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(A) Let the intensity of each individual source be $I_0$. The maximum intensity is $I_{max} = 4I_0$.Given that the intensity at a point is $75 \%$ of

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$1.0$

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(A) Let the intensity of each individual source be $I_0$. The maximum intensity is $I_{max} = 4I_0$.Given that the intensity at a point is $75 \%$ of the maximum intensity,we have $I = 0.75 	imes 4I_0 = 3I_0$.Using the formula for intensity in interference,$I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference:$3I_0 = 4I_0 \cos^2(\phi/2) \implies \cos^2(\phi/2) = 3/4 \implies \cos(\phi/2) = \sqrt{3}/2$.Thus,$\phi/2 = \pi/6$,which gives the phase difference $\phi = \pi/3$.The relation between phase difference $\phi$ and path difference $\Delta x$ is $\phi = (2\pi/\lambda) \Delta x$. Therefore,$\Delta x = (\lambda/2\pi) 	imes \phi = (\lambda/2\pi) 	imes (\pi/3) = \lambda/6$.In Young's double slit experiment,the path difference at a distance $y$ from the central fringe is $\Delta x = yd/D$.Equating the two expressions for path difference: $yd/D = \lambda/6$.Solving for $y$: $y = (\lambda D) / (6d)$.Substituting the given values: $\lambda = 600 	imes 10^{-9} \,m$,$D = 1 \,m$,$d = 0.1 	imes 10^{-3} \,m$.$y = (600 	imes 10^{-9} 	imes 1) / (6 	imes 0.1 	imes 10^{-3}) = (600 	imes 10^{-9}) / (0.6 	imes 10^{-3}) = 1000 	imes 10^{-6} \,m = 1 	imes 10^{-3} \,m = 1 \,mm$.

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