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Homework Help: Interference in thin microscope films

  1. May 28, 2010 #1
    1. The problem statement, all variables and given/known data
    "One microscope slide is placed on top of another with
    their left edges in contact and a human hair under the right edge of the upper slide.
    As a result, a wedge of air exists between the slides. An interference pattern results
    when monochromatic light is incident on the wedge. At the left edges of the slides,
    there is (a) a dark fringe (b) a bright fringe (c) impossible to determine."


    2. Relevant equations



    3. The attempt at a solution
    So I know the answer is a, a dark fringe, but I don't understand why. First of all, when they say "is incident on the wedge", do they simply mean light is falling directly into the wedge? Or falling directly onto the upper slide, reaching the wedge and then hitting the bottom layer. I presume the latter.

    So at the left they're touching each other. In my book it says there is a dark fringe because the only interference is the fact that the light hitting the upper surface has a 180° phase shift. This makes no sense, what would it interfere with? The fact the light gets a phase shift is not enough to cause blackness, right? I'm very confused...
     
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  3. May 28, 2010 #2

    cepheid

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    Yeah it's the latter. It's a given that the light is incident on slides in the direction perpendicular to their surfaces. Otherwise this wouldn't really be a thin film interference problem. The only way to pass through the wedge without passing through the slides would be to come in from the side (is that what you mean with your first suggestion?).

    The reflected wave is interfering with the incident wave. Since there is a phase shift of 180 degrees, the two waves are exactly out of phase and interfere destructively.
     
  4. May 28, 2010 #3
    but the incident wave is travelling in the opposite direction, why would they interfere?

    Or is it that first a wave reflects inside the upper slab and a bit passes through and hits an infinitesimal piece of air and then reflects from the upper side of the lower slab with a 180° phase shift and then they cancel each other? Seems like a very hard way of saying: the wave passes unhindered through the slabs
     
  5. May 29, 2010 #4

    cepheid

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    Well the two waves are overlapping, right? If I shine a flashlight straight at a mirror, the reflected beam will travel back exactly along the same path through space as the incident one, and the two waves will pass through each other (Pop quiz: why doesn't any noticeable interference occur in this example?).

    In the case of the situation presented in your problem: any time light encounters an interface between two different optical media, part of the light will be reflected back from the interface, and part of it will be transmitted. So, some of the light will be reflected back from the top surface of the top slide, because here there is an interface between air and glass. However, because there is no gap between the slides, there is NO interface at the junction between the two slides (it's the same medium on both sides). Hence, the light will pass cleanly from the top slide to the bottom slide. I suppose that if the light then passes through the bottom surface of the bottom slide back into air or some other medium, then this counts as another interface at which there will be partial reflection and partial transmission. However, I think that this a secondary effect that your problem is ignoring because the beam intensity will have been attenuated quite a bit by then.

    So, in summary, the only interference taking place is between the light incident on the top surface and the light reflected back from the top surface.
     
  6. May 29, 2010 #5
    Huh? I'm baffled. Are you 100% sure? So if I take one pane of glas, there is actually destructive interference all over caused by the light reflecting? But I can't see how they can destructively interfere: the reflected wave is moving in another (namely the opposite) direction as the incident beam... They can't overlap. There might be one time where they completely cancel, but one time instant further and the minima and maxima are not aligned anymore (because they have different directions)... So there are two seperate things I'm confused about that are not related:

    1) a single pane of glas is then actually completely a dark fringe? (as described at the start of my post). How do you explain a piece of glas can function as a (bad/vague) mirror? Purely secondary effects?

    2) how can waves travelling in different directions (one left, one right) cancel each other at all times?
     
  7. May 29, 2010 #6

    cepheid

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    Yeah, my explanation for why you get a dark fringe in this particular situation is wrong, so I'm sorry about that. But I am sure about certain things (including the fact that two superposed waves travelling in opposite directions can interfere :rolleyes:).

