Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 24 - The Wave Nature of Light - Problems - Page 709: 42

Answer

$d = 2.35\times10^{-6}m $; $48.7^{\circ}$

Work Step by Step

Solve Eq. 24–4 for the slit separation d. $$d sin \theta = m \lambda $$ $$d = \frac{m \lambda}{ sin \theta }$$ $$d = \frac{1(589\times10^{-9}m)}{sin 14.5^{\circ}}$$ $$d = 2.35\times10^{-6}m $$ Now set m = 3 and find the angle of the third-order maximum. $$d sin \theta = m \lambda $$ $$ \theta = sin^{-1}(\frac{m \lambda} {d})$$ $$ \theta = sin^{-1}(\frac{3(589\times10^{-9}m)} {2.352\times10^{-6}m})=48.7^{\circ}$$
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