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 1003: 4

Answer

(a) $300m$ (b) $30s$

Work Step by Step

(a) We know that the longest possible wavelength gives the constructive radio wave and it is equal to the path difference of the two sources and the observed point. Thus, $\lambda=450m-150m=300m$ (b) We know that $l_1=\sqrt{(150m)^2+(vt)^2}$....eq(1) and $l_2=\sqrt{(150m)^2+(vt)^2}$...eq(2) adding eq(1) and eq(2), we obtain: $\sqrt{(150m)^2+(vt)^2}+\sqrt{(150m)^2+(vt)^2}=150m$ This simplifies to: $t=29.59s$ $\implies t=30s$
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