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(D) The fringe width $\beta$ in a Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is th

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$0.5$

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(D) The fringe width $\beta$ in a Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the separation between the slits.Given: $\lambda = 600 \, nm = 600 	imes 10^{-9} \, m$,$D = 1 \, m$,and $\beta = 1.2 \, mm = 1.2 	imes 10^{-3} \, m$.Rearranging the formula to solve for $d$: $d = \frac{\lambda D}{\beta}$.Substituting the values: $d = \frac{(600 	imes 10^{-9} \, m) 	imes (1 \, m)}{1.2 	imes 10^{-3} \, m}$.$d = \frac{600 	imes 10^{-9}}{1.2 	imes 10^{-3}} = \frac{600}{1.2} 	imes 10^{-6} = 500 	imes 10^{-6} \, m$.$d = 0.5 	imes 10^{-3} \, m = 0.5 \, mm$.

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

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