Suggested languages for you:

Americas

Europe

19P

Expert-verifiedFound in: Page 1002

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**High-power lasers are used to compress a plasma (a gas of charged particles) by radiation pressure. A laser generating radiation pulses with peak power ${\mathbf{1}}{\mathbf{.5}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{3}}}{\mathbf{\text{MW}}}$**** is focused onto 1.00 mm ^{2} of high-electron-density plasma. Find the pressure exerted on the plasma if the plasma reflects all the light beams directly back along their paths.**

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

Power of radiation $P=1.5\times {10}^{3}MW$

Area A=1.00 mm^{2}

**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_{r }is 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.

$\begin{array}{c}{p}_{r}=\frac{2P}{Ac}\\ =\frac{2(1.5\times {10}^{9}W)}{(1\times {10}^{-6}{m}^{2})(3\times {10}^{8}\text{\hspace{0.17em}m}/\text{s})}\\ =1\times {10}^{7}Pa\end{array}$

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

94% of StudySmarter users get better grades.

Sign up for free