Understanding Light Waves: Reflection, Interference, and Perception

[sophiecentaur]

Light doesn't have its wavelength changed upon reflection. What possible mechanism could account for that? For reflection to occur, there must be phase continuity at the boundary. A change in wavelength (and hence, frequency) would not satisfy this condition for the reflected wave.
Granted, you can get an effect where em waves of one frequency can be absorbed, excite an atom or molecule into high level and that energy can be re-emitted in two or more energy jumps. But that is not reflection, it is scattering (in all directions) and it non coherent.
For reflection to occur, the waves hitting a surface interact with the bulk material and there is a coherent reconstruction of the wavefront according to Snell's Law (this happens even for diffuse reflections from multiple small surfaces but on a small scale).

If broadband 'white' light hits a surface that absorbs some wavelengths (don't call it "colours" because colour is a perceptual thing) then the reflected light will consist of a selected range of wavelengths that the eye will interpret as a colour. This is the principle of 'subtractive' colour synthesis - as opposed to additive synthesis which is how this colour display works. If the incident light is not white (Say it consists of mostly energy at mid and longer wavelengths so it would 'look' yellowish from the greens and reds) then the pigment on a 'green' surface will still absorb the long wavelengths as it did with white light and just leave green light to be reflected. If the surface had a 'cyan' colour (under white light), it would reflect medium to short wavelengths but, if only medium / long wavelength light (yellowish) fell on it, there would only be mid wavelength colours reflected so it would look greenish. Note, I use "ish" to describe the apparent colours because this description is not quantitative.

You are right to say that the interaction involves electrons - but not individual electrons, necessarily. It will be with the whole of the structure and not like the dreaded Hydrogen Atom, which is used as a model by people to 'explain' everything. In solids (metals particularly) the electrons are not only in the field of just one atom but exist in the company of many surrounding nuclei and electrons; their energy level is due to all of their neighbours.

 

[John15]

Have not come across Snells Law.
I do realize we are not talking about individual electrons.
What causes the absorbtion of certain wavelengths.
When we think of reflection we obviously think of mirrors which are built in a certain way.
What reflects the waves, the surface only consists of electrons and atomic neuclei, which do the waves bounce off.
Light does appear to be scattered in all directions.
Take our room shine a red laser pointer on the yellow wall you will see a red dot from whichever direction you look at it, you will not see a red dot on any of the other walls though and this must be because the light is scattered, shine it on a mirror and I believe you will see a red dot on another wall. Although I suppose a rough surface will scatter light in all directions.
Of course another important point may be the fact that to build a smooth picture the brain only needs I think 60 cycles per second (a 60 hertz screen) and light cycles millions of times a second so there are plenty of waves to go round, it should be possible to work out how many enter the eye depending on distance from object.

 

[Drakkith]

John your best bet is to pick up a book on basic optics. I recommend "Optics for Dummies". Also, the book "Absolutely Small", a book on Quantum Physics does a good job at explaining much of the QM stuff you are asking. Get these two and I guarantee you will learn a great deal.

 

[sophiecentaur]

Have not come across Snells Law.
I do realize we are not talking about individual electrons.
What causes the absorbtion of certain wavelengths.
When we think of reflection we obviously think of mirrors which are built in a certain way.
What reflects the waves, the surface only consists of electrons and atomic neuclei, which do the waves bounce off.
Light does appear to be scattered in all directions.
Take our room shine a red laser pointer on the yellow wall you will see a red dot from whichever direction you look at it, you will not see a red dot on any of the other walls though and this must be because the light is scattered, shine it on a mirror and I believe you will see a red dot on another wall. Although I suppose a rough surface will scatter light in all directions.
Of course another important point may be the fact that to build a smooth picture the brain only needs I think 60 cycles per second (a 60 hertz screen) and light cycles millions of times a second so there are plenty of waves to go round, it should be possible to work out how many enter the eye depending on distance from object.

Snells Law describes what happens as waves go across a boundary. If you haven't heard of it then I suggest you should start at square one and work towards the clever stuff.
Learn about Coherent and Diffuse reflection to explain why 'shiny' surfaces produce coherent images.

They bounce of ALL OF THEM, as I implied, previously - it is the whole distributed system of charges that is involved when light hits a solid. (As it is, aamof, when light hits an isolated Atom. If it weren't for the nucleus, the electron would behave differently.)

