In YDSE setup,light of wavelength 640 , nm is | vedclass

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Question

(A) The fringe width is given by $\beta = \frac{\lambda D}{d} = \frac{640 	imes 10^{-9} 	imes 1}{0.8 	imes 10^{-3}} = 800 	imes 10^{-6} \, m = 0.8

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Intensity at $y = 0.4 \, mm$ is zero

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(A) The fringe width is given by $\beta = \frac{\lambda D}{d} = \frac{640 	imes 10^{-9} 	imes 1}{0.8 	imes 10^{-3}} = 800 	imes 10^{-6} \, m = 0.8 \, mm$.The path difference at position $y$ is $\Delta x = \frac{yd}{D}$.For $y = 0.4 \, mm$,$\Delta x = \frac{0.4 	imes 10^{-3} 	imes 0.8 	imes 10^{-3}}{1} = 0.32 	imes 10^{-6} \, m$.The phase difference is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{640 	imes 10^{-9}} 	imes 0.32 	imes 10^{-6} = \frac{2\pi}{640} 	imes 320 = \pi$.The intensity at any point is $I = I_0 \cos^2(\phi / 2)$.Substituting $\phi = \pi$,we get $I = I_0 \cos^2(\pi / 2) = I_0 	imes 0 = 0$.

In $YDSE$ setup,light of wavelength $640 \, nm$ is used with $d = 0.8 \, mm$ and $D = 1 \, m$. If intensity at central maximum is $I_0$ and its position is $y = 0$,then:

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