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**yellow?**

Shouldn'tthat be green according to acronym VIBGYOR?

Shouldn't

- Thread starter Raghav Gupta
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Shouldn't

- #2

Quantum Defect

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This depends upon the light source, and how you calculate the "mean", I think.yellow?that be green according to acronym VIBGYOR?

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Drakkith

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I've never heard this before. Can you post a reference? It may help to get some context.How the Mean colour of visible spectrum isyellow?

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DaveC426913

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

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mathman

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

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DaveC426913

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

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

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There was a question in my practical manual-I've never heard this before. Can you post a reference? It may help to get some context.

" 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?

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I've understood the acronym as ROY G BIVyellow?that be green according to acronym VIBGYOR?

Shouldn't

- #9

jbriggs444

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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.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).

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DaveC426913

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Either way it's a fabrication.I've understood the acronym as ROY G BIV

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

http://scitation.aip.org/content/aapt/journal/ajp/40/4/10.1119/1.1986607

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Drakkith

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Violet wavelength - about 400 nmI had a doubt as green colour lies in between the visible spectrum. The mean wavelength and mean frequency of visible spectrum should be green?

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.

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Drakkith

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That it is!

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sophiecentaur

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I think we usually take arithmetic mean in these cases.

Then both have same mean wavelength and mean frequency.

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

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sophiecentaur

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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)I think we usually take arithmetic mean in these cases.

Then both have same mean wavelength and mean frequency.

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.

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sophiecentaur

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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.

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I know arithmetic mean is different from harmonic mean.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 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?

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sophiecentaur

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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.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?

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.

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Okay, now I am getting to understand you.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.

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?

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jbriggs444

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

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Indigo corresponds to that. According to this

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

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

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sophiecentaur

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jbriggs444

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

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

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