Just what do you mean by this statement? it makes no sense at all in terms of em theory, optics or quantum theory.

 

[Drakkith]

Of course another important point may be the fact that to build a smooth picture the brain only needs I think 60 cycles per second (a 60 hertz screen) and light cycles millions of times a second so there are plenty of waves to go round, it should be possible to work out how many enter the eye depending on distance from object.

You are confusing two different things. Hertz is a unit of measurement for periodic phenomenon. On a computer monitor or something similar, the screen refreshes 60 times per second, hence it is 60 hz. However, hertz is also used to describe the number of oscillations per second in a wave. The frequency of an EM wave is measured in hertz and it does not mean that you have a constant wave entering your eyes that oscillates at that frequency. The wave itself does oscillate at that frequency, but it only interacts with things in little bundles, or quanta, called photons. If an EM wave hits your retina, it will transfer ALL of its energy to whatever molecule it interacts with, not just part of it and not energy over time. That's what the quanta thing means, it's an all or nothing thing when it comes to energy transfer via photons. A brighter source of light has a higher intensity, meaning that it has more photons per second being emitted.

 

[cepheid]

The frequency of an EM wave is measured in hertz and it does not mean that you have a constant wave entering your eyes that oscillates at that frequency. The wave itself does oscillate at that frequency, but it only interacts with things in little bundles, or quanta, called photons. If an EM wave hits your retina, it will transfer ALL of its energy to whatever molecule it interacts with, not just part of it and not energy over time. That's what the quanta thing means, it's an all or nothing thing when it comes to energy transfer via photons. A brighter source of light has a higher intensity, meaning that it has more photons per second being emitted.

I think I take issue with this. You seem to be trying to combine a classical and a quantum picture in a strange way. Saying that *all* of the energy of the EM wave is transferred in a single photon is just wrong. Classically you have an EM wave, and its Poynting vector determines the rate of energy transmission per unit area (perpendicular to the propagation direction) and per unit time.

In the quantum picture, you have a stream of photons, and so if anything, the total energy of all the photons that are arriving per unit area and per unit time is equal to the number you would compute in the classical case using the Poynting vector. I'm pretty sure that that has to be the case.

So the energy carried by the wave in the classical picture is determined by the number of photons present in the quantum picture. Turn up the light intensity (EM wave amplitude), and what you are really doing is increasing the number of photons that arrive per unit area and per unit time.

 

[sophiecentaur]

"In the quantum picture, you have a stream of photons,"
Beware the 'little bullets' idea.

 

[Drakkith]

I think I take issue with this. You seem to be trying to combine a classical and a quantum picture in a strange way. Saying that *all* of the energy of the EM wave is transferred in a single photon is just wrong. Classically you have an EM wave, and its Poynting vector determines the rate of energy transmission per unit area (perpendicular to the propagation direction) and per unit time.

Perhaps my terminology is wrong. It's confusing when talking about waves AND photons all at once.(And I don't really know which one John15 means when he says "waves") Which would be better in this case? EM waves or photons? My idea of an EM wave is a single photon. Is that incorrect?

 

[John15]

What do I mean by plenty of waves to go round.
Light is reflected in all directions, so if you shine a beam of light on a wall the light will reflect in approx 1/2 sphere shape, the radius of the sphere is the distance between you and the wall from this you can work out the area covered by the reflected light when it reaches you using the number of waves per second you can then work out how many reflected waves per second will be in each square cm, taking the eye as 1 sq cm and the fact that the brain only needs a certain amount of information, that is what light carries, you can see how much information is reaching the eyes. As you get closer to something you see more detail because the waves are not so spread out, there are more per surface area, you lose colour vision in low light because there is not enough energy to carry the information. I must admit some of this is guesswork and I realize that the more information the better the guess.
Am I right in thinking coherant light is in step so as to speak and difuse more random.
I wish I was in the position to buy books but saying that I am trying to teach myself and finding information in understandable form is difficult. I watched a lecture on relativity and pythagoras theorem was turned into something really complicated, if I hadnt watched the lecture I would not have recognised it. The other problem is the symbols used and trying to work out what they represent and remembering them. Much seems basically fairly simple once you have worked out what is going on.

