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(D) The intensity at any point in a Young's double slit experiment is given by $I = I_{max} \cos^2\left(\frac{\Delta \phi}{2}\right)$,where $\Delta \p

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$200$

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(D) The intensity at any point in a Young's double slit experiment is given by $I = I_{max} \cos^2\left(\frac{\Delta \phi}{2}ight)$,where $\Delta \phi$ is the phase difference.Given that $I = \frac{I_{max}}{4}$,we have $\frac{I_{max}}{4} = I_{max} \cos^2\left(\frac{\Delta \phi}{2}ight)$.This simplifies to $\cos^2\left(\frac{\Delta \phi}{2}ight) = \frac{1}{4}$,which means $\cos\left(\frac{\Delta \phi}{2}ight) = \frac{1}{2}$.Thus,$\frac{\Delta \phi}{2} = \frac{\pi}{3}$,which gives the phase difference $\Delta \phi = \frac{2\pi}{3}$.The phase difference is related to the path difference $\Delta x$ by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$. Since $\Delta x = \frac{yd}{D}$,we have $\Delta \phi = \frac{2\pi}{\lambda} \left(\frac{yd}{D}ight)$.Equating the two expressions for $\Delta \phi$: $\frac{2\pi}{\lambda} \left(\frac{yd}{D}ight) = \frac{2\pi}{3}$.Solving for $y$: $y = \frac{\lambda D}{3d}$.Substituting the given values: $y = \frac{600 	imes 10^{-9} \ m 	imes 1.0 \ m}{3 	imes 1.0 	imes 10^{-3} \ m} = \frac{600 	imes 10^{-9}}{3 	imes 10^{-3}} \ m = 200 	imes 10^{-6} \ m$.Since $1 \ \mu m = 10^{-6} \ m$,the distance $y = 200 \ \mu m$.

In a Young's double slit experiment,the intensity at a point is $\left(\frac{1}{4}\right)^{\text{th}}$ of the maximum intensity. The minimum distance of the point from the central maximum is . . . . . . . . $\mu m$.
(Given: $\lambda = 600 \ nm, d = 1.0 \ mm, D = 1.0 \ m$)

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In a Young's double slit experiment,the intensity at a point is $\left(\frac{1}{4}\right)^{\text{th}}$ of the maximum intensity. The minimum distance of the point from the central maximum is . . . . . . . . $\mu m$.(Given: $\lambda = 600 \ nm, d = 1.0 \ mm, D = 1.0 \ m$)