$l=\frac{P}{A}$

Where, A is area measured in the direction perpendicular to the energy and P is the power in watts.

The average force per unit area due to the wave is known as radiation pressure ${\mathit{p}}_{\mathbf{r}\mathbf{a}\mathbf{d}}$. Radiation pressure is the average dp/dtdivided by the absorbing area A.

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

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

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

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

Where, l is the intensity in $W/m^2$and c is the speed of light that is equal to $3.0\times {10}^{8}m/s$.

Given that,

$$\begin{gathered} r=5m \\ {p}_{rad}=9\times {10}^{-6}Pa \end{gathered}$$

Consider the path of the source to be a sphere.

So, the area of source is

$$\begin{gathered} A=4{\mathrm{\pi r}}^2 \\ =4\mathrm{\pi }{\left(5\right)}^2 \\ =314{\mathrm{m}}^2 \end{gathered}$$

Now, the average power output intensity of source is

p = lA .......(1)

Substitute $I=c{p}_{rad}$in equation

$P=\left(c{p}_{rad}\right)A$ .......(2)

Substitute the values of $c,p_{rad}$and in equation

$$\begin{gathered} P=\left[\left(3\times {10}^8\right)\left(9\times {10}^{-6}\right)\right]\left(314\right) \\ =2700\times 314 \\ =8.47\times {10}^{5}W \end{gathered}$$

Hence, the average power is equal to $8.47\times {10}^{5}W$.