Two slits separated by 0.5 ,mm are illuminate | vedclass

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(A) Given: Slit separation $d = 0.5 \,mm = 0.5 	imes 10^{-3} \,m$, wavelength $\lambda = 500 \,nm = 500 	imes 10^{-9} \,m$, distance to screen $D =

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$5 \pi$

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(A) Given: Slit separation $d = 0.5 \,mm = 0.5 	imes 10^{-3} \,m$, wavelength $\lambda = 500 \,nm = 500 	imes 10^{-9} \,m$, distance to screen $D = 120 \,cm = 1.2 \,m$, and position on screen $y = 3 \,mm = 3 	imes 10^{-3} \,m$. The path difference $\Delta x$ is given by $\Delta x = \frac{yd}{D}$. Substituting the values: $\Delta x = \frac{(3 	imes 10^{-3} \,m) 	imes (0.5 	imes 10^{-3} \,m)}{1.2 \,m} = \frac{1.5 	imes 10^{-6}}{1.2} \,m = 1.25 	imes 10^{-6} \,m$. The phase difference $\Delta \phi$ is related to the path difference by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$. Substituting the values: $\Delta \phi = \frac{2 \pi}{500 	imes 10^{-9} \,m} 	imes (1.25 	imes 10^{-6} \,m) = \frac{2.5 \pi 	imes 10^{-6}}{500 	imes 10^{-9}} = \frac{2.5 \pi}{0.5} = 5 \pi$ radians.

Two slits separated by $0.5 \,mm$ are illuminated by light of wavelength $500 \,nm$. The screen is at a distance of $120 \,cm$ from the slits. The phase difference between the interfering waves at a point $3 \,mm$ on the screen from the central bright fringe is ........... .

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