Understanding Light Waves: Reflection, Interference, and Perception

[John15]

Light travels as waves
Waves interfere with each other
We see reflected light.
We can see the source of light and the things illuminated but not the light inbetween Why?
Light is reflected in all directions from all things why does it not interfere between the object we are looking at and our eyes, light is also reflected from the eyes which should also interfere with light coming in.
Example throw a single pebble in a pond you get easily recognisable wave , same with 2 or 3 but throw a handful in and you just get a mess of waves.

 

[JaredJames]

We can see the source of light and the things illuminated but not the light inbetween Why?

To see light, it has to reflect off something.

 

[sophiecentaur]

Light travels as waves
Waves interfere with each other
We see reflected light.
We can see the source of light and the things illuminated but not the light inbetween Why?
Light is reflected in all directions from all things why does it not interfere between the object we are looking at and our eyes, light is also reflected from the eyes which should also interfere with light coming in.
Example throw a single pebble in a pond you get easily recognisable wave , same with 2 or 3 but throw a handful in and you just get a mess of waves.

Space is a 'linear medium' which means that the fields of one wa e fo not affect the fields of a wave it's crossing. So they don't modify each other.

 

[russ_watters]

To see light, it has to reflect off something.

That isn't correct:

To see light, it has to hit your retina.

 

[russ_watters]

Light travels as waves
Waves interfere with each other
We see reflected light.
[snip]
Light is reflected in all directions from all things why does it not interfere between the object we are looking at and our eyes,
[snip]
Example throw a single pebble in a pond you get easily recognisable wave , same with 2 or 3 but throw a handful in and you just get a mess of waves.

You answered your own question. When you throw a handful of pebbles into a pond you get a whole mess of waves instead of all the waves cancelling each other out because waves have to be perfectly timed and aligned to cancel each other out and neither a handful of pebbles, nor reflected light is coherent enough for that.

 

[John15]

Many thanks for the replies but still a bit confused.
why does light have to reflect off something in order to become visible?
In order to see something as a sharp image the light has to come in a straight line without any interference, so are we saying that this light is able to travel through what must be a mess of waves inbetween without interference, am not talking about waves cancelling out but being jumbled up for want of a better phrase.

 

[JaredJames]

why does light have to reflect off something in order to become visible?

My mistake. I was thinking of it along the lines of you either have to look at the source of the light to see it, or it has to reflect off of something.

In both cases, you need the light to enter the eye. It can't "pass near by".

 

[davenn]

That isn't correct:

To see light, it has to hit your retina.

and expanding on that ...

@ OP yes it travels as waves and that's fine to think of it as waves for some applications
but it also travels as particles ... photons... and it those photons entering your eye that get detected.
The more photons, the brighter the light.

Dave

 

[Dremmer]

Light travels as waves
Waves interfere with each other
We see reflected light.
We can see the source of light and the things illuminated but not the light inbetween Why?[\quote]


The same reason as to why we don't see sound waves. Light waves and sound waves can't be detected by our eyes.

 

[cepheid]

Light travels as waves
Waves interfere with each other
We see reflected light.
We can see the source of light and the things illuminated but not the light inbetween Why?[\quote]


The same reason as to why we don't see sound waves. Light waves and sound waves can't be detected by our eyes.

Say what?

Light waves can be detected by our eyes, provided they are within a specific range of wavelengths. That is how we see things.

 

[John15]

So am I right in thinking that light travels as waves from source but acts as particle when reflected? which is what we see.

Still not sure why it does not interfere though.
Think of a wave tank, standing at one end waves from other end are easy to make out, now add waves coming from top, bottom and sides plus from your end and all you get is turbulence and making it very difficult to make out individual waves.

I do have another question about waves traveling from source but let's try to conclude this first.

 

[sophiecentaur]

It also behaves as waves when reflected and as particles when it travels -if that's the way you want to explain it. Waves are certainly good enough for Optics and Radio studies You should stop trying to categorise this sort of thing. You'll find it's like trying to get hold of a bar of soap in the bath.

Interactions between EM and matter can often be thought of, conveniently, is terms of Photons but, otoh, a lens is 'matter' and treating the action of a lens by considering photons would just be madness.

