Thin-Film Interference (2 Wavelengths)

In summary: I don't know. I have no idea what n is. And m is the order of the fringe, right? So, where to go from there?In summary, the problem involves two glass plaques separated by a wire and illuminated by light with two different wavelengths. The task is to determine the distance from the point of contact where the next dark fringe forms. To solve this, the equations 2nt = mλ and 2nt = (m+1/2)λ for destructive and amplified interference, respectively, are used. The two wavelengths provide an additional constraint, as they must both be dark at the same location for a dark fringe to occur. To find the minimum separation at which it is all dark, the
  • #1
Const@ntine
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Homework Statement


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Two glass plaques of length 10.0 cm osculate, on one end, and are separated by a wire of diameter d = 0.0500 mm on the other. Light with two wavelengths (400 nm & 600 nm) falls on one of them, and we can see it gets reflected. At which distance from the point of contact does the next dark fringe form?

Homework Equations



2nt = mλ
2nt = (m+1/2)λ
dsinθlight = mλ
dsinθdark = (m+1/2)λ

The Attempt at a Solution


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My problem here is that I can't really understand the significance of the two wavelengths, how to use that info and generally how to proceed. My book has a small paragraph on TFI, and only explains the basic stuff (the above formulas for a single instance and that's it). Anyway, I figured:

>d = t : So d, the diameter, plays the role of the thin film's thickness.

>The thin film is the air, and the glass plaques have the same ng, which while bigger than n = 1.00 of the air, is still the same from above and bellow the thin film/air, so I don't have to inverse the two formulas.

>So, 2nt = mλ for destructive confluence, and 2nt = (m+1/2)λ for amplified confluence.

Now, the only thing I came up with was: Take the first formula (for destruction) and try it with both λs, to see which one gives me an integer m. 400 gives me n = 250, whereas 600 gives me 166.6666...7. So, I figured I'd continue with the first wavelength. Then I went back to the chapter about light and dark fringes, and looked at the formulas.

Now it gets murky and I'm just throwing stuff at the wall to see what will stick, but my plan was to put in the data in the "dark fringe" equation, find the angle, then put it in tanθ = y/L and find y. But that didn't work out. So I need some help here.

Any help is appreciated!
 
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  • #2
Hmmm ... it does seem as if you are throwing stuff against the wall as you say. You only need your second equation, and the point is t varies from one edge of the plate to the other. You can use the information given to parameterize t as a function of position along the plate.

The two wavelengths give an additional constraint. For a dark fringe they both have to be dark at the same location.
 
  • #3
Cutter Ketch said:
... You only need your second equation ...

Actually, on second thought, the two reflections being interfered have a 180 degree phase change between them. You only need your first equation.
 
  • #4
Cutter Ketch said:
Hmmm ... it does seem as if you are throwing stuff against the wall as you say. You only need your second equation, and the point is t varies from one edge of the plate to the other. You can use the information given to parameterize t as a function of position along the plate.

Ah, so it's t = d at the far let and t ~ 0 at the far right. ...How do I translate that into a function?

Cutter Ketch said:
The two wavelengths give an additional constraint. For a dark fringe they both have to be dark at the same location.

How do I use that though? Do I take the equation two times, once with 400 & once with 600 ? Like I said, I only know the basics (ight fallson thin film, gets reflected) so I'm not that well-versed in the theory behind it all.

Cutter Ketch said:
Actually, on second thought, the two reflections being interfered have a 180 degree phase change between them. You only need your first equation.

You mean 2nt = mλ, right?
 
  • #5
For some extra info, there was anothe exercise which went like this:

Same "construction" as in the OP, but here the light has a wavelength of λ = 600 nm, and we are informed that 30 dark fringes are created. Andwe have to find the diameter d. My logic on that one was:

>2nt = mλ, m = 0, 1, 2, 3...
>30 dark fringes, so m = 29
>t = d
>n = 1.00 (air)
Putting that all together I get d = 8.70 μm, which is the book's answer.

