Light & Sight


by John15
Tags: light, sight
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Feb14-12, 07:59 PM
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Quote Quote by Dadface View Post
The nature of light is such that it enables us to see things but we don't actually see light itself.
I can't say that I agree with this. When you see an object, it is because photons that are being emitted or reflected from the surface of that object are hitting your retina. The quanta of light (or, if you prefer, just the EM waves) are the things that are actually detected by your sensory equipment.
John15
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Feb15-12, 05:10 AM
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Quote Drakkith
This is similar to why sound waves do not normally interfere and keep you from hearing the speakers in your TV.

Of course enougth speakers playing different sounds from different directions and all you get would be noise.

So could it be said then that light travels as a wave but collapses to particle like on observation/ detection.
Wavelength determines colour and energy is linked to wavelength so how does amplitude work? Would I be right in thinking that amplitude enables the wave to carry more energy per wave i.e 2x amplitude gives twice energy or similar to 2 waves travelling in partnership.

Finally, I think, waves in 3d must propagate outwards spherically, as sphere grows energy should be spread out the same way a balloon gets thinner as its blown up, so taking the sun as source by the time light reaches earth we have a sphere with a radius of over 90 million miles, how does light maintain its integrity when spread out over such an area?
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Feb15-12, 06:13 AM
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Quote Quote by John15 View Post
Quote Drakkith
This is similar to why sound waves do not normally interfere and keep you from hearing the speakers in your TV.

Of course enougth speakers playing different sounds from different directions and all you get would be noise.
You might get sensory overload and be unable to distinguish the sounds, but the waves would not necessarily be interfering with each other enough to be noticeable.
So could it be said then that light travels as a wave but collapses to particle like on observation/ detection.
Sure.

Wavelength determines colour and energy is linked to wavelength so how does amplitude work? Would I be right in thinking that amplitude enables the wave to carry more energy per wave i.e 2x amplitude gives twice energy or similar to 2 waves travelling in partnership.
It depends on how you are viewing the EM Radiation. For example, radio waves can be thought of as actual waves and the amplitude as a direct increase in the signal strength. However when talking of individual photons this no longer works. The energy of the photon would be E=hv, where h is Planks constant and v is the frequency.

