As we know that,
$$
\text { Pressure }=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}
$$
If mass $(\mathrm{m})$ of radiant particle for wave of velocity $(\mathrm{C})$ then $E=m C^2$
Let,
$$
\begin{aligned}
\mathrm{E} &=\mathrm{U} ; \\
\mathrm{U} &=(\mathrm{mC}) \mathrm{C} \\
\mathrm{U} &=\mathrm{PC} \quad[\because \mathrm{P}=\mathrm{mC}]
\end{aligned}
$$
differentiating both side w.r.t. (t)
So, $\quad\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)=\mathrm{C}\left(\frac{\mathrm{dP}}{\mathrm{dt}}\right)$
Then, the Force is the rate of change of momentum i.e.,
$$
\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}
$$
By putting the value of $\left(\frac{\mathrm{dp}}{\mathrm{dt}}\right)$ in equation (ii)
So,
$$
\mathrm{F}=\frac{1}{\mathrm{C}}\left[\frac{\mathrm{dv}}{\mathrm{dt}}\right]
$$
From (i) equation, $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$
So, $\quad$ (Pressure) $(\mathrm{P})=\frac{1}{\mathrm{~A}} \cdot \frac{\mathrm{dU}}{\mathrm{C} \cdot \mathrm{dt}}\left[\because \mathrm{I}=\right.$ Intensity $\left.=\frac{\mathrm{dU}}{\mathrm{A} \cdot \mathrm{dt}}\right]$
$$
P=\frac{I}{C}
$$