The interference pattern formed is given by I(θ)=4I0cos2(∏dsin(θ)/λ). For d=1×10-5m 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
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)=4I0cos2(∏dsin(θ)/λ) & plotted this with θ on the x-axis and I(θ) on the y-axis. I'm not sure is that's right but I can't seem to figure out the second part
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