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(C) Given,the ratio of amplitudes of two interfering waves is $4: 3$.Let $A_1$ and $A_2$ be the amplitudes of the two waves,so $\frac{A_1}{A_2} = \fra

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$49: 1$

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(C) Given,the ratio of amplitudes of two interfering waves is $4: 3$.Let $A_1$ and $A_2$ be the amplitudes of the two waves,so $\frac{A_1}{A_2} = \frac{4}{3}$.We know that the intensity $I$ of a wave is proportional to the square of its amplitude,$I \propto A^2$.The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$.Dividing the numerator and denominator inside the bracket by $A_2$,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{\frac{A_1}{A_2} + 1}{\frac{A_1}{A_2} - 1} \right)^2$.Substituting the value $\frac{A_1}{A_2} = \frac{4}{3}$,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{\frac{4}{3} + 1}{\frac{4}{3} - 1} \right)^2 = \left( \frac{\frac{7}{3}}{\frac{1}{3}} \right)^2 = (7)^2 = \frac{49}{1}$.Thus,the ratio of maximum and minimum intensity is $49: 1$.

If the ratio of amplitudes of two interfering waves is $4: 3$,then the ratio of maximum and minimum intensity is

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