The ratio of maximum and minimum intensities | vedclass

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(D) Given the ratio of maximum intensity to minimum intensity is $\frac{I_{\max}}{I_{\min}} = \frac{36}{1}$.We know that $I_{\max} = (a_1 + a_2)^2$ an

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$7: 5$

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(D) Given the ratio of maximum intensity to minimum intensity is $\frac{I_{\max}}{I_{\min}} = \frac{36}{1}$.We know that $I_{\max} = (a_1 + a_2)^2$ and $I_{\min} = (a_1 - a_2)^2$,where $a_1$ and $a_2$ are the amplitudes of the two waves.Therefore,$\frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \frac{36}{1}$.Taking the square root on both sides,we get $\frac{a_1 + a_2}{a_1 - a_2} = \frac{6}{1}$.By cross-multiplying,$a_1 + a_2 = 6(a_1 - a_2) = 6a_1 - 6a_2$.Rearranging the terms,$5a_1 = 7a_2$.Thus,the ratio of the amplitudes is $\frac{a_1}{a_2} = \frac{7}{5}$.

The ratio of maximum and minimum intensities in an interference pattern is $36: 1$. The ratio of the amplitude of the two interfering waves will be

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