Two coherent sources of intensity ratio x^2 i | vedclass

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Question

(C) Let the intensities of the two coherent sources be $I_1$ and $I_2$. Given the ratio $\frac{I_1}{I_2} = x^2$,we can write $I_1 = x^2 I_2$ or $\frac

Vedclass · Accepted Answer

$\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} = \frac{2x}{1 + x^2}$

Vedclass · Answer

(C) Let the intensities of the two coherent sources be $I_1$ and $I_2$. Given the ratio $\frac{I_1}{I_2} = x^2$,we can write $I_1 = x^2 I_2$ or $\frac{\sqrt{I_1}}{\sqrt{I_2}} = x$.The maximum and minimum intensities are given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.We need to evaluate the ratio $R = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$.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,$R = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$.Dividing numerator and denominator by $I_2$:$R = \frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1} = \frac{2\sqrt{x^2}}{x^2 + 1} = \frac{2x}{1 + x^2}$.Therefore,the correct relation is $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} = \frac{2x}{1 + x^2}$.

Two coherent sources of intensity ratio $x^2$ interfere. In the interference pattern,which of the following relations is correct?

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