## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 11 - Problems - Page 431: 50

#### Answer

$I = 79~mW/m^2$

#### Work Step by Step

Let $I_1 = 25~mW/m^2$ Let $I_2 = 15~mW/m^2$ We can find the ratio of $\frac{I_1}{I_2}$: $\frac{I_1}{I_2} = \frac{25~mW/m^2}{15~mW/m^2} = \frac{5}{3}$ The intensity of a wave is proportional to the square of the amplitude, so $\frac{A_1}{A_2} = \sqrt{\frac{5}{3}}$ When two waves interfere constructively, the amplitude of the resulting wave is the sum of the two amplitudes. Therefore, the amplitude of the resulting wave is this case is: $~A_1+A_2 = \sqrt{\frac{5}{3}}~A_2+A_2$ The intensity $I_2$ is proportional to $A_2^2$, while the intensity $I$ of the resulting wave is proportional to $[(1+\sqrt{\frac{5}{3}})~A_2]^2$. We can find the intensity of the resulting wave: $I = (1+\sqrt{\frac{5}{3}})^2~I_2$ $I = (1+\sqrt{\frac{5}{3}})^2~(15~mW/m^2)$ $I = 79~mW/m^2$

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