    Let me address this point first, because here is where my explanation was the most confusing/misleading. Let me go over what happens when light moves from air to glass. Whenever light is incident on an interface between between two optical media, part of the incident light will be reflected, and part of it will be transmitted. If the first (starting) medium has refractive index n1, and the second medium has refractive index n2, then there are two key results. The first is that the reflected wave will be completely out of phase with the incident one if n2 > n1, and they'll be completely in phase if n2 < n1. The second result is that, for normal (meaning perpendicular) incidence, the ratio of the reflected light intensity IR to the incident light intensity II is given by:

    IR/II = [ (n1 - n2) / (n1 + n2) ]2

    These results can be derived from the theory of light as an electromagnetic wave (it's complicated and requires a deeper understanding light than the level at which you are probably studying it). Similarly, the ratio of the transmitted light intensity IT to the incident intensity should be given by:

    IT/II = (4n1n2) / (n1 + n2)2

    These ratios should add up to 1 by the conservation of energy (all of the light is either transmitted or reflected). Plugging in numbers for air (n1 = 1) and glass (n2 = 1.5), you get that:

    IT/II = 0.96

    IR/II = 0.04

    This is why one surface of a pane of glass functions as such a bad mirror. Only 4% of the light is reflected, the rest being transmitted.

    Immediately, you can see that I made a mistaken assertion in my previous post. The dark fringe cannot be explained entirely by the interference of the incident and reflected waves. It's true that they'll be totally out of phase [EDIT: initially], and that they'll destructively interfere [EDIT: at certain instants in time]. But the reflected wave will never completely cancel the incident one, because its intensity is much much less (only 4%). Therefore, the dark fringe cannot be totally explained by the interference of the incident and reflected beams [EDIT: especially since this interference is not always totally destructive at all times, as discussed below]. So what causes it? The only thing I can think of is that the wave reflected off the bottom surface of the bottom slide WON'T undergo a phase shift (because n1 and n2 are switched at that interface), and hence the two reflected beams (one from top surface of top slide and one from bottom surface of bottom slide) will be out of phase and cancel each other. That's the best inkling I have right now.

    Using the same formula, but switching n2 and n1, I get that [(1.5 -1)/ 2.5]2 = 4% of the transmitted light will be reflected at the bottom glass-to-air interface. This is 0.04*0.96 = 3.8% of the incident intensity. Then, this second reflected light has to pass through the glass-to-air interface back out in the direction it came, meaning 0.038*0.96 = 0.037. So, the wave reflecting off the bottom surface of the two slides and coming back out towards the observer will have 3.7% of the incident intensity, which will be comparable in intensity to the wave reflecting directly off the top surface (4% of the incident intensity). Their intensities are comparable and they are out of phase, so they will mostly cancel, leading to almost no reflected light back in the direction of the observer. That's my best guess as to what is happening in this situation.

    Now, as for this (entirely separate) business of two parallel waves going in opposite directions somehow not overlapping, my response is...huh??? What are you talking about? Why would they not occupy exactly the same space as each other?

    Why not? They're parallel and travelling towards each other! It's inevitable that they will end up superposed.

    EDIT: yeah you're right. I was wrong about this aspect as well. If you have two sinusoidal waves of the same frequency travelling at the same speed and passing through each other in OPPOSITE directions, and if they are completely out of phase to start, then the phase relationship between them will continuously change, meaning that the intensity at any point in space will vary sinusoidally in time with an amplitude that depends where you are. That's just a standing wave (as opposed to a travelling one). So, there will be some locations in space where the intensity will always be zero, but that will not be true in general for all locations. I should have known that. :blushing:
     
    Last edited: May 29, 2010
  8. May 29, 2010 #7

    cepheid

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    But the relative phase of these two reflected waves also depends upon the path difference between the two of them when they meet again, which is equal to four microscope slide thicknesses, which may or may not be an integer number of wavelengths. So...I dunno!
     
  9. May 29, 2010 #8
    Thank you for the detailed reply on my first question! As for the second question, it seems there was a mix-up, but I'm glad that was cleared up :) So first for the second question: so the light that reflects from the upper surface will reach my eye and I perceive light. I presumed my book was simply ignoring what was happening on top, actually, because look at the way it states (part of) the question: "An interference pattern results
    when monochromatic light is incident on the wedge." Of course light is bouncing back from the upper surface everywhere, but it seems they're just ignoring that, taking the relative interference in the air wedge as way more important than the upper-surface reflection. So when they see the right answer is a "dark fringe" where the two plates connect, maybe they just mean "a relative dark fringe"? They did the same thing when discussing Newton's rings: completely disregarding reflection from the upper surface with the danger of implying that a place of destructive interference gives a black region when viewed by you. Maybe you have to assume there is some magic powder on the upper surface stopping reflection.

