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(A) Given the intensity ratio $\frac{I_1}{I_2} = n$, so $I_1 = nI_2$.$I_{	ext{Max}} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{nI_2} + \sqrt{I_2})^2 = (\

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$n=1$

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(A) Given the intensity ratio $\frac{I_1}{I_2} = n$, so $I_1 = nI_2$.$I_{	ext{Max}} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{nI_2} + \sqrt{I_2})^2 = (\sqrt{n} + 1)^2 I_2$.$I_{	ext{Min}} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{nI_2} - \sqrt{I_2})^2 = (\sqrt{n} - 1)^2 I_2$.Now, the ratio is:$\frac{I_{	ext{Max}} - I_{	ext{Min}}}{I_{	ext{Max}} + I_{	ext{Min}}} = \frac{(\sqrt{n} + 1)^2 I_2 - (\sqrt{n} - 1)^2 I_2}{(\sqrt{n} + 1)^2 I_2 + (\sqrt{n} - 1)^2 I_2} = \frac{(\sqrt{n} + 1)^2 - (\sqrt{n} - 1)^2}{(\sqrt{n} + 1)^2 + (\sqrt{n} - 1)^2}$.Expanding the terms:$= \frac{(n + 1 + 2\sqrt{n}) - (n + 1 - 2\sqrt{n})}{(n + 1 + 2\sqrt{n}) + (n + 1 - 2\sqrt{n})} = \frac{4\sqrt{n}}{2(n + 1)} = \frac{2\sqrt{n}}{n + 1}$.Let $f(n) = \frac{2\sqrt{n}}{n + 1}$. To find the maximum, we differentiate with respect to $n$ or observe that for $n=1$, $f(1) = \frac{2(1)}{1+1} = 1$, which is the maximum possible value for this expression since $(\sqrt{n}-1)^2 \ge 0$ implies $n+1 \ge 2\sqrt{n}$.

An interference pattern is obtained with two coherent sources of intensity ratio $n:1$. The ratio $\frac{I_{\text{Max}}-I_{\text{Min}}}{I_{\text{Max}}+I_{\text{Min}}}$ will be maximum if

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