Interference pattern for two glass plates closely separated and tilted

  • #1
Homework Statement:
A fine metal foil separates one end of two pieces of optically flat glass, as in the figure(Figure 1). When light of wavelength 640 nm is incident normally, 35 dark lines are observed (with one at each end).
Relevant Equations:
2t/lamba = m/2
hi,
i know you have to use 2t/lambda= m/2 to get the answer of 10.9 um. However i was taught that there would be phase difference caused by reflection, since n1(air)< n2(glass). SO the equation in my head would be 2t/lambda - 1/2= m/2. Can someone explain why i'm wrong?
 
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  • #3
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  • #4
haruspex
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Yes, one reflection will have a 180 degree phase shift.
the equation in my head would be 2t/lambda - 1/2= m/2.
That's for constructive interference.
This https://en.m.wikipedia.org/wiki/Thin-film_interference#Soap_bubble is for the opposite case, with the higher index in the middle, but it's the same result.

Btw, please use thread titles that give some clue as to the branch of physics involved.
 
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  • #5
kuruman
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Homework Statement:: A fine metal foil separates one end of two pieces of optically flat glass, as in the figure(Figure 1). When light of wavelength 640 nm is incident normally, 35 dark lines are observed (with one at each end).
Relevant Equations:: 2t/lamba = m/2

hi,
i know you have to use 2t/lambda= m/2 to get the answer of 10.9 um. However i was taught that there would be phase difference caused by reflection, since n1(air)< n2(glass). SO the equation in my head would be 2t/lambda - 1/2= m/2. Can someone explain why i'm wrong?
Can you post the question that the problem is asking you to answer? The "relevant equation" is not really relevant unless that is known.
 
  • #7
kuruman
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Please explain what you understand now. I want to make sure you got it. There is a way to look at questions of this kind that does not really involve a magic formula to substitute numbers in.
 
  • #8
etotheipi
the equation in my head would be 2t/lambda - 1/2= m/2

That's for constructive interference.

Shouldn't it be ##\frac{2t}{\lambda} = (m+\frac{1}{2})##? You can either derive it by assuming the angle of the wedge is very small so as to neglect refraction of the ray, or by a similar derivation to the thin film interference effect linked by @haruspex (though this is more complicated since you need to calculate the optical path lengths of both paths). Once you do that you can see that the latter reduces to the former in the limit ##\theta_2 \rightarrow 0##.
 
  • #9
kuruman
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Shouldn't it be ##\frac{2t}{\lambda} = (m+\frac{1}{2})##?
No. Just stop and think. There is a dark fringe at the end where the plates touch. At the other end there is also a dark fringe, 35 fringes in all. Starting at the touching end and moving towards the other end, when you cross the next dark fringe the plate separation increases by how much in units of a wavelength? How many such increases are there from one end to the other? :oldsmile:
 
  • #10
etotheipi
No. Just stop and think. There is a dark fringe at the end where the plates touch. At the other end there is also a dark fringe, 35 fringes in all. Starting at the touching end and moving towards the other end, when you cross the next dark fringe the plate separation increases by how much in units of a wavelength? How many such increases are there from one end to the other? :oldsmile:

My equation was for constructive interference, I meant it in the sense that OP's original relation is not quite correct even for the light fringes. For destructive interference you will need ##\frac{2t}{\lambda} = m## with ##m \in \mathbb{Z}##.
 
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  • #11
haruspex
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Shouldn't it be ##\frac{2t}{\lambda} = (m+\frac{1}{2})##? You can either derive it by assuming the angle of the wedge is very small so as to neglect refraction of the ray, or by a similar derivation to the thin film interference effect linked by @haruspex (though this is more complicated since you need to calculate the optical path lengths of both paths). Once you do that you can see that the latter reduces to the former in the limit ##\theta_2 \rightarrow 0##.
You are right. I was fixated on the OP's question as to whether there should be an odd half wavelength and didn't notice the spurious division by 2 of m.
 
  • #12
kuruman
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My equation was for constructive interference, I meant it in the sense that OP's original relation is not quite correct even for the light fringes. For destructive interference you will need ##\frac{2t}{\lambda} = m## with ##m \in \mathbb{Z}##.
Where ##m## is equal to what concerning this problem? Why is constructive interference relevant when dark fringes are mentioned? What is the correct equation relating the wavelength in air, ##\lambda_{air}##, the separation between the ends, ##t##, and the number of dark fringes observed, ##N##? That's what needs to be derived.
 
  • #13
etotheipi
Where ##m## is equal to what concerning this problem? Why is constructive interference relevant when dark fringes are mentioned? What is the correct equation relating the wavelength in air, ##\lambda_{air}##, the separation between the ends, ##t##, and the number of dark fringes observed, ##N##? That's what needs to be derived.

Ah sorry, yes I wasn't very clear at all and I think we're also using the same letters for different variables :wink:. The constructive interference part is as you say not relevant here.

