The ratio of maximum to minimum intensity due | vedclass

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Question

(A) The ratio of maximum to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \frac{(a+b)^2}{(a-b)^2} = \frac{49}{9}$.Taking the square root

Vedclass · Accepted Answer

$6.25$

Vedclass · Answer

(A) The ratio of maximum to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \frac{(a+b)^2}{(a-b)^2} = \frac{49}{9}$.Taking the square root on both sides,we get $\frac{a+b}{a-b} = \frac{7}{3}$.By cross-multiplying,$3(a+b) = 7(a-b)$,which simplifies to $3a + 3b = 7a - 7b$.Rearranging the terms,$4a = 10b$,so the ratio of amplitudes is $\frac{a}{b} = \frac{10}{4} = \frac{5}{2}$.Since intensity $I \propto a^2$,the ratio of intensities of the component waves is $\frac{I_1}{I_2} = \frac{a^2}{b^2} = \left(\frac{a}{b}\right)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25$.

The ratio of maximum to minimum intensity due to the superposition of two waves is $\frac{49}{9}$. Then the ratio of the intensities of the component waves is:

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