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(C) The maximum intensity $I_{\max}$ for the superposition of two coherent beams is given by the formula:$I_{\max} = (\sqrt{I_{1}} + \sqrt{I_{2}})^2$G

Vedclass · Accepted Answer

$9I$ and $I$

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(C) The maximum intensity $I_{\max}$ for the superposition of two coherent beams is given by the formula:$I_{\max} = (\sqrt{I_{1}} + \sqrt{I_{2}})^2$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 the formula:$I_{\min} = (\sqrt{I_{1}} - \sqrt{I_{2}})^2$Substituting the values:$I_{\min} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$Thus,the maximum and minimum intensities are $9I$ and $I$ respectively.

Two coherent monochromatic light beams of intensities $I$ and $4I$ are superposed. The maximum and minimum possible intensities in the resulting beam are:

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