Two periodic waves of intensities I_1 and I_2 | vedclass

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(B) The intensity $I$ of a wave is proportional to the square of its amplitude $A$,i.e.,$I = kA^2$,where $k$ is a constant. Thus,$A = \sqrt{I/k}$.When

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$2(I_1 + I_2)$

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(B) The intensity $I$ of a wave is proportional to the square of its amplitude $A$,i.e.,$I = kA^2$,where $k$ is a constant. Thus,$A = \sqrt{I/k}$.When two waves superpose,the maximum and minimum amplitudes are given by $A_{\max} = A_1 + A_2$ and $A_{\min} = |A_1 - A_2|$.The maximum intensity is $I_{\max} = k(A_1 + A_2)^2 = k(A_1^2 + A_2^2 + 2A_1A_2) = I_1 + I_2 + 2\sqrt{I_1I_2}$.The minimum intensity is $I_{\min} = k(A_1 - A_2)^2 = k(A_1^2 + A_2^2 - 2A_1A_2) = I_1 + I_2 - 2\sqrt{I_1I_2}$.Adding the maximum and minimum intensities:$I_{\max} + I_{\min} = (I_1 + I_2 + 2\sqrt{I_1I_2}) + (I_1 + I_2 - 2\sqrt{I_1I_2}) = 2(I_1 + I_2)$.

Two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

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