Question: What is the radiation pressure 1.5maway from a 500wlight bulb? Assume that the surface on which the pressure is exerted faces the bulb and is perfectly absorbing and that the bulb radiates uniformly in all directions.

Distance, $\text{r}=1.5\text{m}$

Power, P=500w

The radiation pressure formula is for the total absorption of radiation. First, find the force on the surface due to radiation, and substituting this value and area of the surface, calculate the radiation pressure of the light bulb. When electromagnetic radiation of intensityis incident on a perfectly reflective surface, the radiation pressure$\mathbf{P}$exerted on the surface is given by,, where$\mathit{c}$is the speed of light.

If the radiation is completely absorbed by the surface, the force is,

$F=\frac{IA}{C}$

is the intensity of radiation, and $\text{A}$is the area of the surface perpendicular to the path of the radiation.

The radiation radiated by the bulb is uniform in all directions.

The intensity $\text{I}$is related to the power of the bulb as,

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$

where, $A=4{\mathrm{\pi r}}^2$is the surface area of the bulb.

$I=\frac{P}{4{\mathrm{\pi r}}^2}$

The radiation pressure ${\text{p}}_{\text{r}}$is the force per unit area.

$p_r=\frac{F}{A}$

The radiation pressure ${\text{p}}_{\text{r}}$is the force per unit area.

$$\begin{gathered} P_r=\frac{P}{4{\mathrm{\pi r}}^2\mathrm{c}} \\ =\frac{500w}{4\mathrm{\pi }{\left(1.5\mathrm{m}\right)}^2\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)} \\ =5.9\times {10}^{-8}Pa \end{gathered}$$

Therefore, the radiation pressure is ${\mathrm{p}}_{\mathrm{r}}=5.9\times {10}^{-8}\text{Pa}$.