 

[Drakkith]

As you get closer to something you see more detail because the waves are not so spread out, there are more per surface area, you lose colour vision in low light because there is not enough energy to carry the information. I must admit some of this is guesswork and I realize that the more information the better the guess.

The increase in detail when you get closer to an object is simply because our eyes resolution is only so high. I won't go into detail on why, as that's a whole other thread in itself. Adding more light, by increasing the brightness, doesn't help unless you are in low light levels.

Am I right in thinking coherant light is in step so as to speak and difuse more random.

Coherent light is in phase and the same frequency. I guess you could say it is "in step".

I wish I was in the position to buy books but saying that I am trying to teach myself and finding information in understandable form is difficult. I watched a lecture on relativity and pythagoras theorem was turned into something really complicated, if I hadnt watched the lecture I would not have recognised it. The other problem is the symbols used and trying to work out what they represent and remembering them. Much seems basically fairly simple once you have worked out what is going on.

No worries. Wikipedia is a fantastic source of information for myself. The key is learning to piece things from all the different articles together in a way you understand.

 

[sophiecentaur]

Perhaps my terminology is wrong. It's confusing when talking about waves AND photons all at once.(And I don't really know which one John15 means when he says "waves") Which would be better in this case? EM waves or photons? My idea of an EM wave is a single photon. Is that incorrect?

That is totally incorrect. (Unbelievably so, in fact. Owch Owch - sorry)

If you want to keep both in your head at once, you could say that the wave picture can tell you where you are most likely to encounter a photon - e.g. the peaks and nulls of an interference pattern. A photon is the amount of energy that will be absorbed or emitted by 'anything' (an atom, radio receiver), in one go. That amount of energy, or Quantum, is equal to hf, where h is the Planck constant and f is the frequency.
If you read this thread you will see that people talk in terms of intensity relating to the actual number of photons arriving. They don't mention wavelength in the same breath.

 

[Drakkith]

That is totally incorrect. (Unbelievably so, in fact. Owch Owch - sorry)

And here I thought I was finally getting to know and understand my friendly neighborhood photons!

 

[cepheid]

That is totally incorrect. (Unbelievably so, in fact. Owch Owch - sorry)

If you want to keep both in your head at once, you could say that the wave picture can tell you where you are most likely to encounter a photon - e.g. the peaks and nulls of an interference pattern. A photon is the amount of energy that will be absorbed or emitted by 'anything' (an atom, radio receiver), in one go. That amount of energy, or Quantum, is equal to hf, where h is the Planck constant and f is the frequency.
If you read this thread you will see that people talk in terms of intensity relating to the actual number of photons arriving. They don't mention wavelength in the same breath.

Isn't this essentially what I said? Especially the part about intensity being proportional to photon flux, which was my main point. What was wrong with saying that in quantum mechanics, the idea of a continuous electromagnetic wave has been replaced with the idea of a stream of a large but discrete number of individual photons?

 

[sophiecentaur]

What do I mean by plenty of waves to go round.
Light is reflected in all directions, so if you shine a beam of light on a wall the light will reflect in approx 1/2 sphere shape, the radius of the sphere is the distance between you and the wall from this you can work out the area covered by the reflected light when it reaches you using the number of waves per second you can then work out how many reflected waves per second will be in each square cm, taking the eye as 1 sq cm and the fact that the brain only needs a certain amount of information, that is what light carries, you can see how much information is reaching the eyes. As you get closer to something you see more detail because the waves are not so spread out, there are more per surface area, you lose colour vision in low light because there is not enough energy to carry the information. I must admit some of this is guesswork and I realize that the more information the better the guess.
Am I right in thinking coherant light is in step so as to speak and difuse more random.
I wish I was in the position to buy books but saying that I am trying to teach myself and finding information in understandable form is difficult. I watched a lecture on relativity and pythagoras theorem was turned into something really complicated, if I hadnt watched the lecture I would not have recognised it. The other problem is the symbols used and trying to work out what they represent and remembering them. Much seems basically fairly simple once you have worked out what is going on.

