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 955: 8

Answer

$\lambda = 573~nm$

Work Step by Step

The spacing between these two minima is $\Delta y = \frac{4~\lambda~L}{d}$ We can find the wavelength of the light: $\Delta y = \frac{4~\lambda~L}{d}$ $\lambda = \frac{\Delta y~d}{4~\lambda}$ $\lambda = \frac{(5.5\times 10^{-3}~m)(0.25\times 10^{-3}~m)}{(4)(0.60~m)}$ $\lambda = 5.73\times 10^{-7}~m$ $\lambda = 573~nm$
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