How do Mirrors work on nano scales ?

[Chaos' lil bro Order]

Hello,

I have a question about mirrors.
Wikipedia says, 'Most modern mirrors consist of a thin layer of aluminium deposited at the back of a sheet of glass.'

My question is, how does the aluminum at the atomic scale reflect the incident light so precisely? In other words, IF incident photons excite the aluminum's electrons, why do the deexcitations send photons incident at the same angle to the orthogonal as the original?
Please keep all discussion to microscopic levels as I know how mirrors work macroscopically.

Thanks.

 

[pervect]

One can come up with microscopic examples where the angle of incidence is not equal to the angle of reflection, i.e. the diffraction grating, as discussed by Feynman in his popular physics book, "QED". The angle of reflection in this case depends on the angle of incidence, the wavelength, and the spacing and size of the grating.

A diffraction grating is just a mirror with "missing pieces".

One can therefore conclude from the example of the diffraction grating that reflection process cannot be fully modeled on the mircroscopic level. (Otherwise the absence of certain distant sections of the mirror would have no effect, but we see from the example of the diffraction grating that it does.)

 

[rbj]

A diffraction grating is just a mirror with "missing pieces".

missing pieces at regularly spaced intervals, no?

 

[pervect]

missing pieces at regularly spaced intervals, no?

Yep. The fact that the missing pieces are at regular interval are not particularly relevant to the details of Feynman's argument, though. It's just a partiularly convenient special case of "mirrors missing pieces" where it is particularly easy to calculate the results.

The results clearly indicate that one must consider all pieces of the mirror to get the correct reflection rules, because, on the microscopic scale at least, the absence of particular pieces can affect the results.

 

[inha]

Read the FAQ in this forum which deals with the visible light photon energy range and transmission through glass first. Electronic excitations are of the keV order in most materials whereas the visible light photons are at the eV order of magnitude.

A mirror is "missing pieces at regularly places intervals" anyways I think. This is based on my guess that the deposited aluminum assumes a crystalline structure on the glass substrate. The diffraction grating comparison seems sort of pointless as normal lattice spacings of metals are of the order of 1Å while visible light has wavelengths of the order of 1000Å.

 

[ZapperZ]

I agree with inha.

I hesitated in replying to this thread for a number of reasons, one of which is that I see no end in the discussion because this thing CAN get very complex.

A mirror being nothing more than a diffraction grating (with missing pieces) is not accurate. This is more of an analogy to Bragg reflection, NOT mirror reflection. I have attempted to describe the difficulty in this thread:

https://www.physicsforums.com/showthread.php?t=116678

Mirror reflection within the typical visible range on metallic surfaces (this IS the premise in the OP, isn't it?) has nothing to do with the ions that form the metal. It has everything to do with the conduction electrons. The explanation involves the knowledge of Fermi surface, Brillouin zones, lattice structure, and the electronic band structure. In other words, a lot of solid state physics. This is because it requires the conduction electron to make a transition from one conduction band to the next conduction band that is separated by eactly one reciprocal lattice vector G. This is what preserves the in-plane momentum.

Zz.

 

[Physics Monkey]

I believe we are talking about two different and equally important aspects of the reflection phenomenon.

ZapperZ and inha have pointed out quite correctly that the microscopic details tell you a lot about reflection and transmission. These details allow you to calculate amplitudes for transmission and reflection, as well as modifications to the laws of propagation as encoded in the index of refraction, say. In metals, for example, the dominant players may be conduction electrons, where as in insulators, it is the phonons or point defects that may be most important. In an atomic vapor, it is the individual atomic transitions that are relevant.

pervect has drawn our attention to the other side of the issue, namely that being a mirror has to do with "global" properties too. In the language of photons, there is an amplitude for a photon to go from one place to another and these amplitudes must all be added to calculate the probability of reflection. If one thinks about free space, the path integral produces the rectilinear propagation we all know from ray optics. You can then say diffraction occurs when you restrict the allowed photon paths so much that the cancellation which gives rectilinear propagation becomes noticeably incomplete. Going back to mirrors, we can cut little uniform strips out of the mirror in just the right way (which turns out, of course, to be nothing but the Bragg condition), and the mirror stops acting like a ordinary mirror and becomes a diffraction grating.

