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 707: 6

Answer

651 nm.

Work Step by Step

$$d sin\theta=m\lambda$$ For small angles, the tangent and sine are approximately equal. Consider m = 1. For 2-slit interference, the distance between bright fringes equals the position of the first m=1 bright fringe from the centerline. The tangent is the spacing from the centerline, x, divided by the distance from the slits to the screen, L. $$d\frac{x}{L}=(1) \lambda$$ $$x=\frac{\lambda L}{d}$$ Let the red laser be subscript 1, and the laser pointer be subscript 2. In the 2 cases, the slit spacing d and the distance from the slits to the screen, L, stay the same. $$x_1=\frac{\lambda_1 L}{d}$$ $$x_2=\frac{\lambda_2 L}{d}$$ $$\lambda_2=\frac{\lambda_1}{x_1}x_2=\frac{632.8nm}{5.00mm}5.14mm\approx 651 nm$$
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