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(C) The maximum intensity $I_{	ext{max}}$ for two waves of equal intensity $I$ is given by $I_{	ext{max}} = (\sqrt{I} + \sqrt{I})^2 = 4I = 100$.Thus

Vedclass · Accepted Answer

$89$

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(C) The maximum intensity $I_{	ext{max}}$ for two waves of equal intensity $I$ is given by $I_{	ext{max}} = (\sqrt{I} + \sqrt{I})^2 = 4I = 100$.Thus,$I = 25$ units. The amplitude $a$ is proportional to $\sqrt{I}$,so $a = \sqrt{25} = 5$ units.If the intensity of one source is reduced by $20\%$,the new intensity $I'$ becomes $I' = I - 0.20I = 0.80I = 0.80 	imes 25 = 20$.The new amplitude $a'$ is $\sqrt{20} = 2\sqrt{5}$.The other source remains at amplitude $a = 5$.The new maximum intensity $I'_{	ext{max}}$ is $(a + a')^2 = (5 + \sqrt{20})^2 = (5 + 4.472)^2 \approx (9.472)^2 \approx 89.7$.However,using the standard interpretation where intensity is proportional to the square of amplitude,if $I_1 = 25$ and $I_2 = 20$,then $I_{	ext{max}} = (\sqrt{25} + \sqrt{20})^2 = (5 + 4.47)^2 \approx 89.7$. Given the options,$89$ is the closest integer value.

Two coherent waves of equal intensity produce a maximum intensity of $100$ units. If the intensity of one of the sources is reduced by $20\%$,then the new maximum intensity will be:

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