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Homework Statement
The interference pattern formed is given by I(θ)=4I_{0}cos^{2}(∏dsin(θ)/λ). For d=1×10^{5}m and λ=500nm plot the intensity pattern as a function of θ for small θ. How would this change if a block of material of thickness 500nm and refractive index n=1.5 were placed over one slit (without altering the intensity from that slit)?
Please see the attachment
Homework Equations
sin(θ)≈θ
I(θ)=4I_{0}cos^{2}(∏dsin(θ)/λ)
maybe these:
Maxima at d sin(θ)=mλ for m=0,1,2, ...
Minima at d sin(θ)=(m+0.5)λ for m=0,1,2, ...
The Attempt at a Solution
1st maxima at θ=sin^{1}(0)=0 radians
1st minima at θ=sin^{1}(λ/2d)=0.025 radians
2nd maxima at θ=sin^{1}(λ/d)=0.05 radians
2nd minima at θ=sin^{1}(3λ/2d)=0.075 radians
3rd maxima at θ=sin^{1}(2λ/d)=0.1 radians
Then I subbed these into I(θ)=4I_{0}cos^{2}(∏dsin(θ)/λ) & plotted this with θ on the xaxis and I(θ) on the yaxis. I'm not sure is that's right but I can't seem to figure out the second part
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