Mean colour of visible spectrum?

[Raghav Gupta]

How the Mean colour of visible spectrum is yellow?
Shouldn't that be green according to acronym VIBGYOR?

 

[Quantum Defect]

How the Mean colour of visible spectrum is yellow?
Shouldn't that be green according to acronym VIBGYOR?

This depends upon the light source, and how you calculate the "mean", I think.

 

[Drakkith]

How the Mean colour of visible spectrum is yellow?

I've never heard this before. Can you post a reference? It may help to get some context.

 

[DaveC426913]

The mean is going to be based on wavelength - which ranges between 350 and 800 nm.

It depends on what source you use for "visible light", as there seems to be a quite a range of opinions.

 

[mathman]

https://www.google.com/search?q=sun..._mxdQslKLnbL5FEbOTJ_jhUHzQyo=&biw=937&bih=668

Above shows solar spectrum. Peak is in the green. It is asymmetric, higher at red end than at purple end.

 

[DaveC426913]

Above shows solar spectrum. Peak is in the green. It is asymmetric, higher at red end than at purple end.

Sure but
a] question was not about solar spectrum, just visible spectrum, and
b] question was not about average (i.e. weighted), it is about mean (i.e. highest minus lowest).

In this example, mean is 600nm

 

[Raghav Gupta]

I've never heard this before. Can you post a reference? It may help to get some context.

There was a question in my practical manual-
" In general for which colour we take the refractive index of a material in lens and glass slabs.""
The answer was given
Yellow colour. Since it is the mean colour of visible spectrum.

I had a doubt as green colour lies in between the visible spectrum. The mean wavelength and mean frequency of visible spectrum should be green?

 

[2milehi]

How the Mean colour of visible spectrum is yellow?
Shouldn't that be green according to acronym VIBGYOR?

I've understood the acronym as ROY G BIV

 

[jbriggs444]

Sure but
a] question was not about solar spectrum, just visible spectrum, and
b] question was not about average (i.e. weighted), it is about mean (i.e. highest minus lowest).

The statement was vague. The boundaries of the visible spectrum are not exact. The statement was not about mean wavelength or mean frequency but about mean "color". All averages are weighted. It's just that the weights are often assumed to be uniform. But uniform by what measure? A uniform weight by frequency will give a different mean than a uniform weight by wavelength.

 

[DaveC426913]

I've understood the acronym as ROY G BIV

Either way it's a fabrication.

Newton saw 6 colours but he felt strongly that 7 was a divine number, so he added indigo.

I will see if I can find a reliable reference for this. There're plenty of not-so-reliable references to it.

"It has been suggested that, at the time, Newton was trying make some anology with the musical scale and the octave (with its seven intervals) and hence was keen to identify seven colours in the rainbow or visible spectrum. "
http://colourware.org/2009/07/20/indigo-a-colour-of-the-rainbow/

"Newton probably had other, very good reasons to define the Rainbow as a function of the favored magical number of seven,..."
http://naturalmagickshop.com/articles/The-Myth-Magic-and-Science-of-the-Rainbow.html

Here's one in the American Journal of Physics:

"The author hypothesizes that Newton saw seven reasonably distinct colors in the artist's paint mixture color circle (red, orange, yellow, green, blue, violet, and purple) and therefore assumed he could also see seven distinct colors in his crude spectral projections."
http://scitation.aip.org/content/aapt/journal/ajp/40/4/10.1119/1.1986607

 

[Drakkith]

I had a doubt as green colour lies in between the visible spectrum. The mean wavelength and mean frequency of visible spectrum should be green?

Violet wavelength - about 400 nm
Red Wavelength - about 650 nm
Mean Wavelength - 525 nm

Light with a wavelength of 525 nm lies in the green area.

I would guess that the the standard for measuring the refractive index using yellow light is due to historical reasons, probably something to do with the sodium spectral line at 589 nm.

 

[Raghav Gupta]

Okay, got it . Thanks to all. Sorry for asking a last off topic question but isn't that avatar of you Drakkith is a Doom game hero? I really liked that game in my childhood

 

[Drakkith]

That it is!

