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 1213: 36.37

Answer

92 cm.

Work Step by Step

The angular size of the first dark ring is given by $sin \theta_1=1.22\frac{\lambda}{D}$. $$sin \theta_1=1.22\frac{620\times10^{-9}m }{7.4\times10^{-6}m }=0.1022$$ $$ \theta_1=5.867^{\circ}$$ The radius of the ring on the screen is $(4.5m)(tan \theta_1)=0.462 m$, and the diameter is twice that.
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