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(D) Given the ratio of intensities $I_1 : I_2 = 9 : 4$.Let $I_1 = 9k$ and $I_2 = 4k$.The amplitudes are proportional to the square root of intensities

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$25: 1$

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(D) Given the ratio of intensities $I_1 : I_2 = 9 : 4$.Let $I_1 = 9k$ and $I_2 = 4k$.The amplitudes are proportional to the square root of intensities,so $a_1 = \sqrt{I_1} = 3\sqrt{k}$ and $a_2 = \sqrt{I_2} = 2\sqrt{k}$.The ratio of maximum intensity to minimum intensity is given by the formula:$\frac{I_{max}}{I_{min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.Substituting the values:$\frac{I_{max}}{I_{min}} = \left( \frac{3\sqrt{k} + 2\sqrt{k}}{3\sqrt{k} - 2\sqrt{k}} \right)^2 = \left( \frac{5\sqrt{k}}{1\sqrt{k}} \right)^2 = 5^2 = 25$.Thus,the ratio is $25 : 1$.

Two light beams of intensities in the ratio of $9: 4$ are allowed to interfere. The ratio of the intensity of maxima and minima will be

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