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No matter what thickness a piece of glass is wouldn't its optical thickness be close to an integer multiple of a wavelength such that it could create interference effects? I feel like I am missing something here.

sTr8_Struggin
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    Do you think it is practical to control the thickness of your glass window to within, say, less than 10nm across your entire window? Compare that with using a controlled deposition process for the thin film stack (which can be done on an industrial scale). – Jon Custer Jun 22 '16 at 13:44

2 Answers2

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You are completely right in stating that the same effect should occur for thicker slabs. There are however at least 3 practical reasons why the effect is more easily observed in thin films.

  1. Light sources are typically not completely monochromatic. They emit slightly different colours at the same time. Imagine a light source whose wavelength varies by 0.1 percent. For a layer which is a couple of wavelengths thick, all colours will interfere destructively under the same angle. However when the layer is 1000 wavelengths thick, one colour will interfere constructively, while the other interferes destructively. The interference pattern will thus be lost. BTW, the length over which a light source can interfere is called the coherence length.

  2. Light sources are typically not infinitesimally small. Instead of being a point source, the source has a certain width. This means that the light effectively arrives under various angles at a certain point. For thin samples this does not matter as all these angles will interfere constructively at the same time. However for a thick layer, light under certain angles will interfere constructively, while other angles interfere destructively and the interference pattern is lost.

  3. Light sources are typically not completely spatially coherent. This means that the wave front is not completely neat and flat. The distance over which the wavefront is still flat (~up to half a wavelength) is called the coherence width. The result is that when a part of the beam interferes with another part of the beam, which is further away than the coherence width, the interference pattern is lost. For a thick layer, when light is applied under an angle, the light typically interferes with another part of the beam further than the coherence width. Note that mathematically reasons 2 & 3 are actually the same.

  4. It is difficult to make thick flat films. However not as impossible as it might seem. Wafers are typically polished up to atomic flatness over micrometer distances and vary only hundreds of nanometers over millimeter distances.

Glorfindel
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Crimson
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  • Regarding the point 1, you should still see splitting of colours right? – Yashas Jun 22 '16 at 14:57
  • @YashasSamaga I think all 4 interference-killers are meant to act simultaneously. So if the effect "survives" #1, the #2 will neglect it effectively. – Crowley Jun 22 '16 at 17:20
  • @YashasSamaga For small thicknesses yes, but above a certain thickness, the interference effect for the different colors is so different that everything averages out to something gray, see for example this image. As mentioned in the answer, the distance beyond which this happens depends on the coherence length, which can be as short as just a few micrometers for broadband white light. – Bas Swinckels Jun 22 '16 at 18:42
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    desdiscardtructively?.. – Ruslan Jun 22 '16 at 18:51
  • #1 does not actually apply. Even with thin films, certain colors interact constructively while other colors interact destructively. This is why you get a rainbow effect on thin films; thick films are no different in this respect. Also #2 doesn't really apply either - it just means that, when viewed at different angles, the interference pattern will change. – Timothy Smith Jun 22 '16 at 19:46
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Thin layer diffraction.

Ignoring for simplicity's sake the usual refraction that takes place at the interfaces of the media, if:

$$|OA|+|AB|=D\big(\frac{1}{\cos \theta}+\tan \theta\big)=n\lambda$$

with $n$ an integer and $\lambda$ the wave length, then we have positive interference.

But as we increase $D$, the distance $\Delta$ also increases, so for high values of $D$ these rays can no longer interfere.

Gert
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  • I suspect that a different ray reflecting off the bottom surface COULD cause interference. Jon Custer gives an excellent example of the difficulty involved in achieving the precision required to get a thick piece of glass to show interference effects. – David White Jun 22 '16 at 13:57
  • @DavidWhite: at very low $\theta$, perhaps? – Gert Jun 22 '16 at 14:04