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 708: 8

Answer

$6.4\times10^{-7}m$

Work Step by Step

The spacing on the screen between bright fringes is addressed in Example 24-1. For light shining through 2 slits separated by a distance d, the spacing between successive fringes on a screen a distance $\mathcal{l}$ away is $\Delta x = \frac{\lambda \mathcal{l}}{d}$. By using a ruler on Figure 24–9a, we measure that the spacing $\Delta x$ is about 1.4 mm. Solve for the wavelength. $$\lambda=\frac{d\Delta x}{\mathcal{l}}=\frac{(0.17\times10^{-3}m)(0.0014m)}{0.37m}$$ $$\lambda=6.4\times10^{-7}m$$
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