Finally, I think, waves in 3d must propagate outwards spherically, as sphere grows energy should be spread out the same way a balloon gets thinner as its blown up, so taking the sun as source by the time light reaches earth we have a sphere with a radius of over 90 million miles, how does light maintain its integrity when spread out over such an area?
Again, it depends on how you are looking at the light. As a wave model the light simply spreads out like any wave does and the intensity of the light drops off as distance from the emitting objects increases. In terms of photons, the Sun emits uncountable numbers of photons which then move straight out away from the Sun. The intensity drops off as distance increases because the photons spread out into space, similar to birdshot from a shotgun spreading out in the air. The energy of each photon is set and does not change and the photon does not spread out as it travels.
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Feb15-12, 08:34 AM
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Quote Quote by cepheid View Post
I can't say that I agree with this. When you see an object, it is because photons that are being emitted or reflected from the surface of that object are hitting your retina. The quanta of light (or, if you prefer, just the EM waves) are the things that are actually detected by your sensory equipment.
True our retinal cells detect the waves /photons incident upon them and this is a part of the process that gives us the sense of vision,but it is the object that we see and not the photons.We see by means of photons but how can we actually see a photon?
We use expressions such as "we can see the light rays" or "the light we see is the light contained within the visible region of the EM spectrum".Such expressions are usually good and understandable within the context they are used but there may be times when a slightly different phraseology would add some useful extra detail.
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Feb15-12, 09:13 AM
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"Seeing" can be regarded as the whole perception process which involves receiving the light right through to placing an object in our brain's model of our surroundings. Or as our eye reacting to light that enters it. Take your pick.
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Feb16-12, 07:04 AM
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Drakkith a fine answer I think I may be getting there.
One thing you say that a photons energy is fixed and the photon does not spread out as it travels so could it be regarded as each photon ( I am assuming that 1 photon = 1 wavelength ) to be moving with a virtual box around it, if so is this where planks quanta came from.
Would it be possible to expand a bit on amplitude? is it natural or something we add to an EM signal for instance.
It seems to me that EM waves behave strangely. I have found many formula for the relationship between wavelength, frequency and energy, all include either h, c or both, what they seem to say is that waves travel with a constant momentum which seems to be hc as in wavelength x energy = hc . e.g. comparing it to matter short wavelength acts like a bullet while in comparison long wavelength acts like a sponge ball, with both having the same mass and move at the same speed.
One reason for these questions is that depending on colours I see some 2d surfaces as 3d where red seems further away and blue seems closer especially where black is either in the forgroud or background, e.g. I have a book with a black cover the yellow writing seems flat on the black but the blue seems to stand out above the black yet others say it all looks flat. I am trying to relate this to the colour wavelengths.
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Quote Quote by John15 View Post
( I am assuming that 1 photon = 1 wavelength )
Owch!!! Where did that come from? I think you made too big a jump somewhere along the line. In fact, if you associate a photon with the wavelength of a wave you get things completely the wrong way round. A long wavelength radiation like Radio has millions (literally) more photons than a short wavelength radiation like X Rays, for the same flux of energy. Radio Frequency Photons have too little energy to detect individually (they come in vast numbers and interact with particles at a very low energy but high energy (like gamma) photons can be detected by the individual clicks of a Geiger Muller detector. The spatial extent of a photon is very debatable, in fact, and depends very much on what experiment you are considering.

This also relates to the momentum of photons, because the momentum of high frequency photons is high.

Your post also strays into the psychology of perception of colours and of scenes. That is a very complex topic and is too far away from these fundamental matters to be treated similarly. It's just asking to get bogged down almost before you've started.
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Feb17-12, 10:34 AM
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I am wondering if light is the right term to use. Would "we see radiated energy within certain wavelengths" be more accurate?
Also what form does that energy take? I ask this because we see a red hot iron bar in a darkened room so we are not seeing reflected light in this case but radiation in the form of heat, this could also explain why we see the source - because its hot - and not the energy/photons inbetween (see 1st post) perhaps because they act as closed systems and so cannot radiate energy so cannot be detected whilst moving. So are we seeing a heat signature, light in the form of photons or a mixture? remembering that infrared shows heat so why should visible wavelengths be any different.
What is special about black regarding energy and wavelength?
How do we define a photon?
Looking at the above answer a photon is defined by time rather than wavelength or energy ie taking 1 second as a start a wave with a length of 1 second will have a given amount of energy (call this 1 photon) and a wave of 10 seconds length will have energy of 1/10th per second or contain 10 photons that add up in total to the shorter wave, this would appear to give the shorter wavelength higher momentum. Or perhaps a photon or quanta could be described as the shortest wavelength highest energy wave that can exist. Or am I still barking up the wrong tree.
Still interested in an answer to amplitude.
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Quote Quote by John15 View Post
I am wondering if light is the right term to use. Would "we see radiated energy within certain wavelengths" be more accurate?
"Light" is just a term for electromagnetic radiation whose wavelengths fall within the visible range, i.e. the range of wavelengths to which our eyes are sensitive. In fact, in some contexts, people use the term "light" to mean electromagnetic radiation in general. In other words, they are not restricting the term to just visible light: they speak of ultraviolet light, x-ray light, infrared light, etc. So I really see no distinction between using the term "light" and using the term "radiated energy at certain wavelengths." The latter is just a description of what light IS.

