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(C) Given the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = 2x$. We know that $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\m

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$\frac{2 \sqrt{2 x}}{2 x+1}$

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(C) Given the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = 2x$. We know that $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$. The expression to evaluate is $V = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$. Substituting the expressions for $I_{\max}$ and $I_{\min}$: $V = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2}$. Using the algebraic identities $(a+b)^2 - (a-b)^2 = 4ab$ and $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$: $V = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$. Dividing the numerator and denominator by $I_2$: $V = \frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1}$. Substituting $\frac{I_1}{I_2} = 2x$: $V = \frac{2\sqrt{2x}}{2x + 1}$.

Two coherent light sources having intensity in the ratio $2x$ produce an interference pattern. Then the value of $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$ will be

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