Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 22 - Electromagnetic Waves - Problems - Page 641: 30

Answer

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Work Step by Step

a. The energy delivered in each pulse is the laser’s average power output multiplied by the time of the pulse. $$P=\frac{E}{t}$$ $$E=Pt=(1.5\times10^{11}W)(1.0\times10^{-9}s)=150J$$ b. The intensity is the power per unit area. The laser power is distributed uniformly over the cross-sectional area of the beam. The energy per unit area per unit time is the magnitude of the average intensity, equation 22–8. Find the rms value of the electric field. $$\overline{I}=\frac{P}{A}=\frac{1}{2}\epsilon_ocE_o^2=\frac{1}{2}\epsilon_oc(\sqrt{2}E_{rms})^2=\epsilon_ocE_{rms}^2$$ $$E_{rms}=\sqrt{\frac{\overline{I}}{\epsilon_oc }}=\sqrt{\frac{P}{A\epsilon_oc }}$$ $$E_{rms}=\sqrt{\frac{1.5\times10^{11}W }{(\pi (0.0022m)^2) (8.85\times10^{-12} C^2/(N\cdot m^2))(3.00\times10^8m/s) }}$$ $$\approx 1.9\times10^9 V/m$$
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