# Mean colour of visible spectrum?

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

Quantum Defect
Homework Helper
Gold Member
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
Staff Emeritus
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
Gold Member
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.

DaveC426913
Gold Member
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 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.""
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?

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
Homework Helper
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
Gold Member
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

Last edited:
Drakkith
Staff Emeritus
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.

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

• mp3car
Drakkith
Staff Emeritus
That it is!

• mp3car
sophiecentaur
Gold Member
Why consider the mean wavelength when you could just as easily consider the mean frequency? Would you get the same answer? :) (Harmonic mean)

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.

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
Gold Member
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
Gold Member
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.

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?

Last edited:
sophiecentaur
Gold Member
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.

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?

jbriggs444
Homework Helper
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?

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

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
Gold Member
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
Homework Helper
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.

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.""
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.

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
Gold Member
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.

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.

Last edited: