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(C) The intensity at any point in a Young's double-slit experiment is given by $I = 4I_0 \cos^2(\frac{\Delta \phi}{2})$,where $I_0$ is the intensity o

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$\frac{\beta}{4}$

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(C) The intensity at any point in a Young's double-slit experiment is given by $I = 4I_0 \cos^2(\frac{\Delta \phi}{2})$,where $I_0$ is the intensity of each individual slit and $\Delta \phi$ is the phase difference.At the central maximum,the intensity is $I_{max} = 4I_0$.We want to find the position $y$ where the intensity $I$ falls to half of the maximum intensity,i.e.,$I = \frac{I_{max}}{2} = 2I_0$.Substituting this into the intensity formula: $2I_0 = 4I_0 \cos^2(\frac{\Delta \phi}{2}) \implies \cos^2(\frac{\Delta \phi}{2}) = \frac{1}{2} \implies \cos(\frac{\Delta \phi}{2}) = \frac{1}{\sqrt{2}}$.This implies $\frac{\Delta \phi}{2} = \frac{\pi}{4}$,so the phase difference is $\Delta \phi = \frac{\pi}{2}$.Since the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$,we have $\frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x$,which gives the path difference $\Delta x = \frac{\lambda}{4}$.For small angles,the path difference is $\Delta x = \frac{dy}{D}$.Equating the two: $\frac{dy}{D} = \frac{\lambda}{4} \implies y = \frac{\lambda D}{4d}$.Since the fringe width is $\beta = \frac{\lambda D}{d}$,we substitute this to get $y = \frac{\beta}{4}$.

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