 

[jim hardy]

"""We can see the source of light and the things illuminated but not the light inbetween """

how about a hologram ?

 

[sophiecentaur]

How ABOUT a Hologram?
Do you know how they work, Jim?

 

[Drakkith]

So am I right in thinking that light travels as waves from source but acts as particle when reflected? which is what we see.

Light can be explained as a wave in almost all aspects. Diffraction, reflection, and interference are all perfectly described by treating light as an electromagnetic wave. However, when light finally gets to something and is absorbed is can only be treated as a particle that gives all of its energy up and no longer exists. This is how the molecule in your eye's cone and rod cells works, it absorbs energy from the photon and changes shape, setting off a chain of events that ends with you "seeing" something. The key is that the light has to enter and interact with your eye. The reason we cannot see the light in between is that two waves don't "bounce" off of each other. One light wave passing by another does not reflect or refract off of it. (Other than normal interference effects which is not the same) Think of your water waves. If you make two waves they don't bounce off of each other and go in completely different directions or create new waves at the point they interact.

Still not sure why it does not interfere though.

Light can only interfere when the wavelength and such are almost exactly alike, meaning it is "Coherent". Normally light entering your eye does not meet this requirement. This is similar to why sound waves do not normally interfere and keep you from hearing the speakers in your TV.

Think of a wave tank, standing at one end waves from other end are easy to make out, now add waves coming from top, bottom and sides plus from your end and all you get is turbulence and making it very difficult to make out individual waves.

It may make it difficult, but it by no means makes it impossible to determine which wave is which. Your ears do this all day long.

I recommend picking up a book on optics. I have several, including Optics for Dummies which is pretty nice. But any book about basic optics will be able to explain it all.

 

[Naty1]

Try reading this online article and note the illustrations:


http://en.wikipedia.org/wiki/Light_interference#Optical_interference

You'll find a lot the relates to your questions like:

Conventional light sources emit waves of differing frequencies and at different times from different points in the source. If the light is split into two waves and then re-combined, each individual light wave may generate an interference pattern with its other half, but the individual fringe patterns generated will have different phases and spacings, and normally no overall fringe pattern will be observable.

"different points in the source" could be the gas molecules in a fluoresescent light, the filament in an incandescent light, or our local star, the sun...

This relates to various 'points' having different energies, so the amplitude and frequency [color] also has variations...it's not easy to get monochromatic (single frequency) light...hence we see versions of 'white light' meaning light appears sort of white visually but is actually a combination of colors:

http://en.wikipedia.org/wiki/White_light

 

[Dadface]

The nature of light is such that it enables us to see things but we don't actually see light itself.

 

[sophiecentaur]

This relates to various 'points' having different energies, so the amplitude and frequency [color] also has variations...it's not easy to get monochromatic (single frequency) light...hence we see versions of 'white light' meaning light appears sort of white visually but is actually a combination of colors:

http://en.wikipedia.org/wiki/White_light

You rattled my cage a bit here. It would be much more accurate to say that white light (or, in fact many of the colours we see) consists of a combination of different WAVELENGTHS (or frequency, as you wish). i.e a SPECTRUM. There is so much confusion between colour and wavelength and this can lead to further confusion when it's taken further. 'Colour' is something that we allocate to single wavelengths or combinations of light with different wavelengths. 'Wavelength' means just one thing so it is worthwhile using that word when that's what is meant.
Inhabitants of Planet Zog would agree 100% with our spectral analysis of a light source but, even allowing for any translation of actual words, they would not agree about colours (and neither would a Dog or Cow!) Not being picky, btw. It's far more important than that.

 

[cepheid]

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]

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 traveling 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?

 

[Drakkith]

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 traveling 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.

 

[Dadface]

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.

 

[sophiecentaur]

"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.

 

[John15]

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.

 

[sophiecentaur]

( 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.

 

[John15]

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.

 

[cepheid]

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.

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.

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.

 

[John15]

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 energy 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.

 

[Drakkith]

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/Frequency#Frequency_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 energy 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.

 

[John15]

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.