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(C) Let the intensities of the two coherent sources be $I_1$ and $I_2$. Given the ratio $I_1/I_2 = \beta/1$,we have $I_1 = \beta I$ and $I_2 = I$.The

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

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(C) Let the intensities of the two coherent sources be $I_1$ and $I_2$. Given the ratio $I_1/I_2 = \beta/1$,we have $I_1 = \beta I$ and $I_2 = I$.The fringe visibility (or contrast) $V$ is defined as the ratio of the difference between maximum and minimum intensity to their sum:$V = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$We know that $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.Substituting these into the formula:$V = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2} = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$Substituting $I_1 = \beta I$ and $I_2 = I$:$V = \frac{2\sqrt{(\beta I)(I)}}{\beta I + I} = \frac{2\sqrt{\beta} I}{I(\beta + 1)} = \frac{2\sqrt{\beta}}{1 + \beta}$

Two coherent sources with intensity ratio $\beta : 1$ produce interference. Fringe visibility will be

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