In a standard YDSE setup,two coherent sources | vedclass

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(B) Let the intensities of the two coherent sources be $I_1$ and $I_2$. Given the ratio $\beta = \frac{I_1}{I_2}$.Maximum intensity $I_{\max} = (\sqrt

Vedclass · Accepted Answer

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

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(B) Let the intensities of the two coherent sources be $I_1$ and $I_2$. Given the ratio $\beta = \frac{I_1}{I_2}$.Maximum intensity $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$.Minimum intensity $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.We need to calculate $X = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$.Substituting the expressions:$I_{\max} - I_{\min} = (\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2 = 4\sqrt{I_1 I_2}$.$I_{\max} + I_{\min} = (\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2 = 2(I_1 + I_2)$.Thus,$X = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$.Dividing numerator and denominator by $I_2$:$X = \frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1} = \frac{2\sqrt{\beta}}{\beta + 1}$.

In a standard $YDSE$ setup,two coherent sources of light of intensity ratio $\beta$ produce an interference pattern. The value of $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$ is equal to (where $I_{\max}$ and $I_{\min}$ are the maximum and minimum intensities of the resultant wave).

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