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Show that the radiation pressure exerted by an $EM$ wave of intensity $I$ on a surface kept in vacuum is $\frac{I}{c}$.
Solution
$\text { Pressure }=\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{F}}{\mathrm{A}} \quad \therefore \mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$
Rate of change of momentum is force,
$\therefore \mathrm{F}=\frac{d p}{d t}$ Now $\mathrm{E}=m c^{2}$ $\therefore \mathrm{U}=(m c) c \quad[\because \mathrm{E}=\mathrm{U}]$ $\therefore \mathrm{U}=\mathrm{P} c$ By taking differentiation both side w.r.t time, $\quad \frac{d \mathrm{U}}{d t}=c \frac{d \mathrm{P}}{d t}$ $\therefore \frac{d \mathrm{U}}{d t} \times \frac{1}{c}=\mathrm{F} \quad\left[\because \frac{d \mathrm{P}}{d t}=\mathrm{F}\right]$ Now $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}=\frac{d \mathrm{~V}}{d t} \times \frac{1}{\mathrm{Ac}}$ $\therefore \mathrm{P}=\frac{\mathrm{I}}{c}$
