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If the ratio of the intensities of two waves producing interference is $49: 16$, then the ratio of the resultant maximum intensity to minimum intensity will be
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The correct answer is:
121:9
Let the intensities of the two waves be $I_1$ and $I_2$.
Given: $I_1: I_2=49: 16$
Therefore, the ratio of maximum and minimum intensities is given by,
$\frac{I_{\max }}{I_{\min }}=\left(\frac{A_1+A_2}{A_1-A_2}\right)^2=\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{7+4}{7-4}\right)^2=\frac{121}{9}$
Given: $I_1: I_2=49: 16$
Therefore, the ratio of maximum and minimum intensities is given by,
$\frac{I_{\max }}{I_{\min }}=\left(\frac{A_1+A_2}{A_1-A_2}\right)^2=\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{7+4}{7-4}\right)^2=\frac{121}{9}$
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