I think both parts of the picture are fascinating and important.

 

[pervect]

I agree with inha.

I hesitated in replying to this thread for a number of reasons, one of which is that I see no end in the discussion because this thing CAN get very complex.

A mirror being nothing more than a diffraction grating (with missing pieces) is not accurate. This is more of an analogy to Bragg reflection, NOT mirror reflection. I have attempted to describe the difficulty in this thread:

https://www.physicsforums.com/showthread.php?t=116678

I'll agree that things can get very complex.

However, I have to disagree that a diffraction grating is not a mirror with missing pieces.

I.e. if you take the following pattternxxxxxxxxxx
ooooooooo
xxxxxxxxxx
ooooooooo
xxxxxxxxxx
ooooooooo
xxxxxxxxxx

you physically form a diffraction grating by removing the pieces of the mirror marked oooooooooooo from the solid mirror.The purpose of the example is to illustrate the complexity of the phenomon of reflection. (And, I should add, it's not my original idea, it's straight from Feynman).

One of the simplest analyses I'm aware of is Feynman's popularization in QED, where one considers the photons to be point particles which somehow take all possible paths. The example of the diffraction grating is brought up to illustrate why something like this is necessary.

 

[ZapperZ]

I'll agree that things can get very complex.

However, I have to disagree that a diffraction grating is not a mirror with missing pieces.

I.e. if you take the following patttern


xxxxxxxxxx
ooooooooo
xxxxxxxxxx
ooooooooo
xxxxxxxxxx
ooooooooo
xxxxxxxxxx

you physically form a diffraction grating by removing the pieces of the mirror marked oooooooooooo from the solid mirror.


The purpose of the example is to illustrate the complexity of the phenomon of reflection. (And, I should add, it's not my original idea, it's straight from Feynman).

One of the simplest analyses I'm aware of is Feynman's popularization in QED, where one considers the photons to be point particles which somehow take all possible paths. The example of the diffraction grating is brought up to illustrate why something like this is necessary.

Do you get a "reflection", or an interference pattern?

Again, I mentioned earlier that this is an example of a BRAGG reflection. I get the same thing when I do this onto crystal planes.

But this is not the same as a reflection of a mirrored surface. The physics is different. You don't get ordered interference.

Zz.

 

[pervect]

The physics is what we are trying to explore, the interaction of light with a mirrored surface.

The point I'm trying to make (borrowed from Feynman - have you read QED? He probably expains it better than I do) is that we can't consider an arbitrarily small mirror act in such a manner that the angle of reflection equal to an angle of incidence. The diffraction grating (a series of equally spaced small mirrors) illustrates this via example. The rule "angle of incidence = angle of reflection" is thus only an approximation to the actual physics.

 

[ZapperZ]

The physics is what we are trying to explore, the interaction of light with a mirrored surface.

The point I'm trying to make (borrowed from Feynman - have you read QED? He probably expains it better than I do) is that we can't consider an arbitrarily small mirror act in such a manner that the angle of reflection equal to an angle of incidence. The diffraction grating (a series of equally spaced small mirrors) illustrates this via example. The rule "angle of incidence = angle of reflection" is thus only an approximation to the actual physics.

Yes I have, and that is why I brought up the Bragg reflection.

A diffraction grating isn't made up of "mirrors". It can be made up of anything as long as you put many, many slits into it. All you need to do is LOOK at the reflected image. For example, put a discharge tube near it and look at the "reflected" light. Do you see the image of the discharge tube? Now replace it with an ordinary mirror. Is there a distinct difference? I'll bet you $100 there is. Then why is the analogy between them valid?

