Two coherent monochromatic light beams of int | vedclass

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(C) The maximum intensity in an interference pattern is given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$.Substituting the given values,$I_{\max} = (\

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$24$

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(C) The maximum intensity in an interference pattern is given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$.Substituting the given values,$I_{\max} = (\sqrt{4I} + \sqrt{9I})^2 = (2\sqrt{I} + 3\sqrt{I})^2 = (5\sqrt{I})^2 = 25I$.The minimum intensity in an interference pattern is given by $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.Substituting the given values,$I_{\min} = (\sqrt{4I} - \sqrt{9I})^2 = (2\sqrt{I} - 3\sqrt{I})^2 = (-\sqrt{I})^2 = I$.The difference between the maximum and minimum intensities is $I_{\max} - I_{\min} = 25I - I = 24I$.Comparing this with $xI$,we get $x = 24$.

Two coherent monochromatic light beams of intensities $4 I$ and $9I$ are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is $xI$. The value of $x$ is . . . . . . .

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