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: 15

Answer

14 mm.

Work Step by Step

$$d sin\theta=m\lambda$$ For small angles, the tangent and sine are approximately equal. The tangent is the spacing from the centerline, x, divided by the distance from the slits to the screen, L. $$d\frac{x}{L}=m \lambda$$ $$x=\frac{m \lambda L}{d}$$ Let the 480 nm light be subscript 1, and the 650 nm light be subscript 2. In the 2 cases, the slit spacing d and the distance from the slits to the screen, L, stay the same. $$x_1=\frac{m_1\lambda_1 L}{d}$$ $$x_2=\frac{m_2 \lambda_2 L}{d}$$ Solve for $x_2$. $$x_2=\frac{\lambda_2 m_2}{\lambda_1 m_1}x_1=\frac{(650nm)(2)}{(480 nm)(3)}16mm\approx 14 mm$$
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