Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 21 - Superposition - Exercises and Problems - Page 624: 63

Answer

$20$

Work Step by Step

We know, for maxima for two identical sources, that the path length difference between the two waves must be an integer number of wavelengths. Thus, we need to find the point on the 10-m circular path around the two antennas at which $\Delta r=m\lambda$. But, we need first to find $\lambda$ which is given by $v=\lambda f$, $$\lambda=\dfrac{c}{f}=\dfrac{3\times 10^8}{750\times 10^6}=\bf 0.4 \;\rm m$$ So, the separation distance $d$ between the two antennas is $$d=\dfrac{2}{0.4}\lambda=5\lambda$$ On diameter that the passes the two antennas, we have a path length difference for the two waves, (see the figure below), $$\Delta r=6-4=2\;\rm m$$ which is also $5\lambda$. So we have two maxima on this diameter [see point 1]. Now the path length difference on the diameter that passes in the middle between the two antennas (which is north-south line in the figure below), $$\Delta r=\sqrt{5^2+1^2}-\sqrt{5^2+1^2}=0$$ so we have here also 2 maxima [see point 2]. Now at point 1 we have $\Delta r=5\lambda$, and at point 2 $ \Delta r=0\lambda=0 $, so between them, in the first quarter, we can find $\Delta r,\lambda,2\lambda,3\lambda,4\lambda$. This means that in each quarter we have 5 maxima, and hence in the whole circle we got 20 maxima.
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