A black, totally absorbing piece of cardboard of area $\mathbf{\text{A = 2.0}}\mathbf{}{\mathbf{\text{cm}}}^{\mathbf{\text{2}}}$intercepts light with an intensity of$\mathbf{110}\mathbf{}{\mathbf{\text{W/m}}}^{\mathbf{\text{2}}}$ from a camera strobe light. What radiation pressure is produced on the cardboard by the light?

Area of cardboard, $\mathrm{A}=2{\text{cm}}^{\text{2}}$.

Intensity, I = 10 W/m².

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 the area of the surface, calculate the radiation pressure of the light bulb. When electromagnetic radiation of intensity $\mathrm{I}$is incident on a perfectly reflective surface, the radiation pressure$\mathrm{P}$exerted on the surface is given by,, where$\mathrm{c}$is the speed of light.

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

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

Here I is the intensity of radiation, and $A$is the area of the surface perpendicular to the path of the radiation.

The intensity $I$is related to the power of the bulb P as,

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

Where, Ais the area of the cardboard.

The radiation pressure P_r is the force per unit area, and it can be calculated as,

$$\begin{gathered} p_r=\frac{IA}{AC} \\ {p}_r=\frac{I}{C} \end{gathered}$$

Substitute the values in the expression, and we get,

$$\begin{gathered} P_r=\frac{10w}{3\times {10}^{8}m/s} \\ =3.3\times {10}^{-8}Pa \end{gathered}$$

Therefore, the radiation pressure ${\mathrm{p}}_{\mathrm{r}}$is $3.3\times {10}^{-8}\text{Pa}$.