Quote Quote by John15 View Post
Also what form does that energy take? I ask this because we see a red hot iron bar in a darkened room so we are not seeing reflected light in this case but radiation in the form of heat, this could also explain why we see the source - because its hot - and not the energy/photons inbetween (see 1st post) perhaps because they act as closed systems and so cannot radiate energy so cannot be detected whilst moving. So are we seeing a heat signature, light in the form of photons or a mixture? remembering that infrared shows heat so why should visible wavelengths be any different.
What is special about black regarding energy and wavelength?
Sigh. You seem very confused. Let me try to help. ANYTHING that has a temperature above absolute zero will "glow." What I mean by "glow" is that it will emit electromagnetic radiation i.e. light. We call this type of emission thermal radiation, and in physics, there is an idealized type of radiator called a blackbody. An ideal blackbody absorbs all EM radiation that is incident on it, and re-emits that radiation with a characteristic spectrum that depends on the equilibrium temperature it has reached. In fact, an ideal blackbody has the property that its thermal emission spectrum depends ONLY on the temperature of the object. It does not depend on anything else e.g. the surface material/composition of the object. Real objects obey this ideal case to a greater or lesser degree, depending on the object. The properties of a blackbody emission spectrum are 1. The hotter the object is, the more TOTAL radiation it will emit (when summed up over all wavelengths). 2. The wavelength at which the amount of radiation PEAKS is inversely proportional to the temperature. In other words, the emission from hotter objects peaks at a shorter wavelength in the EM spectrum than the emission from cooler objects. Properties 1 and 2 are generally obeyed by most things, even though maybe not in the strict blackbody sense. Here's practical example of that: you and I are currently at room temperature, or maybe just slightly above that. At that temperature, our thermal emission spectrum peaks in the *infrared* portion of the spectrum (meaning that the majority of the light that we emit is at infrared wavelengths). The shape of the spectrum at this temperature is such that there is negligible emission at visible wavelengths or shorter. In other words, you and I do not emit much visible light, certainly not enough to be detectable. However, if you take an object at room temperature (like an iron rod) and heat it up to higher and higher temperatures, then its thermal emission will obey both properties 1 and 2. Property 1 says that it will emit MORE overall radiation across the entire spectrum as its temperature increases, and property 2 says that it will emit more of its radiation at shorter wavelengths i.e. the peak of the spectrum will shift to shorter wavelengths. So if you heat the object enough, eventually it will get hot enough that most of its emission shifts from infrared to visible wavelengths, and it starts to "glow" visibly." At lower temperatures, this peak emission is still at the longer visible wavelengths, which is why it appears to glow red. As the temperature increases, the peak of the spectrum shifts to shorter and shorter wavelengths, and hence the colour shifts to orange, yellow, and eventually white. The reason why the object appears white is that, at this temperature, the shape of the blackbody emisson spectrum is such that the amount of emission happens to be mostly constant across the visible range (so you're seeing equal amounts of all the visible wavelengths). If you could continue to heat the object, eventually you'd reach the point where its emission spectrum would actually peak in the UV, and shorter still. There are stars whose surface temperatures are hot enough for that to be the case. I hope that this example has helped to illustrate that the emission from hot objects is electromagnetic radiation, or light. It is not some sort of "special" form of energy or radiation that is only associated with heat. Any object with a temperature will emit EM radiation, and the wavelengths at which most of that radiation is emitted will depend on the temperature of the object.