    Back to your answer on question one: oh by the way, you made a typo when putting in the numbers, "IT/II = 0.96; IT/II = 0.04", you have the same subscripts twice. It didn't matter though, as it was clear from your explanation what you meant, but maybe you'd like to know.

    You say "so they will mostly cancel", that does seem vague, doesn't it? Or are you just presuming the slab width h is h << lambda? I don't see how that's possible at all, and if that is not true, then they could even interfere constructively if the thickness is right.

    I'll type the exact answer from my book (which doesn't clear it up, as I think i have quoted it earlier and it remains vague):

    "At the left edge, the air wedge has zero thickness and the only contribution to the interference is the 180° phase shift as the light reflects from the upper surface of the glass slide."

    So yeah, vague, since it's talking about only one beam and is not even clear where it is reflecting. I'm not English, but it does sound that they're saying it's reflecting from the upper surface of the upper slab. So we have two possible solutions, but the problem is that they're both not compatible with how my book describes it, but the two are:

    1) (suggested by me above in this post) we have to ignore reflection from the most upper surface and then there is a relative dark fringe at the contact region because light is just passing through;
    2) (suggested by you) somehow reflection from the lower surface of the lower slab interferes destructively with the reflection from the upper surface from the upper slab (but this seems ungrounded).

    What I think the book means, and I ask you if it sounds plausible, is:

    the contact point is actually a discontinuity (it's not perfectly one piece) and at other places in the air wedge, there is interference both due to 180° phase shift AND a difference in path lengths, here the path lengths should maybe just be taken zero and there effectively is a piece reflecting from the down side of the upper slab and from the upper side of the lower slab.

    Thanks a lot for your time

    EDIT: saw your latest post after I posted: indeed, that's what I'm trying to say partly in this post, so you can ignore my words on it, as I hadn't read that post yet

    EDIT2: so in my last suggestion (last paragraph), I'm interpreting the "the light reflects from the upper surface of the glass slide" from the book's answer statement as meaning reflection from the upper surface of the bottom slide... (or would that be bad English?)
     
  10. May 29, 2010 #9

    cepheid

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    I don't know what the answer is. Our mutual realization about the path difference seems to invalidate my suggestion (which was your suggestion #2). That leaves suggestion # 1, which is that maybe the left edge is darker *relative* to elsewhere along the slide because most of the light is transmitted at this edge, whereas elsewhere you can get reflection/constructive interference due to the air gap.

    Oh yeah, and I would interpret the statement from your book as meaning that the light is reflecting from the upper surface of the upper slide.

    I'm sorry I couldn't help clear things up.
     
    Last edited: May 29, 2010
  11. May 29, 2010 #10
    But thank you a lot for taking the time :) I suppose in the end the main problem is just the way the book asks the question or gives the vague answer; you made me confident my feeling with the matter is what it should be.
     
  12. May 29, 2010 #11

    vela

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    The interference is caused by light reflecting off the bottom surface of the upper slab and off the upper surface of the bottom slab, just like with Newton's Rings. The 180-degree phase change that occurs when the light reflects off the bottom slab results in the dark fringe at the left end.
     
  13. May 29, 2010 #12

    cepheid

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    In that case, don't you have to assume that the path difference is effectively zero (since the slabs are essentially touching), but that there is still some optically less dense medium in between the two (to get the phase change for the reflection off the top of the lower slab)? Isn't that a contradiction?

    I understand that at an atomic scale, "touching" doesn't really mean anything -- everything has spaces...so are we basically saying that, down at the length scale where the slabs are no longer smooth, there are gaps, and these gaps are still perceived by the light as a transition out of the glass, but are small enough (relative to the wavelength of visible light) to contribute negligibly to the path difference?
     
  14. May 29, 2010 #13
    Indeed cepheid, I think that is what is meant. I admit, it sounds like a bit of hand-waving, but on the other hand, it sounds a lot more evident than (for example) simply the fact itself that light has a phase-shift when hitting a surface with a greater n. (which I'm also presumed to just accept at this point of my course)

    Thank you, vela.
     
