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(D) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2

Vedclass · Accepted Answer

$2$

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(D) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)$.Given $I_1 = I$ and $I_2 = 4I$.At point $A$,$\phi_A = \pi/2$. Thus,$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi/2) = 5I + 4I(0) = 5I$.At point $B$,$\phi_B = \pi/3$. Thus,$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi/3) = 5I + 4I(1/2) = 5I + 2I = 7I$.The difference between the resultant intensities is $I_B - I_A = 7I - 5I = 2I$.Comparing this with $xI$,we get $x = 2$.

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