Young's double slit experiment with one slit covered with block of refractive indexby A9876 Tags: block, covered, double, experiment, index, refractive, slit, young 

#1
Sep112, 12:18 AM

P: 16

1. The problem statement, all variables and given/known data
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 2. Relevant 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, ... 3. 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 



#2
Sep112, 12:24 AM

P: 317

If I(θ)=4I0cos2(∏dsin(θ)/λ), then what do you know about dsin(θ)/λ?
Also, what do you know about the wavelength of light with refractive index n? 



#3
Sep112, 11:55 AM

P: 16

so would optical path difference, δ=r_{2}  (r_{1}t+nt)=(yd/L)(n1)t where r_{2} = distance between slit 2 and P, r_{1} = distance between slit 1 and P & t=thickness of block with refractive index n ? I'm not sure what I'm supposed to know about dsin(θ)/λ. Is it that it's equal to δ/λ? Would the maxima for this situation occur at (yd/L)(n1)t=mλ or (yd/L)(n1)t=(m+0.5)λ? 



#4
Sep112, 11:08 PM

P: 317

Young's double slit experiment with one slit covered with block of refractive index
Actually, your first part is correct, but as far as the second part,
Here's an idea I had in mind: Since we know m=dsin(θ)/λ for all maximas, then you could simply find the intensity at the maximas by replacing I(θ)=4I0cos2(∏dsin(θ)/λ) with I(θ)=4I0cos2(∏m) Just an idea. Also what I had in mind was changing everything from double slit interference to diffraction. When single slit interference occurs, it implies diffraction exists. Have you studied that? It differs from double slit interference by a few concepts of phase changes, if you know what I mean. 



#5
Sep212, 12:53 AM

P: 317




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