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(D) The resultant amplitude $R$ of two interfering waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $R = \sqrt{A_1^2 + A_

Vedclass · Accepted Answer

$19$

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(D) The resultant amplitude $R$ of two interfering waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}$.Given $A_1 = 3 A$,$A_2 = 2 A$,and $\phi = 60^{\circ}$.The intensity $I$ is proportional to the square of the resultant amplitude,$I \propto R^2$.$R^2 = (3 A)^2 + (2 A)^2 + 2(3 A)(2 A) \cos(60^{\circ})$.$R^2 = 9 A^2 + 4 A^2 + 12 A^2 	imes (0.5)$.$R^2 = 13 A^2 + 6 A^2 = 19 A^2$.Therefore,the intensity is proportional to $19 A^2$.

Two monochromatic light waves of amplitudes $3 A$ and $2 A$ interfering at a point have a phase difference of $60^{\circ}$. The intensity at that point will be proportional to (in $A^{2}$)

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