  15. May 29, 2010 #14

    ehild

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    When the glass plates touch each other, they behave as a single slab. The reflectance will be the same as that of a glass slab. Not totally dark: about 7-8 %. If there is a thin air layer in between, the reflectance will increase with respect to that of the glass reaching maximum when the thickness of the air layer is one-quarter of the wavelength. In case of thicker air layer, the reflectance decreases again reaching that of the glass at half-wavelength thickness.



    ehild
     
  16. May 29, 2010 #15
    But ehild, then how do you interpret the book's explanation:

    "At the left edge, the air wedge has zero thickness and the only contribution to the interference is the 180° phase shift as the light reflects from the upper surface of the glass slide."
     
  17. May 29, 2010 #16

    ehild

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    There is no contradiction. The phase difference is 180°, and the reflectance is the same at every case when the cosine of the total phase difference is -1. The reflectance has its minimum value, and this minimum is equal to that of the glass, as there is reflection also from the air-glass interfaces.

    If you had a single thin layer like the wall of a soap bubble, the minimum reflectance would be zero, and it is the same as it were at zero thickness - when the wall would not be present. The reflectance and transmittance of thin layers can be derived by application of Maxwell equations and the boundary conditions for electromagnetic waves, and you find them in any textbook on thin layer optics.

    Here the situation is a bit more complicated as you see the surface of the glass plate. If you were inside the glass and saw the air layer from the glass, :) you would detect a completely dark fringe at the edge of the air film.
     
  18. May 29, 2010 #17

    vela

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    I look at it as a limiting case. As the thickness of the air layer goes to zero, the only phase change left is due to the reflection on the bottom slab.
     
  19. May 29, 2010 #18

    cepheid

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    Both slabs have the same refractive index, so if the air gap is exactly zero, not only is there no phase change, but there is no reflection either!

    As far as I can see, ehild and vela are suggesting two mutually exclusive explanations. With zero air gap, either there is no reflected beam (not counting the one off the top surface), which makes things as dark as they can get (which is what ehild is saying), or there are two reflected beams (according to vela) interfering like in the case of Newton's rings (with an inexplicable phase change* even though the two surfaces in contact have the same refractive index). WHICH explanation is correct, and how could the second one possibly be?




    *This is something I don't understand about the explanation for the central dark spot in Newton's rings either. If you have a glass lens in contact with a glass flat, why the phase change?


    EDIT: Never mind. I think I see what you're saying. Regardless of which explanation you choose, the result is the same. What's the difference between "no reflection" and "total cancellation of the two reflected beams" in terms of what is observed? No difference (as ehild pointed out in post#16). Although the actual explanation is that the gap is zero, leading to n1 = n2 and no reflection, this can be regarded as the result of the Newton ring phenomenon in the LIMIT that the gap is really really small and the phase difference can be regarded as arising entirely from the pi phase shift upon reflection. I'm sorry I was slow to get that.
     
    Last edited: May 29, 2010
  20. May 29, 2010 #19

    ehild

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    Well, is not it the same, two reflected waves which extinct each other or no reflected wave at all?

    Imagine a thin layer of refractive index n1 and thickness d embedded between two media with refractive indices n0 and n2, and a monochromatic wave of wavelength λ is incident perpendicularly at the surface. The reflectance can be expressed with the Fresnel reflection coefficients and the phase change δ when the wave traverses the layer.

    The Fresnel reflection coefficients are

    [tex]r_{01}=\frac{n_0-n_1}{n_0-n_1}[/tex]

    [tex]r_{12}=\frac{n_1-n_2}{n_1-n_2}[/tex]

    A negative reflection coefficient is interpreted as pi phase change at reflection.

    The phase change during travelling across the layer is:

    [tex]\delta=\frac{2\pi}{\lambda} nd [/tex]

    the reflectance is

    [tex]\frac{r_{01}^2+r_{12}^2+2r_{01}r_{12}cos(2\delta)}{1+r_{01}^2r_{12}^2+2r_{01}r_{12}cos(2\delta)}[/tex]

    If the surrounding media are the same, r12=r10=-r01 and the reflectance becomes

    [tex]\frac{2r_{01}^2-2r_{01}^2cos(2\delta)}{1+r_{01}^4-2r_{01}^2cos(2\delta)}[/tex]

    which is zero when d=0 because of the - sign in front the interference term (the cosine term) and this cosine term comes from the pi phase change at one of the two boundaries.

    ehild
     
    Last edited: May 29, 2010
  21. May 29, 2010 #20

    ehild

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    Cepheid, we said the same at the same time :)


    ehild
     
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