In text{YDSE},S_1 and S_2 have the same inten | vedclass

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Question

(A) The intensity at any point $P$ in $	ext{YDSE}$ is given by $I = 4 I_0 \cos^2\left(\frac{\phi}{2}ight)$,where $\phi$ is the phase difference.The

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$(A)-S, (B)-Q, (C)-P, (D)-R$

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(A) The intensity at any point $P$ in $	ext{YDSE}$ is given by $I = 4 I_0 \cos^2\left(\frac{\phi}{2}ight)$,where $\phi$ is the phase difference.The path difference $\Delta x = \frac{xd}{D}$.The phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \cdot \frac{xd}{D}$.Thus,$I = 4 I_0 \cos^2\left(\frac{\pi x d}{\lambda D}ight)$.$(A) x = \frac{D \lambda}{d} \implies I = 4 I_0 \cos^2(\pi) = 4 I_0 \cdot (-1)^2 = 4 I_0$ (Matches $S$).$(B) x = \frac{D \lambda}{4d} \implies I = 4 I_0 \cos^2\left(\frac{\pi}{4}ight) = 4 I_0 \cdot \left(\frac{1}{\sqrt{2}}ight)^2 = 2 I_0$ (Matches $Q$).$(C) x = \frac{D \lambda}{3d} \implies I = 4 I_0 \cos^2\left(\frac{\pi}{3}ight) = 4 I_0 \cdot \left(\frac{1}{2}ight)^2 = I_0$ (Matches $P$).$(D) x = \frac{D \lambda}{6d} \implies I = 4 I_0 \cos^2\left(\frac{\pi}{6}ight) = 4 I_0 \cdot \left(\frac{\sqrt{3}}{2}ight)^2 = 3 I_0$ (Matches $R$).Therefore,the correct matching is $(A)-S, (B)-Q, (C)-P, (D)-R$.

Column-$I$	Column-$II$
$(A) x = \frac{D \lambda}{d}$	$(P) I_0$
$(B) x = \frac{D \lambda}{4d}$	$(Q) 2 I_0$
$(C) x = \frac{D \lambda}{3d}$	$(R) 3 I_0$
$(D) x = \frac{D \lambda}{6d}$	$(S) 4 I_0$

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