91Ó°ÊÓ

Power of radiation $P=1.5×{10}^{3}MW$

Area A=1.00 mm²

Substituting the given values of power, area, and speed of light in the radiation pressure formula, we can find the pressure exerted on the plasma due to the light beams.

Thrust applied on plasma is given as-

$\mathbf{F}\mathbf{=}\frac{\mathbf{2}\mathbf{IA}}{\mathbf{c}}$

Here, I is the intensity of radiation, A is the cross-sectional area and c is the speed of light in vacuum.

The force acting on the plasma, due to radiation, is-

$F=\frac{2IA}{c}$

The radiation pressure P_ris given as-

$\begin{array}{c}{p}_r=\frac{F}{A}\\ =\frac{2IA}{cA}\\ =\frac{2I}{c}\end{array}$

The intensity is $I=\frac{P}{A}$

P is power; A is the total area intercepted by the radiation.

$\begin{array}{c}{p}_r=\frac{2P}{Ac}\\ =\frac{2(1.5×{10}^{9}W)}{(1×{10}^{-6}{m}^2)(3×{10}^8\text{″¾}/\text{s})}\\ =1×{10}^{7}Pa\end{array}$

Pressure exerted on the plasma if the plasma reflects all the light beams directly back along their paths is$1×{10}^{7}Pa$