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(D) The intensity $I$ of a light wave is proportional to the square of its amplitude $A$, i.e.,$I \propto A^2$.Given the ratio of amplitudes $A_1 : A_

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$4 : 1$

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(D) The intensity $I$ of a light wave is proportional to the square of its amplitude $A$, i.e.,$I \propto A^2$.Given the ratio of amplitudes $A_1 : A_2 = 3 : 1$, let $A_1 = 3k$ and $A_2 = k$.The maximum intensity $I_{max}$ is given by $(A_1 + A_2)^2 = (3k + k)^2 = (4k)^2 = 16k^2$.The minimum intensity $I_{min}$ is given by $(A_1 - A_2)^2 = (3k - k)^2 = (2k)^2 = 4k^2$.The ratio of maximum to minimum intensity is $\frac{I_{max}}{I_{min}} = \frac{16k^2}{4k^2} = \frac{16}{4} = 4 : 1$.

Two light waves with amplitudes in the ratio $3 : 1$ produce interference. The ratio of the maximum to minimum intensity is

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