Just because something "reflects" doesn't automatically implies it is the SAME type of reflection via the identical mechanism. A reflection off a smooth dielectric surface is NOT identical to a reflection off a thin film of aluminum.

My prediction in my very first post in this thread has come to fruition.

I hesitated in replying to this thread for a number of reasons, one of which is that I see no end in the discussion because this thing CAN get very complex.

Zz.

 

[pervect]

Yes I have, and that is why I brought up the Bragg reflection.

A diffraction grating isn't made up of "mirrors". It can be made up of anything as long as you put many, many slits into it.

I don't recall claiming that a diffraction grating HAD to be made out of mirrors. For my purposes it is sufficient that a diffraction grating can be made out of mirrors.

I'll add a few quotes direct from Feynman (please excuse the typos) but I too suspect that we have some irreconcilable differences in our views here. I'm not quite sure what the problem is, though.

Fenyman's "arrows" can be more-or-less considered to be the wavefunctions - for more details, please see the original text.

QED, pg 38

We start with a mirror, and the problem of determining how light is refelcted from it (see Fig 19).

S       Q        P

___________________________

fig 19, in ascii.  S is a source, Q is a screen, P is a detector, 
and the horizontal line is a mirror.

At S we have a source that emits light of one color at very low intensity (let's use red light again). The source emits one photon at a time. At P we place a photomultiplier to detect photons. Let's put it at the same height as the source -- drawing arrows will be easier if everything is symmetrical. We want to calculate the chance that the detector will make a click after a photon has been emitted by the source. Since it is possible that a photon could go straight across to the detector, lets's place a screen at Q to prevent that.

Now, we would expect that all the light that reaches the detector reflects of the middle of the mirror, because that's the place where the angle of incidence equals the angle of reflection. Ant it seems fairly obvious that the parts of the mirror out near the two ends have as much to do with the reflection as with the price of cheese, right?

Althoutgh you might think that the parts of the mirror near the two ends have nothing to do with the reflection of the light that goes from the source to the detector, let's look at what quantum theory has to say. Rule: The probability that a particular event occurs is the square of a final arrow that is found by drawing an arrow for each way the event could happen, and then combining ("adding") the arrows. In the experirment measuring the partial reflection of light by two surfaces, there were two ways a photon could get from the source to the detector. In this experiment, there are millions of ways a photon could go:

(examples omitted)

You might think I'm crazy, because for most of the ways I tod you a photon could reflect off the mirror, the angles aren't equal. But I'm not crazy, because that's the way light really goes! How can this be?

Now we skip on a bit, to pg 46...

To test the idea that there is really reflection happening at the ends of the mirror (but it is just cancelling out), we do an experiment with a large piece of mirror that is located in the wrong place for reflection of S to P. This piece of mirror is divided into much smaller sections, so that the timing from one path to the next is not very different. When all the arrows are added, they go nowhere: they go in a circle and add up to nearly nothing.

But let's suppose we carefully scarpe the mirror away in those areas whose arrows have a bias in one direction -- let's say, to the left -- so that only those places whose arrows point generally the other way remain (see fig 26). When we add up only the arrows that point more or less to the right, we get a series of dips and a substantial final arrow -- according to the theory, we should now have a strong reflection! And indeed, we do -- the theory is correct! Such a mirror is called a diffraction grating, and it works like a charm.

As far as the details go: reflection off the front of a dielectric inverts the phase of the wavefunction (i.e. rotates Feynamn's arrows 180 degrees), reflection from the rear of a dielectric does not change the phase. (This is a simplification, actualy, as Feynman explains on pg 16, but it's a simplification that gives the right answer).

It's unclear what the phase is for front reflection from a metal surface from my re-reading of the text.

 

[Physics Monkey]

Just because something "reflects" doesn't automatically implies it is the SAME type of reflection via the identical mechanism. A reflection off a smooth dielectric surface is NOT identical to a reflection off a thin film of aluminum.

