Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 35 - Optical Instruments - Exercises and Problems - Page 1019: 46

Answer

(a) $d = 9.4\times 10^{10}~km$ (b) The separation distance is $~~120~~$ times the distance of Jupiter from the sun.

Work Step by Step

We can write an equation for the resolution: $\theta_{min} = \frac{1.22~\lambda}{D}$ $\theta_{min} = \frac{(1.22)~(650\times 10^{-9}~m)}{2.4~m}$ $\theta_{min} = 3.304\times 10^{-7}~rad$ We can find the separation distance $d$: $\theta = \frac{d}{L}$ $d = L~\theta$ $d = (30,000)(9.46\times 10^{12}~km)(3.304\times 10^{-7}~rad)$ $d = 9.4\times 10^{10}~km$ (b) We can express this distance as a multiple of the distance of Jupiter from the sun: $d = \frac{9.4\times 10^{10}~km}{7.8\times 10^8~km} = 120$ The separation distance is $~~120~~$ times the distance of Jupiter from the sun.
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