Physics: Principles with Applications (7th Edition)

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

Chapter 25 - Optical Instruments - Problems - Page 742: 59

Answer

$f_e=8.5cm$ $\theta=6.21\times10^{-6}rad$ $r=2.4km$

Work Step by Step

$f_e=f_o\frac{d_o}{l}\frac{N}{d}=(2.0m)\frac{6.5km}{384,000km}\frac{250mm}{0.10mm}=8.5cm$ $\theta=1.22\frac{\gamma}{D}=1.22\frac{560\times10^{-9}m}{0.11m}=6.21\times10^{-6}rad$ $r=(384,000km)(6.21\times10^{-6}rad)=2.4km$
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