 

[sophiecentaur]

Why consider the mean wavelength when you could just as easily consider the mean frequency? Would you get the same answer? :) (Harmonic mean)

 

[Raghav Gupta]

Why consider the mean wavelength when you could just as easily consider the mean frequency? Would you get the same answer? :) (Harmonic mean)

I think we usually take arithmetic mean in these cases.
Then both have same mean wavelength and mean frequency.

 

[aroc91]

b] question was not about average (i.e. weighted), it is about mean (i.e. highest minus lowest).

Mean and [arithmetic] average are the same thing. Where are you getting "highest minus lowest" from? That's not any form of averaging as far as I know. If you take the average of 100 and 1 with that method, it comes out to 99, which doesn't make any sense.

 

[sophiecentaur]

I think we usually take arithmetic mean in these cases.
Then both have same mean wavelength and mean frequency.

Arithmetic mean of what? It is only arbitrary and historical that we measure the wavelength of light in preference to the frequency. In fact, the Chemistry of what goes on in our eye receptors will be frequency based and not wavelength based. (i.e. photon energies)
You should try with some different random values before you make a statement like that.
In general, the harmonic mean is not the same as the mean of a set of numbers.
(A +B)/2 is not the same as 1/((1/A + 1/B)/2), which is what you are claiming.

 

[sophiecentaur]

This thread is, as is very common, mixing up the notion of Colour with Wavelength. Our Eyes are not Spectrometers and they (plus brain) see colours, which are usually wavelength combinations. We use just three colour sensors which have very broad overlapping responses. The 'colour' we perceive has nothing necessarily directly to do with the mean or peak of the spectrum of the light. The tristimulus system does work on a mean or centre of gravity of colours on the CIE chart but that is a two dimensional display and not a one dimensional spectrum.
We do not 'see green' when there is a peak in the spectrum in the region of 'spectral green' because we are not designed to. That's all there is to say about it, unless you want to dig much deeper into the whole business of colour perception. It is just not that simple.

 

[Raghav Gupta]

Arithmetic mean of what? It is only arbitrary and historical that we measure the wavelength of light in preference to the frequency. In fact, the Chemistry of what goes on in our eye receptors will be frequency based and not wavelength based. (i.e. photon energies)
You should try with some different random values before you make a statement like that.
In general, the harmonic mean is not the same as the mean of a set of numbers.
(A +B)/2 is not the same as 1/((1/A + 1/B)/2), which is what you are claiming.

I know arithmetic mean is different from harmonic mean.
I am not claiming that both are equal.
I was kind of asking that why we have to take harmonic mean instead of arithmetic mean when we have to find mean frequency of visible spectrum?
I thought till Drakkith's reply I was understanding most of the things.
He gave also the reason that yellow might be the mean colour considered because of sodium history.
So if all in reality is frequency based according to your quote
Then why harmonic mean?
Isn't when we talk about means we usually refer arithmetic mean the most common?

 

[sophiecentaur]

I know arithmetic mean is different from harmonic mean.
I am not claiming that both are equal.
I was kind of asking that why we have to take harmonic mean instead of arithmetic mean when we have to find mean frequency of visible spectrum?
I thought till Drakkith's reply I was understanding most of the things.
He gave also the reason that yellow might be the mean colour considered because of sodium history.
So if all in reality is frequency based according to your quote
Then why harmonic mean?

The frequency is inversely proportional to the wavelength so, if you want the same answer for both, you need to take the harmonic mean for one and the arithmetic mean for the other. I was originally making the point that using a mean wavelength is an arbitrary choice.
Also, there really is no such thing as a Mean Colour (a very fuzzy quantity, at best). There is no future in a conversation that tries to relate what we perceive to the spectrum of incident light unless you are prepared to include how the three sensors will respond, separately, to the black body spectrum and then plot the resultant (processed) signal values onto a CIE chart. But you don't need to do that sum, to know the answer and that is - You Won't See Green. Every day you do that experiment when you look at sunlight reflected on white surfaces.

