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: 37

Answer

490 nm 610 nm 640 nm 670 nm

Work Step by Step

The slit separation is $d=\frac{1}{9800}cm$. Use equation 24–4 to find the wavelength for first order lines. $$d sin \theta = m \lambda $$ $$\lambda=\frac{d sin \theta}{m}$$ $$\lambda=\frac{(\frac{1}{9800}cm)(sin 28.8^{\circ})}{1}$$ $$=4.92\times10^{-5}cm\approx490nm$$ Repeat for the next angle. $$\lambda=\frac{(\frac{1}{9800}cm)(sin 36.7^{\circ})}{1}$$ $$=6.10\times10^{-5}cm=610nm$$ Repeat for the next angle. $$\lambda=\frac{(\frac{1}{9800}cm)(sin 38.6^{\circ})}{1}$$ $$=6.37\times10^{-5}cm\approx640nm$$ Repeat for the next angle. $$\lambda=\frac{(\frac{1}{9800}cm)(sin 41.2^{\circ})}{1}$$ $$=6.72\times10^{-5}cm\approx670nm$$
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