# RGB -> (wavelength, intensity)

Discussion in 'Digital Photography' started by Yeppers, Nov 26, 2006.

1. ### YeppersGuest

Hi all,

Is there a simple straightforward set of equations
for converting normal visible light
* red
* green
* blue
values from my digital camear valuesto
* wavelength
* intensity
?

Second question: how about if I put a IR-pass visible-block filter
such as the Hoya R72 on my digital camera? Can I compute
infrared wavelength and intensity from remaining RGB data?

Thanks!

Yeppers, Nov 26, 2006

2. ### Raphael BustinGuest

On 25 Nov 2006 18:05:05 -0800, "Yeppers" <> wrote:

>Hi all,
>
>Is there a simple straightforward set of equations
>for converting normal visible light
> * red
> * green
> * blue
>values from my digital camear valuesto
> * wavelength
> * intensity

No. Too simplistic a model. What's the wavelenght of
white? Or gray, or black?

Not all visible colors (or RGB combinations) correspond
to a specific wavelength. In the general case, the
perception of color depends on the a spectrum of
illumination (eg., skylight) multiplied by a spectrum of
absorption (eg. the pigments of a leaf) selectively
absorbing portions of the illuminant.

To get really technical, there's no "pure" RGB color
space -- you'd have to specify which one.

rafe b
www.terrapinphoto.com

Raphael Bustin, Nov 26, 2006

3. ### Ray FischerGuest

Yeppers <> wrote:
>Hi all,
>
>Is there a simple straightforward set of equations
>for converting normal visible light
> * red
> * green
> * blue
>values from my digital camear valuesto
> * wavelength
> * intensity

Yes. You want to convert from the RGB to the HLV color space. That
will isolate the color from the luminance and the saturation. Mapping
the color to a wavelength is a straightforward process.

This will tell you more than you want to know.
http://en.wikipedia.org/wiki/HSV_color_space

--
Ray Fischer

Ray Fischer, Nov 26, 2006
4. ### Wayne J. CosshallGuest

Yeppers wrote:
> Hi all,
>
> Is there a simple straightforward set of equations
> for converting normal visible light
> * red
> * green
> * blue
> values from my digital camear valuesto
> * wavelength
> * intensity
> ?
>
> Second question: how about if I put a IR-pass visible-block filter
> such as the Hoya R72 on my digital camera? Can I compute
> infrared wavelength and intensity from remaining RGB data?
>
> Thanks!
>

The answer is probably yes and no, though no in reality. Let me explain.
The RGB filters used in a digital camera have pretty broad spectral
coverage (see the Kodak graph in one of my IR articles:
<http://www.dimagemaker.com/article.php?articleID=466>). What this means
is that the RGB values represent the light intensity over a pretty broad
part of the spectrum. What this means is that, in reality, you cannot
convert this into exact wavelength and intensity values.

However, you could produce a 'simulated' wavelength value by using some
maths but it would not, IMHO, be meaningful in any real way.

Same goes for the R72. See my above mentioned article for how IR
sensitivity works, but the same issue applies. The filter are fairly
broad in their coverage, so you can't pin something down to a particular
wavelength.

Cheers,

Wayne

--
Wayne J. Cosshall
Publisher, The Digital ImageMaker, http://www.dimagemaker.com/
Blog http://www.digitalimagemakerworld.com/

Wayne J. Cosshall, Nov 26, 2006
5. ### Kennedy McEwenGuest

In article <456900c1\$0\$34514\$>, Ray Fischer
<> writes
>Yeppers <> wrote:
>>Hi all,
>>
>>Is there a simple straightforward set of equations
>>for converting normal visible light
>> * red
>> * green
>> * blue
>>values from my digital camear valuesto
>> * wavelength
>> * intensity

>
>Yes. You want to convert from the RGB to the HLV color space. That
>will isolate the color from the luminance and the saturation. Mapping
>the color to a wavelength is a straightforward process.
>

However, that is merely a conversion of the output data and bears almost
no relationship at all to the wavelengths and intensity of the source:
the original scene spectral content.

Say, for example, the green pixels in your camera respond from 500nm to
600nm. Wherever the source spectrum lies in that range you still get
the same output. An almost pure green light at 550nm, or a light at
525nm or even a broad spectrum between 475 and 525 would give the same
RGB output from the camera.

