Light & Sight

John15

Still thinking about reflected light. It must be absorbed by whatever physical thing it interacts with and be re-emitted.
Take a room with different coloured walls say red blue green and yellow, the light is bouncing off all walls in all directions yet we only see the walls in thier individual colours so light bouncing off the red wall onto the yellow wall must have its wavelength changed from red to yellow and so on before it reaches our eyes, we dont see a mix of red and yellow light so is reflection really the right way of looking at it?
Would I be right in thinking the truth lies at the atomic level and is to do with the energy of electrons?

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 realise 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

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

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

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

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

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.

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

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

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

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

cepheid

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

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

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.