 

[Raghav Gupta]

The frequency is inversely proportional to the wavelength so, if you want the same answer for both, you need to take the harmonic mean for one and the arithmetic mean for the other. I was originally making the point that using a mean wavelength is an arbitrary choice.
Also, there really is no such thing as a Mean Colour (a very fuzzy quantity, at best). There is no future in a conversation that tries to relate what we perceive to the spectrum of incident light unless you are prepared to include how the three sensors will respond, separately, to the black body spectrum and then plot the resultant (processed) signal values onto a CIE chart. But you don't need to do that sum, to know the answer and that is - You Won't See Green. Every day you do that experiment when you look at sunlight reflected on white surfaces.

Okay, now I am getting to understand you.
I did some maths
Sorry for using the calculator.:D (as it is a fast way, only for mean freq.)
I took the numbers from Drakkith post 11.
Violet wavelength as 400nm and red wavelength as 650 nm.
When we use arithmetic mean here we get 525 nm as A.M wavelength which is of green.

Now taking the same numbers, calculated the H.M
Got it as 495.238 nm
I guess that taking these numbers are wrong, as the units are coming in nanometers. It should be hertz.
Should I calculate H.M frequency by taking the upper and lower range of visible region frequency?
What is your opinion?

 

[jbriggs444]

What frequency corresponds to a wavelength of 495.238 nm?

How does this frequency compare to the arithmetic mean of the upper and lower frequencies of the visible range?

 

[Raghav Gupta]

What frequency corresponds to a wavelength of 495.238 nm?

How does this frequency compare to the arithmetic mean of the upper and lower frequencies of the visible range?

Indigo corresponds to that. According to this
https://www.google.co.in/#q=frequency+of+visible+light&imgrc=YP-FAEbAa3zJQM%3A;undefined;http%3A%2F%2Fwww.relativitycalculator.com%2Fimages%2FAlbert_Michelson_Part_I%2Fcolor_wavelength_frequency.png;http%3A%2F%2Fwww.relativitycalculator.com%2FAlbert_Michelson_Part_I.shtml;451;162

And blue corresponds by taking into account your 2 para in quote and this link.
Different colours and not green coming, how?
When calculating A.M for frequency
Lowest freq. was 400 Thz
And highest was 790 Thz.
A.M came 595 Thz

 

[sophiecentaur]

You are missing my point here. It was that the concept of a mean wavelength / colour / frequency is not relevant in any way, other than for an exercise in numerology. That is why I pointed out the difference between mean frequency and mean wavelength. You can chose any set of numbers and show that the results of calculations of the means are usually different. Starting with seven letters and choosing the middle one is just an exercise on a Scrabble tray. The questioner asked (implicitly) why we don't see Green Stars. It's a phychovisual thing and nothing to do with means or peaks in the spectrum.

 

[jbriggs444]

Indigo corresponds to that.

What I and others were fishing for was to get you thinking about the relationship between arithmetic mean, harmonic mean, weavelength and frequency.

Violet wavelength as 400nm and red wavelength as 650 nm.
When we use arithmetic mean here we get 525 nm as A.M wavelength which is of green.
Now taking the same numbers, calculated the H.M
Got it as 495.238 nm

What is the frequency corresponding to 400 nm? Divide the speed of light by 400 nm and you get around 750 terahertz.
What is the frequency corresponding to 650 nm? Divide the speed of light by 650 nm and you get around 461 terahertz.

What is the arithmetic mean of those two frequencies? Add and divide by two and you get around 605.5 terahertz.

What wavelength does that frequency correspond to? Divide the speed of light by 605.5 terahertz and you get around 495 nm.

By no coincidence, 495 nm is the harmonic mean of 400 nm and 650 nm.

 

[Raghav Gupta]

There was a question in my practical manual-
" In general for which colour we take the refractive index of a material in lens and glass slabs.""
The answer was given
Yellow colour. Since it is the mean colour of visible spectrum.

So why my practical manual is messing here.
Does it is not upto the standards.
Well I agree it is kind of classifying
And not teaching us to understand.
Please I know that the term mean colour is again coming here.
Just want a comment for this quote.

 

[Raghav Gupta]

What I and others were fishing for was to get you thinking about the relationship between arithmetic mean, harmonic mean, weavelength and frequency.
What is the frequency corresponding to 400 nm? Divide the speed of light by 400 nm and you get around 750 terahertz.
What is the frequency corresponding to 650 nm? Divide the speed of light by 650 nm and you get around 461 terahertz.

