Question: In a Young's double slit experiment a laser of wavelength 600nm is used to obtain a pattern with a distance between the center bright fringe and the first bright fring of 0.5mm. A thin plate of glass (thickness of 100 micrometers and refractive index of 1.5) is then placed over one of the slits. What is the lateral displacement of the central fringe on the screen?
Answer:
We are given:
λ=600 nmw1=0.5 mmΔp=100 μmn1=1.5
In a Young's double slit experiment the spacing of the fringes at a distance z from the slits is given by
w=dzλ, n=0,1,2, …
where λ is the wavelength of the light.
Thus, for first case:
w1=0.5 mm
so:
d=w1zλ
In the second case optical path between slits and plane is increased:
p2=p−Δp+Δp∗n1=dθ+Δp(n1−1)
The interference fringe maxima occur at angles
dsinθ+Δp(n1−1)=nλ, n=0,1,2,3
For small angles:
sinθ=θ=tanθ
For central fringe n=0, thus:
θ=−dΔp(n1−1)
And
tanθ=zy≈θ
So:
zy1=−dΔp(n1−1)=−w1zλΔp(n1−1)zy1=−zλΔp(n1−1)w1y1=−λΔp(n1−1)w1
Calculating:
y1=−λΔp(n1−1)w1=600∗10−9100∗10−6∗(1.5−1)∗0.5∗10−3=0.0416m=4.16cm
See: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html
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