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 956: 29


$d = 0.25~mm$

Work Step by Step

We can find the diameter $d$ of the hole: $w = \frac{2.44~\lambda~L}{d}$ $d = \frac{2.44~\lambda~L}{w}$ $d = \frac{(2.44)~(633\times 10^{-9}~m)(4.0~m)}{2.5\times 10^{-2}~m}$ $d = 2.5\times 10^{-4}~m$ $d = 0.25~mm$
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