You don't have "plenty of waves" except in as far as waves can arrive from "plenty of" sources one wave will spread out from one source. A wave going past you can be the sum of waves at different frequencies and from different sources (it always is, aamof). A radio receiver or an optical receptor will have sensitivity to a certain range of frequencies and it will need a certain minimum rate of them arriving in order to get the wanted information. A colour sensor needs many photons (in the appropriate wavelength range) to arrive before it decides to fire. Rod receptors are thought to respond to just a single photon - after your retina is fully dark adapted. The power gathering of the lens and pupil relates to some of what you say, above - the iris opens and closes in order to admit the optimum power for the sensors to work.

The most familiar source of coherent EM waves is a radio transmitter. The wavefronts in all directions are 'in phase'. Most light sources are not coherent because the atoms that emit each photon are 'going off' at random. This non-coherence makes no difference to forming images in mirrors and lenses (eyes etc) and we can see an object because the power from all the waves arriving from the illumination or the reflection all ends up in the right place on a retina etc.. The non-coherence is, in fact, quite an advantage as it is less confusing. If you see reflected light from a Laser, however, you get an annoying speckled pattern because of interference effects due to the steady phase relationship for the incident waves. Likewise, for radio transmissions, you can get bothersome 'multipath effects' as different coherent signals arrive and produce nulls and peaks; you may have noticed this when driving around in a car.

Wave coherence is not the opposite of diffuse image formation. Diffuse is the opposite to 'specular' reflection or transmission. Specular reflection happens on a 'shiny' surface and the waves all add up to give a meaningful (sometimes referred to as coherent but the word is used differently here) image. In diffuse reflection, each part of the reflecting surface is at a different (random) angle and so neighbouring parts of a rough piece of paper will not give an identifiable image.

On the subject of Book Substitutes. There are a lot of ".edu" sites that give explanations at all levels. You need to do a lot of trawling round to find one that suits you. Warning: there are several sites that are just B S, so check between a number of them for agreement.
before you believe what you read. ".edu" is usually reliable, though - just could be too hard!

One great risk of doing this in 'your way' is that Science learning is easier when you follow some sort of 'programme' of study, which can present you with ideas in a more understandable way. Random dipping into the pool will tend to confuse you.

 

[sophiecentaur]

Isn't this essentially what I said? Especially the part about intensity being proportional to photon flux, which was my main point. What was wrong with saying that in quantum mechanics, the idea of a continuous electromagnetic wave has been replaced with the idea of a stream of a large but discrete number of individual photons?

That's fair enough. I wasn't telling you that was wrong was I? Except that nothing has been "replaced". The two go along in parallel. You can't describe how a lens works using QM, unless you have a great deal of spare time!
Trouble here is that the responses are coming thick and fast and it's not always clear who's arguing with whom.

 

[John15]

Thanks for the input on my education. I have read all the books available in my local library and have been trawling through lectures on youtube. Unfotunately this is all one way with no opportunity to question anything which is where sites like this and people like yourself come in. It is difficult though to find people with the knowledge and patience willing to talk to people like myself, so as you can imagine finding people like yourself and drakkith is rewarding in itself. You my though have difficulty at times in following my line of thinking, I am looking for the whys like why is c constant and the universal speed limit why are the natural constants what they are. Being self taught I have no preconcieved ideas I just look at the info and say what I see not what I have been taught, which may be right or wrong depending on how much data I have to work on.
I believe that the answers are out there, not in string theory but maybe in reinterpreting what we think we know, I notice many coflicting posts on sited like this and from people who seem to know what they are talking about.
Back to my sphere and plenty of waves, it would explain why you lose detail as you move further away ie less energy (waves ) reaching your eyes from a certain point as they spread out and why you see more detail the brighter the light.

 

[sophiecentaur]

As with any measurement / perception, the precision will depend upon signal to noise ratio. In low light levels, when the power is more spread out, there will be less signal power to analyse in the presence of unwanted perturbations. Henct the accuracy / precision will be worse.

 

[Drakkith]

Back to my sphere and plenty of waves, it would explain why you lose detail as you move further away ie less energy (waves ) reaching your eyes from a certain point as they spread out and why you see more detail the brighter the light.

No, it doesn't. Move a certain distance away from an object with lots of fine detail. You will see the detail fade away as you get further away. Repeat with a brighter light source and you will NOT see more detail at the same distance. As long as you didn't start with a very dim light source to begin with this will work.

