# Anti-Reflective coatings

1. Aug 11, 2005

### OPTIDONN

What happens when light encounters a thin film or anti-reflective coating? I know that a/r coatings are based on the principle of constructive and destructive interference. So do some photons turn on each other an cancel each other out and do others team up to increase light transmition? If some one can explain in very, very simple terms.

2. Aug 12, 2005

### Ivan Seeking

Staff Emeritus
If you consider the model in which you get a reflection of a percentage of the incident photons at each surface: If the film thickness is 1/4 of a particular wavelength, then if we take two "photons" having that wavelength and reflect one from the film's front surface, and then take the second photon and reflect it from the lens' surface, since the second photon has traveled 2 X 1/4 of a wavelength farther than the first, the two waves are 180 degrees out of phase, so they cancel and there is no reflection.

The deeper answer involves penetration depths and absorbtion and re-emission of individual photons within the material, but the result is approximated by considering the classical model of reflections at a surface.

Last edited: Aug 12, 2005
3. Aug 12, 2005

### OPTIDONN

Now how exactly do they cancel on another out? Do the photon colide making them dissapear? As for the increased light transmition do how does that happen?

4. Aug 12, 2005

### Staff: Mentor

Interference is not caused by an interaction between different photons. If that were true, we could eliminate interference patterns by making the light weak enough for there to be at most one photon passing through the apparatus at a time, so that there would be no way for another photon to interfere with it.

But when we do this, and let the light shine for a long time so that enough photons eventually accumulate on the film or detector, we still get an interference pattern.

5. Aug 12, 2005

### mezarashi

In this case, it would be most convenient to think of light as a wave. Identical waves that are off-phase by 180 degrees will cancel each other out.

6. Aug 12, 2005

### OPTIDONN

Oh so a wave will turn back on its self and cancel it out, But how does the film increase light transmition? And this might be a stupid question but what happens to the wave when it is put out of phase? Does it disapear?

7. Aug 13, 2005

### Antiphon

In the simplest terms, this is what happens. (I'll be using MKS units).

The light wave consists of an electric field and a magnetic field as
it travels. Wherever the light travels, there is a balence between these
two called the "impedance" of the medium. In free space this ratio
between electric and magnetic fields is about 377. In glass it is smaller,
about 163. This means for a wave in glass, the electric field is only
163 times larger than the magnetic field but in air its about 377 times
larger.

In a material like glass, the electric fields get smaller due to a material
effect called "dielectric polarization". But the magnetic field doesn't
get smaller. This means that at the boundary of the glass, there
is some "leftover" magnetic field which cannot enter the glass and travel
with the wave. It ends up creating a new wave which travels away from
the glass, called a reflection. When you add up the reflection, the
original wave, and the wave that enters the glass everything balences
out in both materials and at the bouandary.

What an AR coating does is cause extra reflections at the new interfaces
so that at the air/film interface the electric and magentic fields just
outside the device are in the same balence as just inside. This eliminates
the reflected wave in the air, and all the energy now makes it all the
way into the glass. The key is that in the AR coating, there is energy
traveling in both directions, while in the glass and in the air (using a
coating) the energy only travels in the one direction you want it to.

Of course the filter (AR coating) works the same way for light coming out
of the glass too.

In electrical engineering this is called impedance matching. Another word
for it is filter design. I have been making such devices for almost 20 years.

Edit: It is true that each individual quantum is filtered as if there were no others present. Photons do not interact [to the first order.]

Last edited: Aug 13, 2005
8. Aug 14, 2005

### Edgardo

Hello OPTIDONN,

I think the best way is to think of waves interfering destructively such that little light is reflected from the surface.

But if you want to explain it with photons, you have to consider quantum mechanical wavefunctions and probabilities. There's the possibility for
the photon to be reflected at the upper surface and at the lower surface.
From this, you could then calculate the probability for the reflection of a photon.

I think the calculations are similar to the ones with the wavemodel, since you are going to use the same phasedifferences. I would calculate it the following way (it's only how I WOULD EXPLAIN IT, I do NOT have any reference for it, thus no garantuee that it is correct):

Since you have two possibilities for the photon to be reflected (upper surface and lower surface), you have two wavefunctions $|\Psi_1>$ and $|\Psi_2>$. The total wavefunction $| \Psi_{total}>$ is then a superposition of those two wavefunctions PLUS the possibility for transmission, thus $|\Psi_3>$

$| \Psi_{total}> = | \Psi_1> + | \Psi_2> + | \Psi_3>$

Let $|\Psi_2> = e^{i \delta} |\Psi_1>$,
where $e^{i \delta}$ is a phase factor. This phase factor depends on the thickness and the material of the antireflection coating.

Then,

$| \Psi _{total}> =| \Psi _1> + e^{i \delta} | \Psi _1> +| \Psi_3>$

$= | \Psi _1> - | \Psi _1> + | \Psi_3> = | \Psi_3>$

where I have set $\delta = \pi$.

The total probability amplitude is
$| \Psi _{total}> = | \Psi_3 >$

If you want to calculate the probability for transmission, you get:

P(transmission) = $|<\Psi_3|\Psi_3>|^2 = 1$

The probability for reflection is:

P(reflection) = 1 - P(transmission) = 1 - 1 = 0.

I hope I haven't confused you, but
I recommend you reading the following texts, where I got the ideas from:

http://departments.colgate.edu/physics/faculty/EGalvez/articles/ajpph.pdf
http://departments.colgate.edu/physics/faculty/EGalvez/articles/ajpbs02.pdf

I can also recommend you Feynman's book ("The Feynman lectures on physics", Volume 3), where he explains in the first two chapters how to obtain the total wavefunction $|\Psi_{total}>$.

Cheers!