I took ##t## to be the width of the air gap at some position ##x## along from where the plates are touching, and ##m## in that equation is just an integer, which should run over ##0## to ##34##, so we have ##2t(x) = m\lambda_{air}## for destructive interference. From then if we know some details about the geometry (I can't find any values in the thread already) it should be straightforward to calculate either the angle between the plates or the total length of the plate (given either one is known).
 
  • #14
kuruman
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Ah sorry, yes I wasn't very clear at all and I think we're also using the same letters for different variables :wink:. The constructive interference part is as you say not relevant here.

I took ##t## to be the width of the air gap at some position ##x## along from where the plates are touching, and ##m## in that equation is just an integer, which should run over ##0## to ##34##, so we have ##2t(x) = m\lambda_{air}## for destructive interference. From then if we know some details about the geometry (I can't find any values in the thread already) it should be straightforward to calculate either the angle between the plates or the total length of the plate (given either one is known).
No angle or geometry is needed. Just read post #9. It gives the method for constructing the correct formula.
 
  • #15
etotheipi
No angle or geometry is needed. Just read post #9. It gives the method for constructing the correct formula.

In that case I'm a little confused as to what the question actually is; if ##d## is the width of the lower plate is known and ##\theta## the angle between them then I would have said something like ##\theta = \arctan{\left(\frac{34\lambda_{air}}{2d}\right)}## in order to solve for the angle. I might have misinterpreted the question!
 
  • #16
kuruman
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In that case I'm a little confused as to what the question actually is; if ##d## is the width of the lower plate is known and ##\theta## the angle between them then I would have said something like ##\theta = \arctan{\left(\frac{34\lambda_{air}}{2d}\right)}## in order to solve for the angle. I might have misinterpreted the question!
OP has not stated what the question is, but that the answer is 10.8 μm. This is consistent with the thickness of the spacer. I will send you a PM later with the solution.
 
  • #17
kuruman
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In that case I'm a little confused as to what the question actually is; if ##d## is the width of the lower plate is known and ##\theta## the angle between them then I would have said something like ##\theta = \arctan{\left(\frac{34\lambda_{air}}{2d}\right)}## in order to solve for the angle. I might have misinterpreted the question!
Actually I think that you may have misinterpreted the question. This is thin film interference wherein the “thin film” is the air gap between plates. The interfering rays are reflections off the bottom of the top plate and off the top of the bottom plate. The plates are thick relative to the wedge-shaped gap.
 
  • #18
etotheipi
Actually I think that you may have misinterpreted the question. This is thin film interference wherein the “thin film” is the air gap between plates. The interfering rays are reflections off the bottom of the top plate and off the top of the bottom plate. The plates are thick relative to the wedge-shaped gap.

That's what I had in mind; I am just confused since I must have answered this exact question about 3 or 4 times throughout my schooling and I'm a little thrown as to what is different in this case :wink:. The first dark fringe occurs where the glass plates join, and the last (35th) dark fringe occurs when the width of the gap of air is ##17\lambda_{air}##. Then it just becomes a trigonometry problem, since the lateral width of the plates becomes a function of ##\theta##.

I could be missing something really obvious, if so please forgive me! I only got like 4 hours of sleep today 😅.
 
  • #19
kuruman
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The first dark fringe occurs where the glass plates join, and the last (35th) dark fringe occurs when the width of the gap of air is ##17\lambda_{air}##. Then it just becomes a trigonometry problem, since the lateral width of the plates becomes a function of ##\theta##.
This part of the reply is directed to @SHAWN JAMES.
Imagine numbering the fringes ), 1, 2, ##\dots## from where the plates touch (fringe 0) to the end (fringe 34). Let ##\Delta y## be the gap at fringe 1. How big is ##\Delta y##? Well, at fringe 0 where we have destructive interference, the two waves are matched so the valleys of one correspond to the peaks of the other. That is because although the path length difference (in meters) is the same for each ray, the phase of only the ray reflected off the bottom plate is shifted by ##\pi##. That is the case for all the dark fringes.

When we move from fringe 0 to fringe 1, again we must match valleys with peaks in the two waves. However the path length difference has increased from zero meters to ##2\Delta y## meters. What does that translate to in units of ##\lambda##? Obviously, to match peaks and values, it must be equal to one wavelength for adjacent fringes. Thus the key equation $$2\Delta y=\lambda.$$ So the gap above fringe 1 is half a wavelength. I leave it up to you to figure out the gap above fringe 34.


This part of the reply is also directed to @etotheipi.
Note that it is not a trigonometry problem. If you know the gap, you cannot figure out the angle because you have only one side of the triangle. You need additional information such as the spacing between adjacent fringes to figure out the angle.

On edit: The statement immediately above is correct in the approximation of neglecting refraction and considering normal incidence on both plates. I have not worked out the details when incidence on the bottom plate is normal but at an angle on the top plate.
 
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