Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 28 - Physical Optics: Interference and Diffraction - Problems and Conceptual Exercises - Page 1004: 7

Answer

$406Hz $; $812Hz $

Work Step by Step

We know that $\Delta l=l_2-l_1$ $\Delta l=(2.55+\frac{0.845}{2})-(2.55-\frac{0.845}{2})m=0.845m $ and $\lambda=\frac{\Delta l}{m}$ $\lambda=\frac{0.845}{m}$ Now $ n=\frac{\nu}{\lambda}$ $ n=\frac{343}{\frac{0.845}{m}}Hz $ $ n=406mHz $ where $ m=1,2,3,........$ for $ m=2$ $ n_2=406(2)=812Hz $ Thus, the two lowest frequencies are $406Hz $ and $812Hz $.
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