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: 19

Answer

9600 $3.5\times10^{-15}s$

Work Step by Step

The length of the pulse is $d = c\Delta t$, where the time is 34 ps. Find the number of wavelengths that fit within this distance. Number of wavelengths that fit = $\frac{c\Delta t}{\lambda}=\frac{(3.00\times10^8m/s)( 34\times10^{-12}s)}{1062\times10^{-9}m}=9600$ If the original 34 picosecond pulse were 9600 times shorter, it would contain only one wavelength. Find out how brief such a pulse would have to be. $$\Delta t’=\frac{34\times10^{-12}s }{9600}=3.5\times10^{-15}s$$
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