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: 43

Answer

7140 slits/cm.

Work Step by Step

Solve Eq. 24–4 for the slit separation d. $$d sin \theta = m \lambda $$ $$d = \frac{m \lambda}{ sin \theta }$$ See Figure 24–26. The angle of the diffracted light increases as the wavelength increases. To be able to view a full order, the largest wavelength in the visible spectrum, 700 nm, must be visible at the maximum angle of diffraction, which is 90 degrees. Use the largest wavelength with that angle to find the minimum slit separation. $$d = \frac{2(700\times10^{-9}m)}{sin 90^{\circ}} = 1.40\times10^{-6}m=1.40\times10^{-4}cm $$ The reciprocal of the minimum slit separation gives the maximum number of slits per cm. $$\frac{1}{d}=\frac{1}{1.40\times10^{-4}cm }=7140\frac{slits}{cm}$$ The answer is reported to 3 significant figures.
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