What is the arithmetic mean of those two frequencies? Add and divide by two and you get around 605.5 terahertz.

What wavelength does that frequency correspond to? Divide the speed of light by 605.5 terahertz and you get around 495 nm.

By no coincidence, 495 nm is the harmonic mean of 400 nm and 650 nm.

Okay got it. So we really get same answers either way. I have not thought of dividing that wavelengths by c . Thanks
So 495 nm is corresponding to green colour.
That done what about my above post in response to SophieCentaur?

 

[sophiecentaur]

So why my practical manual is messing here.
Does it is not upto the standards.
Well I agree it is kind of classifying
And not teaching us to understand.
Please I know that the term mean colour is again coming here.
Just want a comment for this quote.

There is no point in complaining about or getting confused by one book ("practical manual"?). You have to read round other sources and have discussions like this one - but you need to go to a reputable forum like PF. It's a common problem that people who try to write books to simplify things will get facts wrong by over-simplifying. The way to describe a colour that lies somewhere near the middle of the optical spectrum is to call it an 'average' colour, which is a catch-all and non-specific term for 'middle'. If you say "mean" then you imply a particular mathematical operation that cannot be done with written Colour Names.
If you want to 'understand' then (as my signature has said, for a long time) you need to avoid relying on classifying. There is not a Scientific Instrument that will show you a meter needle that will point to 'Green' haha. If an optical instrument is adjusted for minimum distortion in the middle of the optical spectrum then it will be aimed at a particular Wavelength (/frequency) and not a colour. Your manual is, as you say "not up to the standards". I sympathise.

 

[Raghav Gupta]

Thanks to all of you.

 

[mp3car]

This could be calculated a number of different ways with different results. As I think has been said, our eyes do not have a linear response to equal power of different wavelengths with respect to perceived brightness. A mean value would not necessarily be the same as the color that would be observed. For example, of equal optical power, you won't perceive the "mixed" color to be the same as the average wavelength of the sources. Take the example of two lasers. If they both were the same power, same "dot" size and projected on a white surface (I realize saying "white" is oversimplifying), you would not perceive the "dot" to be the color that is equivalent to the average of the two wavelengths. For example, if you mixed an 800nm 5mW laser (which would barely be visible) with a 532nm 5mW laser (pretty "bright"), the difference would be very slight and would essentially appear to be exactly the same as if it were just the 532nm laser, since our eye's response is so much higher for the same given power.

Another example is "cheap" green laser pointers (e.g. $5-$10)... Many of them DO NOT HAVE AN IR CUT FILTER (i.e. can be very dangerous for eyes), and so they emit considerable power in the 808nm and 1064nm wavelengths. 808 is barely perceptible, even at powers as high as 50+mW, and 1064nm is well beyond the visible region at any power. If we arbitrarily say a green laser pointer is emitting 200mw at 808nm, 20mW at 1064, and 5mW at 532nm, you will still see green, even though the unweighted average would be about 800nm. The weighted average would obviously be even higher (longer, further into the IR band).

For those that don't know, most green lasers, especially pointers, are not a green laser diode. I'll keep it high level, but they use an 808nm diode to "pump" another material that lases 1064nm, which is then "cut in half" to 532nm, green. You lose a considerable amount of power in the process... thus why the "pump" IR 808nm diode is a lot higher power level than the final green output, but if there's no IR cut filter, a lot of IR is being emitted, enough to easily cause eye damage, potentially even from stray reflections.

 

[sophiecentaur]

our eyes do not have a linear response to equal power of different wavelengths

The most important aspect of the eye's response is that it has virtually no wavelength resolving facility - never mind its amplitude response. It is not a spectrometer.

 

[fizixfan]

To my simplistic way of thinking, the "mean" value for visible light would be the statistical average between the two extreme wavelengths of visible light, i.e., 400 nm (violet) and 780 nm (red), which would be 590 nm (orange). However, there does not appear to be any standard definition of the shortest or longest visible wavelengths. I took a range of estimates from several different sources from a Google search of "Visible Spectrum" and got an average value of about 560 nm (green-yellow).

