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: 25

Answer

7.38 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. $$sin\theta_1=\frac{\lambda}{D}$$ $$\theta_1=sin^{-1}(\frac{\lambda}{D})$$ $$= sin^{-1}(\frac{558\times10^{-9}m}{0.0348\times10^{-3}m})=0.9187^{\circ}$$ 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. $$x_1=(\mathcal{l})tan\theta_1=(2.30m)tan0.9187^{\circ}=3.69\times10^{-2}m$$ The width of the entire central peak is two times this distance. $$\Delta x = 2 x_1=7.38\times10^{-2}m $$
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