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(C) Let the amplitudes of the two sources be $A_1 = 3x$ and $A_2 = 5x$.The maximum amplitude is $A_{\max} = A_1 + A_2 = 3x + 5x = 8x$.The minimum ampl

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$16 : 1$

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(C) Let the amplitudes of the two sources be $A_1 = 3x$ and $A_2 = 5x$.The maximum amplitude is $A_{\max} = A_1 + A_2 = 3x + 5x = 8x$.The minimum amplitude is $A_{\min} = |A_1 - A_2| = |3x - 5x| = 2x$.The ratio of maximum intensity to minimum intensity is given by the square of the ratio of maximum amplitude to minimum amplitude:$\frac{I_{\max}}{I_{\min}} = \left( \frac{A_{\max}}{A_{\min}} \right)^2 = \left( \frac{8x}{2x} \right)^2 = \left( \frac{4}{1} \right)^2 = \frac{16}{1}$.Thus,the ratio is $16 : 1$.

If the ratio of the amplitudes of two interfering sources is $3 : 5$,then the ratio of the intensity of maxima and minima in the interference pattern will be

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