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(D) Let the intensity of light on the screen due to each slit be $I_0$.The maximum intensity at the centre of the screen is $I_{max} = 4I_0$.The inten

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$125$

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(D) Let the intensity of light on the screen due to each slit be $I_0$.The maximum intensity at the centre of the screen is $I_{max} = 4I_0$.The intensity $I$ at a point where the phase difference is $\phi$ is given by $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 4I_0 \cos^2(\phi/2)$.We are given that the intensity becomes half of the maximum intensity,so $I = \frac{I_{max}}{2} = 2I_0$.Thus,$2I_0 = 4I_0 \cos^2(\phi/2) \implies \cos^2(\phi/2) = 1/2 \implies \cos(\phi/2) = 1/\sqrt{2}$.This gives $\phi/2 = \pi/4$,so the phase difference $\phi = \pi/2$.The phase difference is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.So,$\frac{2\pi}{\lambda} \Delta x = \frac{\pi}{2} \implies \Delta x = \frac{\lambda}{4}$.For small angles,the path difference $\Delta x = d \sin 	heta \approx d 	an 	heta = d(y/D)$.Equating the two,we get $d(y/D) = \lambda/4 \implies y = \frac{\lambda D}{4d}$.Given $\lambda = 5000 \ \mathring{A} = 5 	imes 10^{-7} \ m$,$d = 1.0 \ mm = 10^{-3} \ m$,and $D = 1.0 \ m$.$y = \frac{5 	imes 10^{-7} 	imes 1}{4 	imes 10^{-3}} = 1.25 	imes 10^{-4} \ m = 125 	imes 10^{-6} \ m$.Therefore,the distance is $125$.

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