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 33 - Wave Optics - Exercises and Problems - Page 958: 65


$D = 0.50~m$

Work Step by Step

We can find the required diameter of the aperture: $w = \frac{2.44~\lambda~L}{D}$ $D = \frac{2.44~\lambda~L}{w}$ $D = \frac{(2.44)(532\times 10^{-9}~m)(3.84\times 10^{8}~m)}{10^3~m}$ $D = 0.50~m$
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