In Y.D.S.E., the slits are 0.2 , cm apart and | vedclass

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(A) The formula for fringe width in air is $\beta = \frac{\lambda D}{d}$.Given: Slit separation $d = 0.2 \, cm = 2 	imes 10^{-3} \, m$,wavelength $\l

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$0.225$

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(A) The formula for fringe width in air is $\beta = \frac{\lambda D}{d}$.Given: Slit separation $d = 0.2 \, cm = 2 	imes 10^{-3} \, m$,wavelength $\lambda = 600 \, nm = 600 	imes 10^{-9} \, m$,and screen distance $D = 1 \, m$.First,calculate the fringe width in air: $\beta = \frac{600 	imes 10^{-9} 	imes 1}{2 	imes 10^{-3}} = 300 	imes 10^{-6} \, m = 0.3 \, mm$.When the system is immersed in a medium of refractive index $\mu$,the wavelength changes to $\lambda' = \frac{\lambda}{\mu}$,so the new fringe width is $\beta' = \frac{\beta}{\mu}$.Given $\mu = \frac{4}{3}$,the new fringe width is $\beta' = \frac{0.3}{4/3} = 0.3 	imes \frac{3}{4} = 0.225 \, mm$.

In $Y.D.S.E.,$ the slits are $0.2 \, cm$ apart and are illuminated by yellow light $(\lambda = 600 \, nm).$ What would be the fringe width on a screen placed $1 \, m$ from the plane of the slits if the whole system is immersed in water of refractive index $\mu = \frac{4}{3}$ (in $, mm$)?

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