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: 24

Answer

0.54 cm

Work Step by Step

Use equation 24–3a to calculate the angular width from the middle of the central peak to the first minimum. $$sin\theta_1=\frac{\lambda}{D}$$ $$\theta_1=sin^{-1}(\frac{\lambda}{D})$$ $$= sin^{-1}(\frac{450\times10^{-9}m}{1.0\times10^{-3}m})=0.02578^{\circ}$$ Find the distance (on the screen) between the centerline and the edge of the central maximum, by using the distance to the screen $\mathcal{l}$ and the tangent of the angle. $$x_1=(\mathcal{l})tan\theta_1=(6.0m)tan0.02578^{\circ}=2.70\times10^{-3}m$$ The width of the entire central peak is two times this distance. $$\Delta x = 2 x_1=5.4\times10^{-3}m $$
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