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(D) The maximum intensity $I_{\max}$ for two interfering beams with intensities $I_1$ and $I_2$ is given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$.G

Vedclass · Accepted Answer

$8$

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(D) The maximum intensity $I_{\max}$ for two interfering beams with intensities $I_1$ and $I_2$ is given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$.Given $I_1 = I$ and $I_2 = 4I$,we have $I_{\max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.The minimum intensity $I_{\min}$ is given by $I_{\min} = (\sqrt{I_2} - \sqrt{I_1})^2$.Substituting the values,$I_{\min} = (\sqrt{4I} - \sqrt{I})^2 = (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = I$.The difference between maximum and minimum intensity is $I_{\max} - I_{\min} = 9I - I = 8I$.Comparing this with $xI$,we get $x = 8$.

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

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