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(C) Given the ratio of amplitudes, $\frac{a_1}{a_2} = \frac{3}{2}$.Let $a_1 = 3k$ and $a_2 = 2k$, where $k$ is a constant.The intensity $I$ is proport

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$25: 1$

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(C) Given the ratio of amplitudes, $\frac{a_1}{a_2} = \frac{3}{2}$.Let $a_1 = 3k$ and $a_2 = 2k$, where $k$ is a constant.The intensity $I$ is proportional to the square of the amplitude, $I \propto a^2$.The ratio of maximum intensity to minimum intensity is given by the formula:$\frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2}$.Substituting the values:$\frac{I_{\max}}{I_{\min}} = \frac{(3k + 2k)^2}{(3k - 2k)^2} = \frac{(5k)^2}{(k)^2} = \frac{25k^2}{k^2} = \frac{25}{1}$.Thus, the ratio of maximum and minimum intensity is $25: 1$.

Light waves producing interference have their amplitudes in the ratio $3: 2$. The intensity ratio of maximum and minimum of interference fringes is

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