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

In summary: Use that equation you've derived, and just plot it. Then plot it again for the second part. In summary, the interference pattern formed by two slits with a distance of d=1×10-5m and light with a wavelength of λ=500nm can be plotted as a function of θ. The first maximum occurs at θ=0 radians and the first minimum occurs at θ=0.025 radians. The second maximum occurs at θ=0.05 radians and the second minimum occurs at θ=0.075 radians. The third maximum occurs at θ=0.1 radians. If a block of material with a thickness of 500nm and a refractive index of n=1.
  • #1
A9876
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0

Homework Statement



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

Homework 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, ...

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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  • #2


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


Rayquesto said:
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?
 
  • #4


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


A9876 said:
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.
 

1. How does the presence of a refractive index affect the interference pattern in Young's double slit experiment?

When one of the slits in Young's double slit experiment is covered with a block of refractive index, the presence of the refractive index causes a phase shift in the light passing through it. This results in a shift in the interference pattern, with the bright fringes of the pattern being shifted towards the covered slit.

2. What is the purpose of covering one slit with a block of refractive index in Young's double slit experiment?

The purpose of covering one slit with a block of refractive index is to introduce a phase shift in the light passing through that slit. This allows for the study of the effects of a refractive index on the interference pattern, and can also be used to measure the refractive index of the material used in the block.

3. How does the refractive index of the block affect the interference pattern in Young's double slit experiment?

The refractive index of the block affects the interference pattern by causing a phase shift in the light passing through it. The higher the refractive index of the block, the greater the phase shift and the more noticeable the shift in the interference pattern will be.

4. Does the size of the refractive index block affect the interference pattern in Young's double slit experiment?

Yes, the size of the refractive index block does affect the interference pattern in Young's double slit experiment. A larger block will cause a greater phase shift and therefore a more noticeable shift in the interference pattern. However, the overall pattern will still follow the same basic shape and principles of interference.

5. Can the refractive index block be placed in front of either slit in Young's double slit experiment?

Yes, the refractive index block can be placed in front of either slit in Young's double slit experiment. However, the location of the block will affect the interference pattern differently. Placing the block in front of the slit closer to the light source will result in a larger phase shift and a more noticeable shift in the interference pattern.

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