University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 36 - Diffraction - Problems - Exercises - Page 1214: 36.47

Answer

$3.03\times10^{-5}m $. As suggested, apply the equation for single-slit diffraction, using the thickness of the hair instead of the width of the slit.

Work Step by Step

Use equation 36.3 because the angle is small. $$y_1=\frac{x \lambda}{a}$$ The distance between the two dark fringes on either side of the central maximum is $5.22 cm = 2y_1$. $$y_1=\frac{x \lambda}{a}$$ $$a=\frac{x \lambda}{y_1}$$ $$a=\frac{(1.25m) (632.8\times10^{-9}m)}{2.61\times10^{-2}m}=3.03\times10^{-5}m $$
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