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(D) Let the amplitudes of the two waves be $A_1 = 5x$ and $A_2 = x$.The maximum intensity $I_{max}$ is proportional to the square of the sum of amplit

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$9: 4$

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(D) Let the amplitudes of the two waves be $A_1 = 5x$ and $A_2 = x$.The maximum intensity $I_{max}$ is proportional to the square of the sum of amplitudes: $I_{max} \propto (A_1 + A_2)^2 = (5x + x)^2 = (6x)^2 = 36x^2$.The minimum intensity $I_{min}$ is proportional to the square of the difference of amplitudes: $I_{min} \propto (A_1 - A_2)^2 = (5x - x)^2 = (4x)^2 = 16x^2$.The ratio of maximum to minimum intensity is $\frac{I_{max}}{I_{min}} = \frac{36x^2}{16x^2} = \frac{36}{16} = \frac{9}{4}$.

Two waves having amplitudes in the ratio $5:1$ produce interference. The ratio of the maximum to minimum intensity is

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