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(D) Let the intensity of each source be $I_0$. The resultant intensity is given by $I = I_0 + I_0 + 2I_0 \cos(\phi) = 2I_0(1 + \cos(\phi)) = 4I_0 \cos

Vedclass · Accepted Answer

$2 : 1 : 0$

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(D) Let the intensity of each source be $I_0$. The resultant intensity is given by $I = I_0 + I_0 + 2I_0 \cos(\phi) = 2I_0(1 + \cos(\phi)) = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the total phase difference.Given: Wavelength $\lambda = 20 \, m$,separation $d = 5.0 \, m$,and initial phase difference $\Delta\phi_{initial} = 90^{\circ} = \pi/2$ (with $P$ ahead of $Q$).Path difference $\Delta x = d \sin(	heta)$. Phase difference due to path is $\Delta\phi_{path} = (2\pi/\lambda) \Delta x$.At point $A$ (along the line $PQ$ towards $Q$): $	heta = 90^{\circ}$,$\Delta x = d = 5 \, m$. $\Delta\phi_{path} = (2\pi/20) 	imes 5 = \pi/2$. Since $P$ is ahead,the wave from $P$ travels further,so $\phi_A = \Delta\phi_{initial} - \Delta\phi_{path} = \pi/2 - \pi/2 = 0$. Intensity $I_A = 4I_0 \cos^2(0) = 4I_0$.At point $B$ (perpendicular bisector): $	heta = 0^{\circ}$,$\Delta x = 0$. $\phi_B = \Delta\phi_{initial} = \pi/2$. Intensity $I_B = 4I_0 \cos^2(\pi/4) = 4I_0(1/2) = 2I_0$.At point $C$ (along the line $PQ$ towards $P$): $	heta = -90^{\circ}$,$\Delta x = -d = -5 \, m$. $\Delta\phi_{path} = -\pi/2$. $\phi_C = \Delta\phi_{initial} - (\Delta\phi_{path}) = \pi/2 - (-\pi/2) = \pi$. Intensity $I_C = 4I_0 \cos^2(\pi/2) = 0$.Thus,the ratio $I_A : I_B : I_C = 4I_0 : 2I_0 : 0 = 2 : 1 : 0$.

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