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(A) Let the intensities of the two sources be $I_1$ and $I_2$ such that $\frac{I_1}{I_2} = \alpha$. Since $I \propto A^2$,we have $\frac{A_1}{A_2} = \

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

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(A) Let the intensities of the two sources be $I_1$ and $I_2$ such that $\frac{I_1}{I_2} = \alpha$. Since $I \propto A^2$,we have $\frac{A_1}{A_2} = \sqrt{\alpha}$.The maximum and minimum intensities in an interference pattern are given by $I_{max} = (A_1 + A_2)^2$ and $I_{min} = (A_1 - A_2)^2$.We need to calculate the value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$.Substituting the expressions for $I_{max}$ and $I_{min}$:$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{(A_1 + A_2)^2 - (A_1 - A_2)^2}{(A_1 + A_2)^2 + (A_1 - A_2)^2}$Expanding the squares:$= \frac{(A_1^2 + A_2^2 + 2A_1A_2) - (A_1^2 + A_2^2 - 2A_1A_2)}{(A_1^2 + A_2^2 + 2A_1A_2) + (A_1^2 + A_2^2 - 2A_1A_2)}$$= \frac{4A_1A_2}{2(A_1^2 + A_2^2)} = \frac{2A_1A_2}{A_1^2 + A_2^2}$Dividing the numerator and denominator by $A_2^2$:$= \frac{2(A_1/A_2)}{(A_1/A_2)^2 + 1}$Since $\frac{A_1}{A_2} = \sqrt{\alpha}$,we substitute this into the expression:$= \frac{2\sqrt{\alpha}}{(\sqrt{\alpha})^2 + 1} = \frac{2\sqrt{\alpha}}{\alpha + 1}$.

Two coherent sources of intensity ratio $\alpha$ interfere. The value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is

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