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: 47

Answer

(a) $D = 3.8~cm$ (b) The light from Jupiter would be very hard to detect in comparison with the light from the sun.

Work Step by Step

We can find the required angular resolution: $\theta = \frac{d}{L}$ $\theta = \frac{7.8\times 10^{11}~m}{(4.3)(9.46\times 10^{15}~m)}$ $\theta = 1.9175\times 10^{-5}~rad$ We can find the minimum diameter: $\theta = \frac{1.22~\lambda}{D}$ $D = \frac{1.22~\lambda}{\theta}$ $D = \frac{(1.22)~(600\times 10^{-9}~m)}{1.9175\times 10^{-5}~rad}$ $D = 0.0382~m$ $D = 3.8~cm$ (b) The light from Jupiter would be very hard to detect in comparison with the light from the sun. It would be like trying to detect a firefly on the circumference of a car headlight.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.