The resolution of your eye is based on the minimum spot size it can focus light to and the size and density of the cone/rod cells in your eye. Once the light from two adjacent spots gets under a certain angular separation (how far apart they appear to be from a distance) both spots will be focused onto too few cells or the light from them will be blurred together, rendering them unable to be seen as separate spots. I believe the human eye is limited in resolution mostly by the size and density of the cells, and less from the best spot size it can focus light too. At least until aberration show up and you have things like astigmatism and near/far sightedness appear.

http://en.wikipedia.org/wiki/Visual_acuity
http://en.wikipedia.org/wiki/Optical_resolution

 

[sophiecentaur]

No, it doesn't. Move a certain distance away from an object with lots of fine detail. You will see the detail fade away as you get further away. Repeat with a brighter light source and you will NOT see more detail at the same distance. As long as you didn't start with a very dim light source to begin with this will work.

The resolution of your eye is based on the minimum spot size it can focus light to and the size and density of the cone/rod cells in your eye. Once the light from two adjacent spots gets under a certain angular separation (how far apart they appear to be from a distance) both spots will be focused onto too few cells or the light from them will be blurred together, rendering them unable to be seen as separate spots. I believe the human eye is limited in resolution mostly by the size and density of the cells, and less from the best spot size it can focus light too. At least until aberration show up and you have things like astigmatism and near/far sightedness appear.

http://en.wikipedia.org/wiki/Visual_acuity
http://en.wikipedia.org/wiki/Optical_resolution

The eye is just one example of a system for measurement using light. Its 'design' is based on (you know I don't mean that literally) on diffraction limits and practical requirements (evolution). In general, the resolving power of a system is based on signal strength. The Rayleigh criterion is only a rule of thumb (much to much of a 'round figure' to be taken too seriously) but basic Shannon information theory tells us that the only limit to resolving power is not due to diffraction because increasing signal to noise ratio can take you beyond this limit. Yes it does / would involve more and more complex signal processing but it's already done when TV images are enhanced and in radio telescope arrays.
So you are right where the eye is concerned but it isn't the case in every situation.

 

[Drakkith]

So you are right where the eye is concerned but it isn't the case in every situation.

Just using the eye since that was the OP's example. Believe me, I know about SNR. I do astrophotography.

 

[John15]

I speak about the detail from personal experience, I can read a paper outside in bright sunlight go indoors and I need glasses, the only difference is the light intensity, if this was all down to focus paper would look the same whatever brightness of light.
Have been looking back through thread. A couple of things we talked about thermal radiation, post 27, anything with a temp above absolute 0 will emit EM radiation, of course in order to emit it has to absorb, wavelength also relates to temp so all things emit EM radiation in addition to any reflected light. All things that emit light are hot so all natural light must start as thermal radiation, ignoring chemical light.
Post 41 - 46 relate to photons energy of photon E=hf quote we don't talk about wavelength and photons together, frequency is dependant on wavelength so how do we separate wavelength and photons?
From what I have read it seems to me that quanta are waves with sufficient energy to react with electrons, as electrons have different energy levels the energy of the photon depends on the electron it interacts with.
Take the EM spectrum as a ramp then photons are steps in that ramp a specific places, they cause a form of phase transition in matter, they also join the world of infinite wavelengths with the finite world of matter, all other waves react indirectly as an increase in temp which reacts with all levels of matter.
I have been trying to find out about absorbtion spectra, specifically how much of the EM spectra is absorbed when all the lines are added together.
I have also been trying to find out how individual atoms react to waves that don't qualify as photons, do they just cause vibrations that are re-emitted as heat.

 

[Drakkith]

I speak about the detail from personal experience, I can read a paper outside in bright sunlight go indoors and I need glasses, the only difference is the light intensity, if this was all down to focus paper would look the same whatever brightness of light.

I think that's due to a much greater amount of contrast between the text and the paper thanks to the greater amount of light. This isn't "resolution" in the sense I was talking about, but more of a signal vs noise issue as Sophiecentaur explained, not anything related to "waves spreading out".

I have also been trying to find out how individual atoms react to waves that don't qualify as photons, do they just cause vibrations that are re-emitted as heat.

ALL EM waves qualify as photons. That is, all EM waves transport their energy in quanta.

