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(B) The intensity of a wave is proportional to the square of its amplitude,$I \propto a^2$,so $I = ka^2$. Let $I_1 = ka_1^2$ and $I_2 = ka_2^2$.The ma

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$2(I_1+I_2)$

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(B) The intensity of a wave is proportional to the square of its amplitude,$I \propto a^2$,so $I = ka^2$. Let $I_1 = ka_1^2$ and $I_2 = ka_2^2$.The maximum intensity during interference is given by $I_{\max} = k(a_1 + a_2)^2 = k(a_1^2 + a_2^2 + 2a_1a_2)$.The minimum intensity during interference is given by $I_{\min} = k(a_1 - a_2)^2 = k(a_1^2 + a_2^2 - 2a_1a_2)$.The sum of the maximum and minimum intensities is $I_{\max} + I_{\min} = k(a_1^2 + a_2^2 + 2a_1a_2) + k(a_1^2 + a_2^2 - 2a_1a_2)$.$I_{\max} + I_{\min} = k(2a_1^2 + 2a_2^2) = 2(ka_1^2 + ka_2^2)$.Substituting the intensities back,we get $I_{\max} + I_{\min} = 2(I_1 + I_2)$.

Two light waves of intensities $I_1$ and $I_2$ having the same frequency pass through the same medium at a time in the same direction and interfere. The sum of the minimum and maximum intensities is

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