In a YDSE apparatus,two identical slits are s | vedclass

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(B) The intensity at any point on the screen is given by $I = I_{\max} \cos^2(\phi/2)$.Given $I = 0.75 I_{\max}$,we have $0.75 = \cos^2(\phi/2)$,which

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$0.40$

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(B) The intensity at any point on the screen is given by $I = I_{\max} \cos^2(\phi/2)$.Given $I = 0.75 I_{\max}$,we have $0.75 = \cos^2(\phi/2)$,which implies $\cos(\phi/2) = \pm \sqrt{3}/2$.Thus,$\phi/2 = \pi/6$ or $5\pi/6$,leading to phase differences $\phi = \pi/3$ or $5\pi/3$.The path difference is $\Delta x = (\lambda/2\pi) \phi = yd/D$.For $\phi_1 = \pi/3$,$y_1 = (\lambda D / 2\pi d) 	imes (\pi/3) = \lambda D / 6d$.For $\phi_2 = 5\pi/3$,$y_2 = (\lambda D / 2\pi d) 	imes (5\pi/3) = 5\lambda D / 6d$.Substituting values: $\lambda = 6 	imes 10^{-7} \, m$,$D = 1 \, m$,$d = 10^{-3} \, m$.$y_1 = (6 	imes 10^{-7} 	imes 1) / (6 	imes 10^{-3}) = 10^{-4} \, m = 0.1 \, mm$.$y_2 = 5 	imes 0.1 \, mm = 0.5 \, mm$.The distance between the two points is $\Delta y = y_2 - y_1 = 0.5 - 0.1 = 0.4 \, mm$.

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