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(D) Given the ratio of intensities $I_1 : I_2 = 1 : 9$. Let $I_1 = I$ and $I_2 = 9I$.The formula for the ratio of maximum to minimum intensity in an i

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$4$

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(D) Given the ratio of intensities $I_1 : I_2 = 1 : 9$. Let $I_1 = I$ and $I_2 = 9I$.The formula for the ratio of maximum to minimum intensity in an interference pattern is given by:$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$Substituting the values:$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I} + \sqrt{9I})^2}{(\sqrt{I} - \sqrt{9I})^2}$$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I} + 3\sqrt{I})^2}{(\sqrt{I} - 3\sqrt{I})^2}$$\frac{I_{\max}}{I_{\min}} = \frac{(4\sqrt{I})^2}{(-2\sqrt{I})^2} = \frac{16I}{4I} = \frac{4}{1}$Thus,the ratio is $4 : 1$.

Two monochromatic light beams have intensities in the ratio $1 : 9$. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is (in $: 1$)

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