 

[Shane Kennedy]

It also depends on how you measure it. Can we actually measure the amplitude of the wave, or do we measure by it's power ? Is there a device that can measure light without any colour bias ? I am sure that human eyes vary slightly in their sensitivity too.

 

[Drakkith]

It also depends on how you measure it. Can we actually measure the amplitude of the wave, or do we measure by it's power ?

Not of visible light. The wave alternates too fast for electronics to respond so we can't directly measure the amplitude of each alternation. Instead we measure the power over some amount of time.

Is there a device that can measure light without any colour bias ?

Of course. A properly calibrated spectrometer is one such device.

 

[sophiecentaur]

I wonder what would be the arithmetic mean of Tom, Dick and Harry?

 

[Shane Kennedy]

Not of visible light. The wave alternates too fast for electronics to respond so we can't directly measure the amplitude of each alternation. Instead we measure the power over some amount of time.
Of course. A properly calibrated spectrometer is one such device.

In order to calibrate anything, you need a reference. I was involved in setting up a production test for IR LEDs and sensors. We knew that a block of plastic would attenuate at a certain rate/mm, but where do you get a reference source, or a reference detector ? With light, it seems to be a chicken and egg situation. You need one to get the other ... in both directions.

 

[Shane Kennedy]

I wonder what would be the arithmetic mean of Tom, Dick and Harry?

When at college, learning about matrices, and since, I often wondered how you can get the value of a matrix. What defines what cell does what, and where does the information come from. A practical example would be good.

 

[sophiecentaur]

Not of visible light. The wave alternates too fast for electronics to respond so we can't directly measure the amplitude of each alternation. Instead we measure the power over some amount of time.
Of course. A properly calibrated spectrometer is one such device.

It is quite possible to use 'superheterodyne' (now there's a lovely old word) methods to beat down a CW signal from a laser to a manageable RF frequency and to 'look at' the variations of the fields in the light wave. You may say it's a bit of a cheat but no one (?) uses TRF (tuned radio frequency) receivers these days, so we do the same trick for RF measurement, in any case.
It is true, however, that the THz variations of fields in light waves can neither be detected by thermal sensors nor the output of photochemical or photoelectric devices - which are only aware of numbers of Photons.

Is there a device that can measure light without any colour bias ? I am sure that human eyes vary slightly in their sensitivity too.

All human senses are like this. With the exception of highly trained craftsmen and people with perfect pitch, all quantities have to be measured with instruments, if high accuracy is needed. I think you may still be missing the point that the eye is not a spectrometer. It can make a stab at estimating the wavelength of a single spectral line but it is completely unable to resolve the spectral components of any other light source. All it can do is to 'name the colour' of an object or light source - which is nothing like the spectral content. It is because of this that Colour TV works.

 

[fizixfan]

I am sure that human eyes vary slightly in their sensitivity too.

Good point Shane. I'm what you call "color blind," although "red-green deficient" or better yet "deuteranopia" are more accurate terms.

We have red, green and blue color receptors (cones) in our retinas. I probably have fewer red and green receptors than people with normal vision do.

But - and this is just a theory - color blind and color deficient people may have more rods (for light sensitivity, not color) than non-color blind people, and so may have better night vision. My night vision is pretty good.

I also wonder if the RGB cones distinguish colors and hues via a RGB (additive) or CMYK (subtractive) process. Probably additive, since computer screens use this method, whereas the color printing (pigmenting) process uses the subtractive colors.

 

[blue_leaf77]

Given a spectrum shape $A(\omega)$, the mean frequency is
$$ \omega_{av} = \frac{\int \omega |A(\omega)|^2 d\omega}{\int |A(\omega)|^2 d\omega} $$.
I don't see there is a way to calculate the spectrum if one has not been given a particular spectrum.

And by the way there have been a couple of measurement methods able to resolve electric field oscillation such as FROG, SPIDER, etc.

 

[sophiecentaur]

Good point Shane. I'm what you call "color blind," although "red-green deficient" or better yet "deuteranopia" are more accurate terms.

We have red, green and blue color receptors (cones) in our retinas. I probably have fewer red and green receptors than people with normal vision do.

But - and this is just a theory - color blind and color deficient people may have more rods (for light sensitivity, not color) than non-color blind people, and so may have better night vision. My night vision is pretty good.

