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(C) The intensity of the resultant wave in interference is given by $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.For maximum intensity, $\cos \ph

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$9I$ and $I$

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(C) The intensity of the resultant wave in interference is given by $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.For maximum intensity, $\cos \phi = 1$, so $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$.Given $I_1 = I$ and $I_2 = 4I$, we have $I_{max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.For minimum intensity, $\cos \phi = -1$, so $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.Thus, $I_{min} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$.Therefore, 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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