Quote Quote by John15 View Post
How do we define a photon?
Looking at the above answer a photon is defined by time rather than wavelength or energy ie taking 1 second as a start a wave with a length of 1 second will have a given amount of energy (call this 1 photon) and a wave of 10 seconds length will have energy of 1/10th per second or contain 10 photons that add up in total to the shorter wave, this would appear to give the shorter wavelength higher momentum. Or perhaps a photon or quanta could be described as the shortest wavelength highest energy wave that can exist. Or am I still barking up the wrong tree.
Still interested in an answer to amplitude.
I really have no idea what you are saying here in regards to the "time" of a photon. Classical electrodynamics says that light (meaning EM radiation in general) consists of oscillating electric and magnetic fields that propagate through space as waves. I admit to being not very knowledgeable about *quantum* electrodynamics (QED), but from what I understand, in QED, these electromagnetic fields are "quantized" in the same sense that other physical observables from classical physics are quantized in quantum physics. To be quantized means to be restricted to only being able to occupy a discrete set of energy levels (at least as far as I understand it). Somebody who is more knowledgeable than I am should comment further on what a photon actually is, all I can say is that is is the quantized version of the electromagnetic field. The last thing I can say about photons is that, just as classical electromagnetic waves have a frequency f, so too do photons. It is much more difficult to interpret (or intuit) what this "frequency" means photons, but if we just accept that "frequency" is a property of photons, we can note further that the energy E of a photon obeys the property that E = hf, where h is a constant. This is the discretization of energy that I mentioned before.
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Feb17-12, 04:18 PM
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Many thanks to all who have replied and I hope you will be patient with me.
The blackbody radiation properties mentioned above 1 and 2 I have already worked out,although not the inverse part,(why is everything inverse?) which is why I questioned if we see by temperature difference, in other words the energy from light is absorbed heating the object up slightly and energy is radiated in order to maintain an equilibrium and it is this energy we see. I suspect though that the radiation spectrum depends on the difference in temperature between the hot body and its surroundings, the greater the difference the shorter the wavelength emitted, the shorter the wavelength the quicker the energy loss, the closer you are in temp to your surroundings the longer the wavelength causing lower energy loss.
You say you and I do not emit much visible light, certainly not enough to be detectable so how do we see each other?
So can anyone tell me the difference berween photons and wavelengths. E=hf must be the same as E=hc/wavelength this gives the energy level of a single wavelength so how do you work out the energy of a photon? We can work out the total energy per second, why seconds because c is mts per second, by muliplying energy per wavelength by frequency. Surely an EM field cannot just occupy discrete energy levels as waves occupy all enegy levels discrete energy levels would mean that only certain wavelengths would be allowed.
Sorry I got the maths wrong in my last post the 10 second wave would have an energy level 1/10th the 1 second 1 so energy would be 1/100th per wavelength second.
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Feb17-12, 05:00 PM
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Quote Quote by John15 View Post
I suspect though that the radiation spectrum depends on the difference in temperature between the hot body and its surroundings, the greater the difference the shorter the wavelength emitted, the shorter the wavelength the quicker the energy loss, the closer you are in temp to your surroundings the longer the wavelength causing lower energy loss.
The radiation emitted depends solely on the temperature of the object, not any temperature difference. A block of iron heated to 5,000k inside a box kept at 5k emits exactly the same amount of radiation as another block of iron at 5,000k inside a box at 1,000k. The difference is that the box at 1,000k is emitting much more radiation than the box at 5k, so the block of iron inside the hotter box will be absorbing more radiation and will remain at a higher temperature longer.

You say you and I do not emit much visible light, certainly not enough to be detectable so how do we see each other?
Light from the Sun, light bulbs, and various other sources are reflecting off of you. This is why you cannot see in the dark, there is not enough light being reflected off of objects around you to allow you to see them.

So can anyone tell me the difference berween photons and wavelengths. E=hf must be the same as E=hc/wavelength this gives the energy level of a single wavelength so how do you work out the energy of a photon?
A photon is the name we give to the object that carries energy through space as an electromagnetic wave. E=hf and E=hc/wavelength is the same, because c/wavelength IS the frequency, f. The speed of the wave divided by the wavelength of a wave equals the frequency. See here: http://en.wikipedia.org/wiki/Frequen...uency_of_waves

