Intensities of the two waves of light are I a | vedclass

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(B) The resultant intensity $I_{res}$ of two interfering waves is given by the formula: $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.For maximum

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$9$

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(B) The resultant intensity $I_{res}$ of two interfering waves is given by the formula: $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.For maximum intensity,the phase difference $\phi$ must be $0$,so $\cos \phi = 1$.Thus,the maximum intensity is given by: $I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2$.Given $I_1 = I$ and $I_2 = 4I$.Substituting these values: $I_{max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.

Intensities of the two waves of light are $I$ and $4I$. The maximum intensity of the resultant wave after superposition is (in $I$)

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