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Raghav Gupta
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How the Mean colour of visible spectrum is yellow?
Shouldn't that be green according to acronym VIBGYOR?
Shouldn't that be green according to acronym VIBGYOR?
Raghav Gupta said:How the Mean colour of visible spectrum is yellow?
Shouldn't that be green according to acronym VIBGYOR?
Raghav Gupta said:How the Mean colour of visible spectrum is yellow?
Sure butmathman said:Above shows solar spectrum. Peak is in the green. It is asymmetric, higher at red end than at purple end.
There was a question in my practical manual-Drakkith said:I've never heard this before. Can you post a reference? It may help to get some context.
Raghav Gupta said:How the Mean colour of visible spectrum is yellow?
Shouldn't that be green according to acronym VIBGYOR?
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 said: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).
Either way it's a fabrication.2lmilehi said:I've understood the acronym as ROY G BIV
Raghav Gupta said: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?
sophiecentaur said:Why consider the mean wavelength when you could just as easily consider the mean frequency? Would you get the same answer? :) (Harmonic mean)
DaveC426913 said:b] question was not about average (i.e. weighted), it is about mean (i.e. highest minus lowest).
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)Raghav Gupta said:I think we usually take arithmetic mean in these cases.
Then both have same mean wavelength and mean frequency.
sophiecentaur said: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.
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.Raghav Gupta said: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?
sophiecentaur said: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.
Indigo corresponds to that. According to thisjbriggs444 said: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 I and others were fishing for was to get you thinking about the relationship between arithmetic mean, harmonic mean, weavelength and frequency.Raghav Gupta said:Indigo corresponds to that.
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
So why my practical manual is messing here.Raghav Gupta said: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.
Okay got it. So we really get same answers either way. I have not thought of dividing that wavelengths by c . Thanksjbriggs444 said: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.
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.Raghav Gupta said: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.
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.mp3car said:our eyes do not have a linear response to equal power of different wavelengths
Shane Kennedy said: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 ?