I'm writting that here since both the exercise in the OP and some others use the same construction/basic intel, and I figured it'd help if I posted how I handled the previous one (ie didn't use the "t decreases the further we move to the right" bit).
 
  • #6
Darthkostis said:
How do I translate that into a function?
You don't need to. First find the minimum separation at which it is all dark, then convert to a distance from the end.
Darthkostis said:
Do I take the equation two times, once with 400 & once with 600 ?
Yes. You can find all the separations at which one will be dark, and all the separations at which the other will be dark. Then figure out where both will be.
 
  • #7
haruspex said:
You don't need to. First find the minimum separation at which it is all dark, then convert to a distance from the end.

Yes. You can find all the separations at which one will be dark, and all the separations at which the other will be dark. Then figure out where both will be.
When you say "seperations", what do you mean? I'm asking because due to the translation problem I'm not familiar with the term.

Thanks!
 
  • #8
Darthkostis said:
When you say "seperations", what do you mean? I'm asking because due to the translation problem I'm not familiar with the term.

Thanks!
I mean the separation between the two pieces of glass. Gap, if you prefer.
 
  • #9
haruspex said:
I mean the separation between the two pieces of glass. Gap, if you prefer.
Oh, okay then. I'll look into it and report back after I've tried my hand at the exercise.

Thanks!
 
  • #10
I looked at it again, but I can't come up with something. How do I translate "everything is dark" to the 2nt = mλ formula? λ is the wavelength, t is the gap/seperation, n is a constant (1.00 since the thin film is the air here), which leaves m.
 
  • #11
Darthkostis said:
How do I translate "everything is dark" to the 2nt = mλ
If we just stick with one wavelength for the moment, the formula tells you for which values of t it will be dark. Just plug in m=1, 2, ... and solve.
Are you asking why that is true?
Putting in the other wavelength gives a different set of values of t. The question is asking for the smallest value of t (and hence of d) which is dark for both.
 
  • #12
haruspex said:
If we just stick with one wavelength for the moment, the formula tells you for which values of t it will be dark. Just plug in m=1, 2, ... and solve.

So I just try different values for m, for both wavelengths, and try to find which ones give me the same t? For example, I get t = 600 nm for m = 3 for the λ = 400 nm, and t = 600 nm for m = 2 for the λ = 600 nm. Is that it?

haruspex said:
Are you asking why that is true?
Putting in the other wavelength gives a different set of values of t. The question is asking for the smallest value of t (and hence of d) which is dark for both.

To be honest, I'm not really sure what it means. The "question" itself from the book is: "at which distance, from the point of contact, does the next dark fringe form". I took it to mean as "how far from the right side is the next dark fringe", but what does "next dark fringe" mean?
 
  • #13
Darthkostis said:
I get t = 600 nm for m = 3 for the λ = 400 nm, and t = 600 nm for m = 2 for the λ = 600 nm. Is that it?
Looks right to me. What does that correspond to as a distance from where the plates meet?
Darthkostis said:
what does "next dark fringe" mean?
Because of the 180 degree phase difference, the point where the plates meet will be a dark fringe for both wavelengths. You want the next such along.
 
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  • #14
Darthkostis said:
So I just try different values for m, for both wavelengths, and try to find which ones give me the same t? For example, I get t = 600 nm for m = 3 for the λ = 400 nm, and t = 600 nm for m = 2 for the λ = 600 nm. Is that it?
To be honest, I'm not really sure what it means. The "question" itself from the book is: "at which distance, from the point of contact, does the next dark fringe form". I took it to mean as "how far from the right side is the next dark fringe", but what does "next dark fringe" mean?

Because of the phase change of the ray reflected from the top surface of the bottom plate dark fringes (lines) are centred at places where the path difference (2t) = 0 or an integral number of wavelengths.

Each wavelength produces its own interference pattern and these two patterns overlap.

The resultant pattern can be rather complicated, for example there will be places where a bright fringe from one wavelength overlaps a dark fringe from the second wavelength. The result is that a fairly bright fringe is seen.

