Two sources of light are 0.6 , mm apart and t | vedclass

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(D) Given: Slit separation $d = 0.6 \, mm = 0.6 	imes 10^{-3} \, m$, distance to screen $D = 1.2 \, m$, wavelength $\lambda = 6000 \, \mathring{A} =

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$5 \pi \, 	ext{radian}$

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(D) Given: Slit separation $d = 0.6 \, mm = 0.6 	imes 10^{-3} \, m$, distance to screen $D = 1.2 \, m$, wavelength $\lambda = 6000 \, \mathring{A} = 6000 	imes 10^{-10} \, m$, and position on screen $y = 3 \, mm = 3 	imes 10^{-3} \, m$.The path difference $\Delta x$ at a point $y$ is given by $\Delta x = \frac{d \cdot y}{D}$.Substituting the values: $\Delta x = \frac{(0.6 	imes 10^{-3} \, m) 	imes (3 	imes 10^{-3} \, m)}{1.2 \, m} = \frac{1.8 	imes 10^{-6}}{1.2} \, m = 1.5 	imes 10^{-6} \, m$.The phase difference $\Delta \phi$ is related to path difference by $\Delta \phi = \frac{2 \pi}{\lambda} \cdot \Delta x$.Substituting the values: $\Delta \phi = \frac{2 \pi}{6000 	imes 10^{-10}} 	imes 1.5 	imes 10^{-6} = \frac{2 \pi 	imes 1.5 	imes 10^{-6}}{6 	imes 10^{-7}} = \frac{3 \pi 	imes 10^{-6}}{6 	imes 10^{-7}} = 0.5 \pi 	imes 10 = 5 \pi \, 	ext{radian}$.

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