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OPTIDONN

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OPTIDONN

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- #2

Ivan Seeking

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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.

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.

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- #3

OPTIDONN

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- #4

jtbell

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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

mezarashi

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OPTIDONN said:

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

OPTIDONN

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- #7

Antiphon

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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.]

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

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.]

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- #8

Edgardo

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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 [itex]|\Psi_1> [/itex] and [itex] |\Psi_2>[/itex]. The total wavefunction [itex] | \Psi_{total}> [/itex] is then a superposition of those two wavefunctions PLUS the possibility for transmission, thus [itex] |\Psi_3> [/itex]

[itex] | \Psi_{total}> = | \Psi_1> + | \Psi_2> + | \Psi_3> [/itex]

Let [itex]|\Psi_2> = e^{i \delta} |\Psi_1>[/itex],

where [itex]e^{i \delta}[/itex] is a phase factor. This phase factor depends on the thickness and the material of the antireflection coating.

Then,

[itex] | \Psi _{total}> =| \Psi _1> + e^{i \delta} | \Psi _1> +| \Psi_3>[/itex]

[itex] = | \Psi _1> - | \Psi _1> + | \Psi_3> = | \Psi_3> [/itex]

where I have set [itex]\delta = \pi [/itex].

The total probability amplitude is

[itex] | \Psi _{total}> = | \Psi_3 > [/itex]

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

P(transmission) = [itex] |<\Psi_3|\Psi_3>|^2 = 1 [/itex]

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 [itex]|\Psi_{total}>[/itex].

Cheers!

Edgardo

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 [itex]|\Psi_1> [/itex] and [itex] |\Psi_2>[/itex]. The total wavefunction [itex] | \Psi_{total}> [/itex] is then a superposition of those two wavefunctions PLUS the possibility for transmission, thus [itex] |\Psi_3> [/itex]

[itex] | \Psi_{total}> = | \Psi_1> + | \Psi_2> + | \Psi_3> [/itex]

Let [itex]|\Psi_2> = e^{i \delta} |\Psi_1>[/itex],

where [itex]e^{i \delta}[/itex] is a phase factor. This phase factor depends on the thickness and the material of the antireflection coating.

Then,

[itex] | \Psi _{total}> =| \Psi _1> + e^{i \delta} | \Psi _1> +| \Psi_3>[/itex]

[itex] = | \Psi _1> - | \Psi _1> + | \Psi_3> = | \Psi_3> [/itex]

where I have set [itex]\delta = \pi [/itex].

The total probability amplitude is

[itex] | \Psi _{total}> = | \Psi_3 > [/itex]

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

P(transmission) = [itex] |<\Psi_3|\Psi_3>|^2 = 1 [/itex]

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 [itex]|\Psi_{total}>[/itex].

Cheers!

Edgardo

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- #9

Claude Bile

Science Advisor

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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

Hurkyl

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

?

- #11

Antiphon

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Claude Bile said:

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.

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

Claude Bile

Science Advisor

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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.

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.

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- #13

Antiphon

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Claude Bile said: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.

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.

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- #14

Claude Bile

Science Advisor

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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

Antiphon

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Claude Bile said:

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.

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

described for the infinite periodic structure. That is, there are

for which propagation cannot occur in an AR coating let alone a range of them.

- #16

Claude Bile

Science Advisor

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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

Edgardo

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Hurkyl said:

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

?

In my above calculations, to be honest, I haven't really thought about the correct coefficients. You will note that [tex]|\Psi_3>[/tex] actually must have

a coefficient that also depends on [tex] \delta [/tex], 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 [tex]|\Psi_2>[/tex] actually stands for [tex]e^{i \delta} | \Psi _1>[/tex] according to my notation. The only "basis vectors" are [tex]|\Psi_1>[/tex] and [tex]|\Psi_3>[/tex], with

[tex]|\Psi_1>[/tex] reflection

[tex]|\Psi_3>[/tex] transmission

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Righeira

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