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

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(A) The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$.Substituting the values: $\beta = \frac{640 	imes 10^{-9} \, m 	imes 1 \, m}{

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

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

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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