But there will also be places where two bright fringes overlap to make a brighter fringe.

And there will be places where two dark fringes overlap resulting in a dark fringe.

The question is about the dark fringes. Along the line of contact the path difference is zero for both wavelengths and that's where the first dark fringe is observed.

As you move to the left you can reach places where both wavelengths produces other dark fringes, the resultant fringe being dark.

Basically the question is asking you to find the second place at which this happens, the first place being along the line of contact.

But remember m, as in your equation, will have different values for each wavelength. I suggest that you use m for one wavelength, as you have done already, and a different symbol, for example n, for the second wavelength.
 
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  • #15
haruspex said:
Looks right to me. What does that correspond to as a distance from where the plates meet?

Because of the 180 degree phase difference, the point where the plates meet will be a dark fringe for both wavelengths. You want the next such along.

Dadface said:
Because of the phase change of the ray reflected from the top surface of the bottom plate dark fringes (lines) are centred at places where the path difference (2t) = 0 or an integral number of wavelengths.

Each wavelength produces its own interference pattern and these two patterns overlap.

The resultant pattern can be rather complicated, for example there will be places where a bright fringe from one wavelength overlaps a dark fringe from the second wavelength. The result is that a fairly bright fringe is seen.

But there will also be places where two bright fringes overlap to make a brighter fringe.

And there will be places where two dark fringes overlap resulting in a dark fringe.

The question is about the dark fringes. Along the line of contact the path difference is zero for both wavelengths and that's where the first dark fringe is observed.

As you move to the left you can reach places where both wavelengths produces other dark fringes, the resultant fringe being dark.

Basically the question is asking you to find the second place at which this happens, the first place being along the line of contact.

But remember m, as in your equation, will have different values for each wavelength. I suggest that you use m for one wavelength, as you have done already, and a different symbol, for example n, for the second wavelength.

Thanks a lot for the info and help! It seems the wording of the questiong messed me up a bit. Here's the completed exercise:

>Same kind of materialabove and bellow the air.
>Thus 2nt = mλ is the formula for the dark fringes.
>We're looking for the minimal "t" (height/gap/seperation between the two plaques) for which both wavelengths produce a dark fringe.

λ1: t = m*200 nm

For m = 1: t = 200 nm
For m = 2: t = 400 nm
For m = 3: t = 600 nm

λ2: t = n*300 nm

For m = 1: t = 300 nm
For m = 2: t = 600 nm

>So t = 600 nm is the minimal one.

Now, the two plaques create an tiny angle between them, close to the point of contact, which we can find.

tanθ = d/L <=> ... <=> θ = 5*10-4 rad

The angle remains the same, and using it and the minimal t, we can find how faritis from the point of contact.

tanθ = t/l <=> ... <=> t = 1.20 mm (which is the book's answer)

Thanks a ton for the help and patience everybody!
 

Related to Thin-Film Interference (2 Wavelengths)

1. What is thin-film interference?

Thin-film interference occurs when light waves reflect off of two different surfaces of a thin film, causing them to interfere with each other. This interference can result in changes in the color or brightness of the reflected light.

2. How are two wavelengths involved in thin-film interference?

In thin-film interference, two wavelengths of light are involved because the light waves reflect off of two different surfaces of the thin film. This causes the two waves to interfere with each other, resulting in changes in the color or brightness of the reflected light.

3. What is the equation for calculating the path difference in thin-film interference?

The equation for calculating the path difference in thin-film interference is given by 2nt cosθ, where n is the refractive index of the film, t is the thickness of the film, and θ is the angle of incidence.

4. How does the thickness of the thin film affect the interference pattern?

The thickness of the thin film affects the interference pattern by changing the path difference between the two light waves. A thicker film will result in a larger path difference, leading to a different interference pattern compared to a thinner film.

5. What are some real-life applications of thin-film interference?

Thin-film interference has many practical applications, such as in anti-reflective coatings for glasses and camera lenses, interference filters for optical devices, and in the colorful patterns seen on soap bubbles and oil slicks.

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