In Young's double slit experiment the wavelength of light used is $6000 Å$, the screen is 40 cm from the slits and the fringe width is 0.012 cm , the distance between two slits is

The fringe width in Young's double-slit experiment is determined by the relation $β = \frac{\lambda D}{d}$, where $λ$ is the wavelength, $D$ is the screen distance, and $d$ is the slit separation.

Given values are wavelength $λ = 6000 \text{\AA} = 6 \times 10^{-7} \text{m}$, screen distance $D = 40 \text{cm} = 0.40 \text{m}$, and fringe width $β = 0.012 \text{cm} = 1.2 \times 10^{-4} \text{m}$.

Solving for the slit separation yields $d = \frac{\lambda D}{β} = \frac{(6 \times 10^{-7}) \times 0.40}{1.2 \times 10^{-4}} = 2 \times 10^{-3} \text{m}$.

Converting to centimeters gives $d = 0.2 \text{cm}$, which corresponds to option D.