Hi ZapperZ,

I'm fairly certain that no one is suggesting that all kinds of reflection happen by the same mechanism. That's clearly not right. The idea that pervect nicely pointed out (and that I tried to support) is that without changing the microscopic details , you can convert a mirror into a diffraction grating. That's all the point I was making.

My prediction in my very first post in this thread has come to fruition.
Zz.

That's not a bad thing is it? I mean, this is cool stuff, so I would have thought that a nice discussion would be fun.

EDIT: I see pervect beat me to it, and with a better discussion too.

 

[ZapperZ]

But what you are describing does not describe a "mirror" reflection. It is describing Feynman's idea of path integral. And this has nothing to do with a "diffraction grating". Again, if you don't believe me, look at the "reflection" off a diffraction grating. You'll see that this is nowhere near the ordinary reflection off a mirror surface. So I do not see how it can be used as the microscopic explanation.

Zz.

 

[pervect]

For example, put a discharge tube near it and look at the "reflected" light. Do you see the image of the discharge tube? Now replace it with an ordinary mirror. Is there a distinct difference? I'll bet you $100 there is. Then why is the analogy between them valid?
Zz.

I have to go for a bit, but I think your point may be that the diffraction grating can't create a good image because the coefficient of reflection depends on the angle, and that you need the reflection to be uniform vs angle in order to create a good image?

 

[ZapperZ]

I see somehow that either people are missing the question, or I'm reading way too much into it.

When someone asks "how does a photon gets reflected", it tell me that the question asked for the microscopic mechanism of how a photon, hitting a surface, will come OUT afterwards. The feynman quote above tells you NOTHING about this. All it says is why, AFTER the photon comes off the surface, that it goes that one particular direction.

Did I get this right?

My approach was the former - WHY did that photon comes out in the first place, and WHY did it come out in THAT direction? If the photon is absorbed, Feynman's explanation is MOOT!

Did I get THAT right?

Zz.

 

[Physics Monkey]

But what you are describing does not describe a "mirror" reflection. It is describing Feynman's idea of path integral. And this has nothing to do with a "diffraction grating". Again, if you don't believe me, look at the "reflection" off a diffraction grating. You'll see that this is nowhere near the ordinary reflection off a mirror surface. So I do not see how it can be used as the microscopic explanation.

Zz.

I am not saying a mirror is the same as a diffraction grating. Quite the opposite in fact. A mirror can be changed into a diffraction grating by removing pieces. This is not a microscopic explanation, it is saying that there is more to being a mirror than microscopics. Also, I don't think you are correct to say that mirror with pieces cut out has nothing to do with a diffraction grating. It is a diffraction grating.

 

[Physics Monkey]

I see somehow that either people are missing the question, or I'm reading way too much into it.

I think we are all talking around each other. I tried to indicate in my first post that various parties are answering two different aspects of the same question. That's all.

 

[ZapperZ]

I am not saying a mirror is the same as a diffraction grating. Quite the opposite in fact. A mirror can be changed into a diffraction grating by removing pieces. This is not a microscopic explanation, it is saying that there is more to being a mirror than microscopics. Also, I don't think you are correct to say that mirror with pieces cut out has nothing to do with a diffraction grating. It is a diffraction grating.

I don't understand since your statements appear to contradict each other. You're not saying a mirror is the same as a diffraction grating, but then it IS a diffraction grating.

You can always prove me wrong when I say that a diffraction grating reflection is not the same as the mirror reflection by LOOKING at the reflected image. If it looks the same to you, then I quit.

Zz.

 

[Physics Monkey]

I don't understand since your statements appear to contradict each other. You're not saying a mirror is the same as a diffraction grating, but then it IS a diffraction grating.

I don't understand, how am I contradicting myself? I said take a mirror, a nice continuous mirror like in your bathroom, and change it . Specifically, cut regular strips out of the mirror. What you have left over is a diffraction grating that behaves in a physically different way from the original mirror even though the microscopic details have not changed.

