College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 11 - Problems - Page 431: 51

Answer

$I = 0.085~mW/m^2$

Work Step by Step

Let $I_1 = 25~mW/m^2$ Let $I_2 = 28~mW/m^2$ We can find the ratio of $\frac{I_1}{I_2}$: $\frac{I_1}{I_2} = \frac{25~mW/m^2}{28~mW/m^2} = \frac{25}{28}$ The intensity of a wave is proportional to the square of the amplitude, so $\frac{A_1}{A_2} = \sqrt{\frac{25}{28}}$ When two waves interfere destructively, the amplitude of the resulting wave is the difference of the two amplitudes. Therefore, the amplitude of the resulting wave is this case is: $~A_2-A_1 = A_2 - \sqrt{\frac{25}{28}}~A_2$ The intensity $I_2$ is proportional to $A_2^2$, while the intensity $I$ of the resulting wave is proportional to $[(1-\sqrt{\frac{25}{28}})~A_2]^2$. We can find the intensity of the resulting wave: $I = (1-\sqrt{\frac{25}{28}})^2~I_2$ $I = (1-\sqrt{\frac{25}{28}})^2~(28~mW/m^2)$ $I = 0.085~mW/m^2$
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