Question

of Textbook: In a two-slit interference experiment of the Young type, the aperture-to-screen distance is m and the wavelength is 600 nm. If it is desired to have a fringe 2spacing of mm, what is the required slit separation? 1

Answer 1

Answer:

The required slit separation is 0.6mm

Explanation:

The wavelength:  ∧=600nm, ∧=$600*10^{-9}$m

The fringe spacing: Δy=1mm, Δ$y=1*10^{-3}$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

$d=\frac{1*600*10^{-9} }{1*10^{-3} }$

$d=6*10^{-4}m \\d=0.6mm$