The key is that unlike a classical wave like a water wave, the energy that light carries with it is only transferred in little bits at a time. An example is the photoelectric effect. In this effect light is shined on a metal surface in a vacuum and electrons are detected coming off the surface of the metal. The light is transferring energy to the electrons in the metal and "kicking" some of them out of the metal. The key is that turning up the intensity of the light, aka making it brighter, only determines the NUMBER of electrons per second coming off the metal. The speed at which they travel when ejected, depends on the amount of energy they absorb from the light. With a classical wave you would expect a higher intensity to impart more energy to each electron. But this isn't the case. It turns out that the frequency of the light is what determines the speed of the ejected electrons, not the intensity. If you drop the frequency of the light into the deep red area, NO electrons are ejected no matter how bright the light is! http://en.wikipedia.org/wiki/Photoelectric_effect

Surely an EM field cannot just occupy discrete energy levels as waves occupy all enegy levels discrete energy levels would mean that only certain wavelengths would be allowed.
You are correct. An EM wave can have any amount of energy. The discrete energy levels only come into play when elementary particles form into composite ones, protons and electrons forming atoms. However, as I mentioned above, energy is carried in little bits called Quanta, not smoothly over time as you would expect from a classical wave.
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Feb20-12, 05:25 AM
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Tried a little experiment with a mirror. Place mirror parallel to face touching nose and you can see your chin reflected, this shows that light reflects at seemingly impossible angles, in fact in did not matter what angle the mirror was as long as the reflection was in line of sight, all light in the room was also reflected light and had to reflect at least 3 x to get to my eyes. So all this light must be moving as waves as particles could not reflect at such angles or can they. So how do we see these wave, the distance between face and mirror is about 20mm so the waves have to travel up this narrow gap to the eyes, if the waves have to collapse to become detectable the why do they wait till the furthest point. Are they detected as energy fluctuations via direct interaction with atoms/molecules or as EM fields. Not to sure if I can really explain where I am trying to go with this.
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Feb20-12, 06:15 AM
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Quote Quote by John15 View Post

So all this light must be moving as waves as particles could not reflect at such angles or can they.
If you think of a photon as a tiny little "ball", classically it can. However we know that it is not a tiny little ball. For a while back in the 1700's light was thought to be made up of "corpuscles", aka particles. Isaac Newton himself held this view.

So how do we see these wave, the distance between face and mirror is about 20mm so the waves have to travel up this narrow gap to the eyes, if the waves have to collapse to become detectable the why do they wait till the furthest point.
The wavelength of visible light is about 700-350 nanometers, a nanometer being 1 billionth of a meter. As long as the distance between two things is larger than this the light can easily pass through. Don't worry about the "collapse", just realize that once the light enters your eyes and hits your retina it interacts with the molecules there and allows you to see.

Are they detected as energy fluctuations via direct interaction with atoms/molecules or as EM fields. Not to sure if I can really explain where I am trying to go with this.
Yeah, not really sure what you mean either.
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Feb20-12, 12:29 PM
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Of course do we really know how are brains are dealing with the information recieved through our eyes, for a start it rotates the image through 180 degrees without us knowing it so what else could be going on without us realizing it?
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Feb20-12, 12:30 PM
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Yes. The word "collapse" is very much a 'higher Physics" term and, as with a lot of QM,, should not be taken literally because it is not explainable in conventional terms. It is fruitless to try to impose your view of what 'stands to reason' -'cos it doesn't.
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Feb20-12, 07:50 PM
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Quote Quote by John15 View Post
Of course do we really know how are brains are dealing with the information recieved through our eyes, for a start it rotates the image through 180 degrees without us knowing it so what else could be going on without us realizing it?
I don't see how this is relevant to the discussion on how light works. Asking do we "really" know something is a bad question, as there is no good way to describe what something "really" is other than to describe it's properties and how it interacts per known laws of science. So do we "really" know how our brain deals with visual information? I say yes to the best of our ability. There is a large amount of information on how our visual system works, and while we don't know everything, we do know a lot.
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Feb21-12, 08:10 AM
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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?
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Feb21-12, 09:33 AM
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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.


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