I also wonder if the RGB cones distinguish colors and hues via a RGB (additive) or CMYK (subtractive) process. Probably additive, since computer screens use this method, whereas the color printing (pigmenting) process uses the subtractive colors.

You are using terms in colour synthesis here but colour analysis is not suited to those descriptions. It s a bit simplistic to describe the receptors as 'red', 'green' and 'blue' receptors because they are all sensitive to more or less the whole visual spectrum. This is essential for the way they work.
This link (and dozens others from Google) tells you the main points about the tristimulus colour vision theory. That theory works well enough for Colour TV and other displays to work very well. Those displays work on additive mixing and give pretty good colour fidelity. within their gamut. Subtractive mixing (colour film and colour printing) is not so good if you can only use three primaries on their own. Spot colours can be used in printing to improve reproducibility (e.g. the Red in the CocaCola adverts would never be done 'right' with a dot matrix printer)

 

[klimatos]

The wavelength of maximum intensity for insolation (incoming solar radiation) is generally given as 475 nanometers in most solar radiation studies. This wavelength is toward the green end of the visible spectrum. However, the scattering of a portion of the blue component of that greenish radiation by the intervening atmosphere (Rayleigh scattering) allows us to perceive sunlight as somewhat yellow. This was my standard explanation in my atmospheric science classes.

 

[mp3car]

The most important aspect of the eye's response is that it has virtually no wavelength resolving facility - never mind its amplitude response. It is not a spectrometer.

No wavelength resolving facility? No, it's not a spectrometer, but the various cone cells respond differently when we view one wavelength compared to another. No, we can't tell a difference between 532nm and 533nm, nor a difference between a surface illuminated with 70W/m^2 vs. 71W/m^2...

However, if a surface is emitting monochromatic 400nm, I am going to have a different cone response than I would it were 600nm. And as has been said, the perceived brightness depends not only on power, but also the wavelength. I am not sure what you mean by "no wavelength resolving facility" considering we can easily resolve a difference between the two different wavelengths in this example. As I said, of course we can't tell 532nm vs. 533 and maybe even 532 and 540nm. But in the big picture, taking the entire system into account of the eye and brain, 500nm and 600nm are "resolved" very differently (I am using the term "resolve" to mean separate, distinguish, discern).

 

[sophiecentaur]

No wavelength resolving facility? No, it's not a spectrometer, but the various cone cells respond differently when we view one wavelength compared to another. No, we can't tell a difference between 532nm and 533nm, nor a difference between a surface illuminated with 70W/m^2 vs. 71W/m^2...

However, if a surface is emitting monochromatic 400nm, I am going to have a different cone response than I would it were 600nm. And as has been said, the perceived brightness depends not only on power, but also the wavelength. I am not sure what you mean by "no wavelength resolving facility" considering we can easily resolve a difference between the two different wavelengths in this example. As I said, of course we can't tell 532nm vs. 533 and maybe even 532 and 540nm. But in the big picture, taking the entire system into account of the eye and brain, 500nm and 600nm are "resolved" very differently (I am using the term "resolve" to mean separate, distinguish, discern).

There is no way that the eye can distinguish between a Yellow Sodium line and an appropriate mix of Green and Red monochromatic light. I thing that demonstrates pretty well that the eye cannot resolve different wavelengths. I can't think why people feel it necessary to defend the abilities of the eye by suggesting it can do things that it can't Why should it matter?

 

[blue_leaf77]

All materials must exhibit dispersion (different response to frequency), this is guaranteed by Kramers-Kronig relation. The only way a "medium" behavior can be independent of frequency is that to have unit refractive index (again proved by Kramers-Kronig relation) which means vacuum, so the eye being sensitive to different wavelength is a must. Probably the more appropriate way to get around this discrepancy is to introduce resolution for the eye. Anyway what we call colors is actually how our brain gives its response to incoming stimulus, it doesn't output numbers like spectrometers. On the other hand spectrometers see numbers, not colors. Therefore we don't have to associate eye to spectrometers in all ways, only in some ways that it's sensitive to different wavelength.