Edgardo

Last edited: Aug 14, 2005
9. Aug 14, 2005

### Claude Bile

AR coatings are often considered to be 1 dimensional photonic crystals. Essentially, photons at the design wavelength cannot propagate through the coatings at normal incidence because of the presence of a photonic bandgap in that direction.

A photonic bandgap is a gap in the dispersion relation of the material, a wavelength interval where no photon can propagate. This is analogous to the electronic bandgaps found in bulk materials.

Photonic bandgaps arise due to the periodic modulation of a refractive index. This creates anti-resonances between the interfaces (through interferance effects explained previously in this thread), essentially forbidding the propagation of light through the structure.

Claude.

10. Aug 14, 2005

### Hurkyl

Staff Emeritus
Why is the superposition not, for instance,

$$| \Psi_{total}> = (1/\sqrt{2}) | \Psi_1> + (1/2) | \Psi_2> + (1/2)| \Psi_3>$$

?

11. Aug 15, 2005

### Antiphon

Claude, I'm sorry but this is wrong. AR coatings are designed and constructed
via classical analysis. A quantum treatment does not shed any light on how
they work.

12. Aug 15, 2005

### Claude Bile

Not true, AR coatings and Bragg grating structures are frequently cited as an example of a 1D photonic bandgap structure.

This quote;

"Photonic Crystals are periodic dielectric structures that have a bandgap which forbids propagation of certain frequencies of light. This bandgap arises due to backscattering of light from the material when the periodic structure is of the order of the wavelength of incident light."

was taken from the following site;

http://www.optics.ee.dal.ca/sheshidher.html [Broken]

AR coatings are periodic dielectric structures of the order of the wavelength of the incident light. While they are not commonly analysed in terms of photonic bandgaps, the principle nevertheless is the same.

Claude.

Last edited by a moderator: May 2, 2017
13. Aug 16, 2005

### Antiphon

A single atom cannot properly be regarded as a crystal. Neither can a
single-material AR coating.

You might make the photonic crystal case for a many-layer periodic
structure but those are generally not AR coatings.

The article is stretching the word "bandgap" to the breaking point.
The bandgap concept has no classical analog. In a bandgap, you must
typically meet a minimum energy threshold to cause a transition- hence
the "gap". The photoelectric effect is an example. There is no classical
description for this process.

This is not the case in any AR coating. The situation is described accurately
in purely classical terms. The use of the word "bandgap" is a misapplication
of terminology.

Last edited by a moderator: May 2, 2017
14. Aug 16, 2005

### Claude Bile

The use of the word bandgap is perfectly acceptable in this instance. The dispersion relation of structures with a periodically varying dielectric constant contain frequency intervals which are forbidden for all wavevectors (the photonic bandgap). Such a relation is derived from the classical Maxwell's equations. Such a derivation can be found here;

http://ab-initio.mit.edu/photons/tutorial/

The reason for using this sort of terminology is because it can be generalised to include 2D and 3D periodic structures.

Claude.

15. Aug 17, 2005

### Antiphon

I see now why the term bandgap is being used here. On page 7 of the
introduction it states "The dielectric/air bands are analogous to the
valence/conduction bands in a semiconductor." So this is not being
treated as a quantum process but as an energy propagation stopband.

I'm willing to live with this as an analogy, but it still doesn't have anything
to do with AR coatings. The analysis carried out in the paper applies
rigourously to infinite structures. Very large (many layers) but finite
structure are well approximated by the analysis.

Single layer films are not well approximated by it at all. This is why I
maintain that a single layer AR coating should not be viewed as a crystal.
To illustrate this, an AR coating does not display the photonic bandgap
described for the infinite periodic structure. That is, there are no wavenumbers
for which propagation cannot occur in an AR coating let alone a range of them.

16. Aug 17, 2005

### Claude Bile

I agree, a single layer AR coating would not constitute a crystal. When I referred to AR coatings in my previous post, I meant multilayer stacks (as most AR coatings are multilayer stacks on the optics I work with).

Applying this theory to AR coatings (multilayer) is done usually from an educational standpoint to highlight the features of the 1D case before moving onto 2D and 3D cases. I don't think this theory has ever been used to design AR coatings. AR coatings (multilayer) also provide a good starting point when learning this theory as they are familiar to people working in optics.

I have seen a modified form this theory (one that includes truncated structures) applied to a single layer (then double layer, triple layer and so on) and seen how the bandgap slowly evolves as each new layer is added, though this was for a fibre Bragg grating, not an AR coating.

Claude.

17. Aug 18, 2005

### Edgardo

In my above calculations, to be honest, I haven't really thought about the correct coefficients. You will note that $$|\Psi_3>$$ actually must have
a coefficient that also depends on $$\delta$$, otherwise one has always the probability equal 1 of having transmission

As I said, I was not sure about my calculations and they are rather a quick sketch. But you're welcome to correct them

As for you wavefunction, I think it's not normalized, since $$|\Psi_2>$$ actually stands for $$e^{i \delta} | \Psi _1>$$ according to my notation. The only "basis vectors" are $$|\Psi_1>$$ and $$|\Psi_3>$$, with

$$|\Psi_1>$$ reflection
$$|\Psi_3>$$ transmission

Last edited: Aug 18, 2005
18. Apr 13, 2010

### Righeira

hi I am new to this forum. this question might be simple. but i am confused a lot..i am a project student working on AR multilayer coatings.... coatings are done by dip coater... my doubt is that there are coatings on either side...how does the light pass through one side and come along the other when their refractive index is reverse on the other side of the coating... one side it is n of air,coating1,coating2 glass as the substrate...but on the other side it is reverse...where can i find the details...i find very difficult to understand this...i understood the concept of interferences why that particular thickness is selected and all that....please help me...where can i clarify the doubts for this