 

[ZapperZ]

I don't understand, how am I contradicting myself? I said take a mirror, a nice continuous mirror like in your bathroom, and change it . Specifically, cut regular strips out of the mirror. What you have left over is a diffraction grating that behaves in a physically different way from the original mirror even though the microscopic details have not changed.

If that's the case, then I don't see what this argument is all about, i.e. why is the diffraction grating brought up in the first place. After all, the original question DID ask for the microscopic details, no? If a diffraction grating adds nothing to the microscopic explanation, what was its use again?

I still stand by my original explanation. Whether you use a grating, a mirror, a dielectric, a jello, etc... you HAVE to explain how the incoming photon hits a surface, and THEN a photon comes out of that surface, and THEN explain why it comes out in THAT particular direction. Ignoring the mechanism of the material involved in the reflection process misses a BIG factor.

Zz.

 

[Physics Monkey]

If that's the case, then I don't see what this argument is all about, i.e. why is the diffraction grating brought up in the first place. After all, the original question DID ask for the microscopic details, no? If a diffraction grating adds nothing to the microscopic explanation, what was its use again?

Again, I'm not sure why were arguing either since we're answering different parts of the question. The OP did ask about microscopic details, specifically why the angle of incidence equals the angle of reflection. The point of bringing up the diffraction grating was to note that it is not just the microscopic details that determine this fact. pervect didn't talk about microscopics, but you and inha both gave nice sketches. However, and I can only speak for myself, the tone of the original post seemed to suggest that the OP thought that it was only the microscopic details that mattered. I first posted in an attempt to emphasize the importance of both aspects of the problem.

 

[pervect]

If that's the case, then I don't see what this argument is all about, i.e. why is the diffraction grating brought up in the first place. After all, the original question DID ask for the microscopic details, no? If a diffraction grating adds nothing to the microscopic explanation, what was its use again?

I still stand by my original explanation. Whether you use a grating, a mirror, a dielectric, a jello, etc... you HAVE to explain how the incoming photon hits a surface, and THEN a photon comes out of that surface, and THEN explain why it comes out in THAT particular direction. Ignoring the mechanism of the material involved in the reflection process misses a BIG factor.

Zz.

Let's see if I can fill in some of the logic chains that prompted my answer.

The OP asks about the "microscopic details". What does that mean? Well, I would guess he might be asking about a microscopic mirror. What is "microscopic"? In this context, it seems logical to guess that he's thinking of a mirror that's smaller than the wavelength of light.

Well, such a microscopic mirror would radiate equally in all directions, because of diffraction.

Now, we know that large mirros don't do this, that for large mirrors we have the angle of incidence equal to the angle of reflection. And the OP was asking specifically about this point (why the angle of incidence was equal to the angle of reflection).

Feynman's answer in QED (which I've already talked about enough, I think) seems like the perfect explanation for why this happens - how one smoothly goes from an array of point mirors, each of which reflects equally in all directions, to the expected behavior of a relatively large uniform mirror (large, in this context, meaning much greater than 1 wavelength) where the angle of incidence is equal to the angle of reflection.

 

[Chaos' lil bro Order]

Can we answer a simple question?

Hello,

I have a question about mirrors.
Wikipedia says, 'Most modern mirrors consist of a thin layer of aluminium deposited at the back of a sheet of glass.'

My question is, how does the aluminum at the atomic scale reflect the incident light so precisely? In other words, IF incident photons excite the aluminum's electrons, why do the deexcitations send photons incident at the same angle to the orthogonal as the original?
Please keep all discussion to microscopic levels as I know how mirrors work macroscopically.

Thanks.

THIS IS FRUSTRATING THE HELL OUT OF ME. Why must be spend 20 posts arguing whether a diffraction grating can be made out of a mirror? READ THE FREAKING QUESTION I POSTED AND PUT ASIDE EGOS and let's work this out together.