 

[sophiecentaur]

All materials must exhibit dispersion (different response to frequency), this is guaranteed by Kramers-Kronig relation. The only way a "medium" behavior can be independent of frequency is that to have unit refractive index (again proved by Kramers-Kronig relation) which means vacuum, so the eye being sensitive to different wavelength is a must. Probably the more appropriate way to get around this discrepancy is to introduce resolution for the eye. Anyway what we call colors is actually how our brain gives its response to incoming stimulus, it doesn't output numbers like spectrometers. Therefore we don't have to associate eye to spectrometers in all ways, only in some ways that it's sensitive to different wavelength.

I'm not sure what you are implying about the way the sensors work but the accepted theory is that each of the three sensors have a very wide band sensitivity. Taken on its own, a sensor will not 'know' whether it is receiving low level light of wavelength near its peak of sensitivity or high level light way off its peak. It is only when the brain has information about all three sensors that a sense of 'Colour' (not necessarily wavelength) can be deduced by comparing the relative output levels of the sensors. Dispersion doesn't come into it. Indeed, how can it with non spectral incident light?
Have you read about the tristimulus theory of colour vision (links given earlier in this thread or Google it). Look at the graphs of the three responses and then read how the three signals are processed together. Colour TV mimics the process very well and can synthesise a display colour with a combination of Three Primary Phosphors, producing an entirely different spectrum from the original object that will give a near-perfect match to the eye. You may need to suspend your disbelief until the end of what you read as it seems to fly in the face of what you are saying.
Not only is the eye not a spectrometer, it is not an uncalibrated spectroscope either. The ear, on the other hand - but one thing at a time.

 

[mp3car]

No disagreement that we can't isolate specific lines in a multi-frequency composition. My example I gave most recently specifically said monochromatic, because I thought you were even saying we don't differentiate wavelengths, which obviously isn't what you were saying! I think I'm on the same page with you now... And I think I know what you're alluding to with sound... Two or more audio frequencies together is, as far as I know, impossible to recreate with a single frequency, but with light, your eye can be "tricked" into thinking two or more are a single wavelength...

 

[fizixfan]

You are using terms in colour synthesis here but colour analysis is not suited to those descriptions. It s a bit simplistic to describe the receptors as 'red', 'green' and 'blue' receptors because they are all sensitive to more or less the whole visual spectrum. This is essential for the way they work.
This link (and dozens others from Google) tells you the main points about the tristimulus colour vision theory. That theory works well enough for Colour TV and other displays to work very well. Those displays work on additive mixing and give pretty good colour fidelity. within their gamut. Subtractive mixing (colour film and colour printing) is not so good if you can only use three primaries on their own. Spot colours can be used in printing to improve reproducibility (e.g. the Red in the CocaCola adverts would never be done 'right' with a dot matrix printer)

Thanks for this link. It helped me to understand that the three different types of cone cells in our retina respond to a range of wavelengths, and actually overlap, but have fairly discrete peaks. This has broadened my understanding of the subject.

It's interesting to note that some women (and possibly even some men) may actually have tetrachromatic vision. http://en.wikipedia.org/wiki/Tetrachromacy

 

[sophiecentaur]

The peaked shape of the responses makes it possible to position the UV co ordinates of a colour on the CIE chart (which is a simplified description of what the brain does). If the responses were flat topped, it would not be possible to solve the equation with the given output signal levels. The peak is a necessary feature and is not there so that we have special sensitivity at three particular wavelengths.
On the subject of variations on the basic tristumulus system, it would be interesting to know whether it is racially differentiated, too.

 

[fizixfan]

On the subject of variations on the basic tristumulus system, it would be interesting to know whether it is racially differentiated, too.

Here's an study which finds that "Caucasian Boys Show Highest Prevalence of Color Blindness Among Preschoolers" http://www.aao.org/newsroom/release/color-blindness-among-preschoolers-ophthalmology-journal-study.cfm

"Researchers from the Multi-Ethnic Pediatric Eye Disease Study Group tested 4,005 California preschool children age 3 to 6 in Los Angeles and Riverside counties for color blindness. They found the following prevalence by ethnicity for boys:

• 5.6 percent of Caucasian boys
• 3.1 percent of Asian boys
• 2.6 percent for Hispanic boys
• 1.4 percent of African-American boys"