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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_

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$19$

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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 $R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.Given $A_1 = 3A$,$A_2 = 2A$,and $\phi = 60^{\circ}$.Substituting these values: $R = \sqrt{(3A)^2 + (2A)^2 + 2(3A)(2A) \cos 60^{\circ}}$.Since $\cos 60^{\circ} = 0.5$,we have $R = \sqrt{9A^2 + 4A^2 + 12A^2(0.5)} = \sqrt{9A^2 + 4A^2 + 6A^2} = \sqrt{19A^2} = A\sqrt{19}$.Intensity $I$ is proportional to the square of the resultant amplitude: $I \propto R^2$.Therefore,$I \propto (A\sqrt{19})^2 = 19A^2$.

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

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