Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 22 - Wave Optics - Exercises and Problems - Page 649: 19

Answer

$w = 4.0~mm$

Work Step by Step

We can find the width of the central maximum: $w = \frac{2~\lambda~L}{d}$ $w = \frac{(2)(500\times 10^{-9}~m)(2.0~m)}{0.50\times 10^{-3}~m}$ $w = 4.0\times 10^{-3}~m$ $w = 4.0~mm$
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