Answer:
The required slit separation is 0.6mm
Explanation:
The wavelength: ∧=600nm, ∧=[tex]600*10^{-9}[/tex]m
The fringe spacing: Δy=1mm, Δ[tex]y=1*10^{-3}[/tex]m
The aperture-to-screen distance: L=1m
Now we can apply fringe width formula to find silt separation
Δ=L∧/d
cross multiply to find d
d=L*∧/Δ
substitute
[tex]d=\frac{1*600*10^{-9} }{1*10^{-3} }[/tex]
[tex]d=6*10^{-4}m \\d=0.6mm[/tex]
