University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 36 - Diffraction - Problems - Exercises - Page 1213: 36.39


1.88 m.

Work Step by Step

The angle $\theta=\frac{28 km}{1200km}$. Use Rayleigh’s criterion, $sin \theta_1=1.22\frac{\lambda}{D}$. $$D =1.22\frac{\lambda}{ sin \theta_1}$$ The angle is small, so $sin \theta \approx \theta$. $$D =1.22 (3.6 cm) \frac{1200km}{28 km}=188 cm$$
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