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(C) The intensity at any point in a Young's double-slit experiment is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase differen

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$1.25 	imes 10^{-4}\, m$

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(C) The intensity at any point in a Young's double-slit experiment is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.Given $I = \frac{I_{max}}{2}$,we have $\cos^2(\frac{\phi}{2}) = \frac{1}{2}$,which implies $\frac{\phi}{2} = \frac{\pi}{4}$,so $\phi = \frac{\pi}{2}$.The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$. Thus,$\frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x$,which gives $\Delta x = \frac{\lambda}{4}$.For small angles,the path difference $\Delta x = \frac{yd}{D}$.Equating the two,we get $\frac{yd}{D} = \frac{\lambda}{4}$,so $y = \frac{\lambda D}{4d}$.Substituting the given values: $\lambda = 5 	imes 10^{-7}\, m$,$D = 1\, m$,and $d = 10^{-3}\, m$.$y = \frac{5 	imes 10^{-7} 	imes 1}{4 	imes 10^{-3}} = 1.25 	imes 10^{-4}\, m$.

In Young's double-slit experiment,the value of $\lambda = 500\, nm$. The value of $d = 1\, mm$ and $D = 1\, m$. Then the minimum distance from the central maximum for which the intensity is half the maximum intensity will be:

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