I specified that we are talking about a mirror with 'aluminum powder' backing which is responsible for the reflection. Assume that 1 cm of Crown Glass is set atop the backing and that the mirror is planar and perfectly uniform (contains no bumps or curvatures). If I send photons incident at the mirror at 40 degrees to the orthogonal, the reflected image is seen at 40 degrees on the other side of the orthogonal. THAT'S IT, plain and simple. Where did diffraction gratings come into this? If you don't know the answer to my question, please don't rearrange to topics that you DO know about to save face. Let's learn together as a team, its more fun that way.

I think we can safely say that the Crown Glass plays a minor factor in our reflection and I could write a couple sentences on its role, but let's just focus on the 'powdered aluminum' since its our primary reflecting surface and therefore our primary interest for discussion.

If anyone knows the atomic structure of powdered aluminum, this would seem like a logical place to start in my opinion. As in, what is the crystalline structure of each powder fragment and how do all the fragments combined produce a reflected image.

I suspect the answer may have something to do with an incident photon striking the aluminum powder particles and due to their crystalline structure (guessing), the photon ejects an electron from somewhere in the matrix, whereupon there is only one path the ejected electron can go after all interactions in the matrix are complete, that path being perfectly reflected out at the same angle of incidence. Basically what I'm saying is, the structure of the crystal dictates the property of the aluminum powder that makes it reflect at the same angle of incidence.

Please offer opinions, corrections and add-ons.

Please let's stick to the properties of the powdered aluminum 'backing'.Thanks, excitedly awaiting...

 

[ZapperZ]

THIS IS FRUSTRATING THE HELL OUT OF ME. Why must be spend 20 posts arguing whether a diffraction grating can be made out of a mirror? READ THE FREAKING QUESTION I POSTED AND PUT ASIDE EGOS and let's work this out together.

And people wonder WHY I didn't want to respond to this thread in the first place...

Zz.

 

[inha]

Debating how things really are isn't what you'd working your question out?

 

[pervect]

THIS IS FRUSTRATING THE HELL OUT OF ME. Why must be spend 20 posts arguing whether a diffraction grating can be made out of a mirror? READ THE FREAKING QUESTION I POSTED AND PUT ASIDE EGOS and let's work this out together.

So let's go back to the original quesiton...

I have a question about mirrors.
Wikipedia says, 'Most modern mirrors consist of a thin layer of aluminium deposited at the back of a sheet of glass.'

My question is, how does the aluminum at the atomic scale reflect the incident light so precisely? In other words, IF incident photons excite the aluminum's electrons, why do the deexcitations send photons incident at the same angle to the orthogonal as the original?
Please keep all discussion to microscopic levels as I know how mirrors work macroscopically.

The sticking point here is what you mean by "microscopic details". We got the part about the mirror.

I attempted to answer the question in my post #12, the longish quote from a popular book, "QED" by Feynman. I'm still convinced this is the right answer, even after the input from other posters. (Of course, I'm not infallible, unfortunately).

The short version of the answer, to repeat the same point I made earlier, is that the photons reflect off the mirror at all angles, and interference effects due to the large scale structure of the mirror (not the microscopic structure!) are what cause the angle of incidence to equal the angle of refraction.

This brief argument is necessarily short and does not have all the details and diagrams of the book. Feynman really does explain it better than I do, he has diagrams to help, it's proofread, and it's much longer and better quality than anthing I can duplicate in a short post.

I would therefore strongly recommend reading the book if my brief descritpion has not been very clear. I think it will answer your question, and if it does not, it's still a very good book :-).

It's quite cheap, and should be readily available from libraries, too.

 

[Chaos' lil bro Order]

Hmm, guess I'll read the book.

 

[Chaos' lil bro Order]

By the way, Feynman is a personal hero of mine and I just so happen to have a rare 45 minute video interview with him on my harddrive. Is there a place on PF where I can upload it for people to get? I'm sure a lot of you would really like it.