In the figure below,P and Q are two equally i | vedclass

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(D) Let the intensity of each source be $I$. The resultant intensity is given by $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) = 2I + 2I \c

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$2:1:0$

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(D) Let the intensity of each source be $I$. The resultant intensity is given by $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) = 2I + 2I \cos(\Delta \phi) = 4I \cos^2(\Delta \phi / 2)$,where $\Delta \phi$ is the total phase difference.$1$. For point $A$: The path difference $\Delta x = x_P - x_Q = 5\, m$. The phase difference due to path is $\Delta \phi_{path} = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{20} 	imes 5 = \frac{\pi}{2}$. Since $P$ is ahead of $Q$ by $90^{\circ}$ $(\pi/2)$,the wave from $Q$ travels a shorter distance to $A$. The total phase difference is $\Delta \phi_A = \Delta \phi_{source} - \Delta \phi_{path} = \frac{\pi}{2} - \frac{\pi}{2} = 0$. Thus,$I_A = 4I \cos^2(0) = 4I$.$2$. For point $B$: The path difference $\Delta x = 0$ as $B$ is on the perpendicular bisector. The total phase difference is $\Delta \phi_B = \Delta \phi_{source} = \frac{\pi}{2}$. Thus,$I_B = 4I \cos^2(\pi/4) = 4I(1/2) = 2I$.$3$. For point $C$: The path difference $\Delta x = x_Q - x_P = 5\, m$. The phase difference due to path is $\Delta \phi_{path} = \frac{\pi}{2}$. The total phase difference is $\Delta \phi_C = \Delta \phi_{source} + \Delta \phi_{path} = \frac{\pi}{2} + \frac{\pi}{2} = \pi$. Thus,$I_C = 4I \cos^2(\pi/2) = 0$.Therefore,the ratio $I_A : I_B : I_C = 4I : 2I : 0 = 2 : 1 : 0$.

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