 

[sophiecentaur]

I speak about the detail from personal experience, I can read a paper outside in bright sunlight go indoors and I need glasses, the only difference is the light intensity, if this was all down to focus paper would look the same whatever brightness of light.
Have been looking back through thread. A couple of things we talked about thermal radiation, post 27, anything with a temp above absolute 0 will emit EM radiation, of course in order to emit it has to absorb, wavelength also relates to temp so all things emit EM radiation in addition to any reflected light. All things that emit light are hot so all natural light must start as thermal radiation, ignoring chemical light.
Post 41 - 46 relate to photons energy of photon E=hf quote we don't talk about wavelength and photons together, frequency is dependant on wavelength so how do we separate wavelength and photons?
From what I have read it seems to me that quanta are waves with sufficient energy to react with electrons, as electrons have different energy levels the energy of the photon depends on the electron it interacts with.
Take the EM spectrum as a ramp then photons are steps in that ramp a specific places, they cause a form of phase transition in matter, they also join the world of infinite wavelengths with the finite world of matter, all other waves react indirectly as an increase in temp which reacts with all levels of matter.
I have been trying to find out about absorbtion spectra, specifically how much of the EM spectra is absorbed when all the lines are added together.
I have also been trying to find out how individual atoms react to waves that don't qualify as photons, do they just cause vibrations that are re-emitted as heat.

It has to be pointed out, from the start, that the whole business of reflection, scattering and the interaction of waves with solids is very complex. You won't get it until you have the basics all sorted out and understand what is meant by all the terms. You then need to get familiar with what happens when EM waves interact with single atoms. If you still have the time and energy, you can then move on to solids! (No such thing a s free lunch).

I suggest that you haven't actually read that but you are paraphrasing in an inaccurate way. Quanta are not waves and no one says they are. You could say that a photon is 'part of' a wave, in as far as that is the Quantum of the wave's energy that can be absorbed by a charge system. (No more and no less energy can be involved)
You can look upon a wave as a map, in space, of how the energy is being transported. The way the E and H fields vary in space and time follows the equation of the wave (the wiggly graph, if you like). That wave is the same, irrespective of the intensity / total energy flux.


This makes no sense at all. Waves don't "qualify as photons". Whenever there is an EM wave, there will be photons and they are all identical, it the wave just consists of one frequency (monochromatic). The only way that a wave can interact is via photons. When light hits a solid, you can get interactions with the system as a whole (in which case you are dealing with a system with large number of atoms which will have a concerted effect and this effect will be on the wave as a whole (a 'macroscopic effect, if you like). This will give you a reflection. If the surface is smooth, the resulting wave shape will be like that of the incident ray (wavefront) and you'll get a nice, specular reflection. If the surface is rough , there may be many reflections at different angles that will give a diffuse beam. A wave can also go through a solid without any great level of interaction. It can just get slowed down by the bulk effect. This accounts for refraction. Solids often absorb certain wavelengths, of course (glass tends to absorb IR and UV, for instance) but pass others. But impurities can be put into glass which will interact at the atomic level with certain energies of the light and absorb certain wavelengths (pigments).

Here, again, you have read some stuff and are trying to regurgitate in your own words. It's not really working. There is a continuous range of possible frequencies. Each frequency is associated with a different photon energy. Photons aren't "steps". This would be like saying that a piece of string is 'made up of inches'. Yes, you can say that it is possible to give a length to the nearest inch but that's all. The term 'spectrum' just refers to the relative intensity levels over a (continuous) range of frequencies.

All things emit some EM radiation - not necessarily light, as it depends on their temperature. Plenty of natural sources of light are not 'hot'. Stars etc are hot enough but cold clouds of gas can absorb and re-emit light of specific frequencies in our direction. And why 'ignore chemical light'? It follows the same rules as 'thermally generated light'. You are trying to over simplify to no real purpose. As for the appearance of paper in the light or in the shade- the light that comes from it varies under different circumstance. Your BRAIN works overtime to give you the impression that it is still the same piece of paper and it even does a fair job of colour correction to allow for time of day, or varying illumination. What you 'see' is not a measurement. For that, look at what you get on a digital camera that has all its controls set to 'manual'. That is more of a proper measurement.

 

[jim hardy]

if this was all down to focus paper would look the same whatever brightness of light.

it should be noted that in bright light our iris closes down a few f-stops and just as in a camera lens we experience the better optics of the smaller aperture.