Young's double slit experiment with one slit covered with block of refractive index


by A9876
Tags: block, covered, double, experiment, index, refractive, slit, young
A9876
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#1
Sep1-12, 12:18 AM
P: 16
1. The problem statement, all variables and given/known data

The interference pattern formed is given by I(θ)=4I0cos2(∏dsin(θ)/λ). For d=110-5m 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

2. Relevant equations

sin(θ)≈θ
I(θ)=4I0cos2(∏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)=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
Attached Thumbnails
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Rayquesto
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#2
Sep1-12, 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?
A9876
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#3
Sep1-12, 11:55 AM
P: 16
Quote Quote by Rayquesto View Post
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?
Wavelength in medium, λ = λ0/n where λ0 = wavelength in vacuum

so would optical path difference, δ=r2 - (r1-t+nt)=(yd/L)-(n-1)t
where r2 = distance between slit 2 and P, r1 = 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)-(n-1)t=mλ or (yd/L)-(n-1)t=(m+0.5)λ?

Rayquesto
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#4
Sep1-12, 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.
Rayquesto
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#5
Sep2-12, 12:53 AM
P: 317
Quote Quote by A9876 View Post
Wavelength in medium, λ = λ0/n where λ0 = wavelength in vacuum

so would optical path difference, δ=r2 - (r1-t+nt)=(yd/L)-(n-1)t
where r2 = distance between slit 2 and P, r1 = 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)-(n-1)t=mλ or (yd/L)-(n-1)t=(m+0.5)λ?
I think doing this is the way to go, because the path difference depends on the difference in lamda between two light waves that interfere with one another. Forget the diffraction idea. I'm so used to thinking about these experimental type concepts with the assumption that the wavelength has to be the same no matter what. I never thought about it in this way.


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