The camera convolves the source spectral content with its own filter
bandwidths before the signal is sampled by the sensor. Consequently
there is no way to get back from the spectral content (wavelength &
intensity for each pixel) with any more accuracy than the filter
bandwidths themselves.

The error in the original question is the assumption that the original
light can be uniquely defined by a set of red, green and blue values, it
can't. Even worse, those red green and blue values will only produce a
certain range of wavelengths and intensity if they are input to a
display with a particular spectral output of each of the red green and
blue channels. In most cases the rgb output is merely an approximation
achieved by calibration, but doesn't need to be any better because
another spectral convolution occurs in your eye.
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's pissed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)

Kennedy McEwen, Nov 26, 2006
6. ### AESGuest

In article <45695002\$0\$23531\$>,
"Wayne J. Cosshall" <> wrote:

> The RGB filters used in a digital camera have pretty broad spectral
> coverage (see the Kodak graph in one of my IR articles:
> <http://www.dimagemaker.com/article.php?articleID=466>). What this

And this leads to an accidental experimental confirmation of your
statement that I once encountered.

The graph on your page shows that the green and blue filters actually
"turn on" again (and the red filter stays on, and the underlying sensor
retains some sensitivity) as you move beyond the visible, well out into
the near IR.

As a result, sending 1.06 micron YAG laser radiation into a digital
camera produces a bright *purple* image, at least in the digital camera
I was involved with -- and it's not due to any harmonic generation.

AES, Nov 26, 2006
7. ### Ray FischerGuest

Kennedy McEwen <> wrote:
> Ray Fischer
>>Yeppers <> wrote:
>>>Hi all,
>>>
>>>Is there a simple straightforward set of equations
>>>for converting normal visible light
>>> * red
>>> * green
>>> * blue
>>>values from my digital camear valuesto
>>> * wavelength
>>> * intensity

>>
>>Yes. You want to convert from the RGB to the HLV color space. That
>>will isolate the color from the luminance and the saturation. Mapping
>>the color to a wavelength is a straightforward process.
>>

>However, that is merely a conversion of the output data and bears almost
>no relationship at all to the wavelengths and intensity of the source:
>the original scene spectral content.

True enough. Finding the original scene's spectral content is not
possible without a spectrograph. You need more than just three
different color sensors.

[...]
>The error in the original question is the assumption that the original
>light can be uniquely defined by a set of red, green and blue values, it
>can't. Even worse, those red green and blue values will only produce a
>certain range of wavelengths and intensity if they are input to a
>display with a particular spectral output of each of the red green and
>blue channels. In most cases the rgb output is merely an approximation
>achieved by calibration, but doesn't need to be any better because
>another spectral convolution occurs in your eye.

What the RGB->HLV conversion will do is give you an approximation of a
single wavelength that will approximate the perception of the color of
the original scene.

--
Ray Fischer

Ray Fischer, Nov 26, 2006
8. ### Wayne J. CosshallGuest

AES wrote:
> In article <45695002\$0\$23531\$>,
> "Wayne J. Cosshall" <> wrote:
>
>> The RGB filters used in a digital camera have pretty broad spectral
>> coverage (see the Kodak graph in one of my IR articles:
>> <http://www.dimagemaker.com/article.php?articleID=466>). What this

>
> And this leads to an accidental experimental confirmation of your
> statement that I once encountered.
>
> The graph on your page shows that the green and blue filters actually
> "turn on" again (and the red filter stays on, and the underlying sensor
> retains some sensitivity) as you move beyond the visible, well out into
> the near IR.
>
> As a result, sending 1.06 micron YAG laser radiation into a digital
> camera produces a bright *purple* image, at least in the digital camera
> I was involved with -- and it's not due to any harmonic generation.

Yup, and it is that rise in transmission that allows IR photography with
digital cameras, providing the separate IR blocking filter is not too
strong.

Cheers,

Wayne

--
Wayne J. Cosshall
Publisher, The Digital ImageMaker, http://www.dimagemaker.com/
Blog http://www.digitalimagemakerworld.com/

Wayne J. Cosshall, Nov 26, 2006
9. ### Kennedy McEwenGuest

In article <4569de25\$0\$34500\$>, Ray Fischer
<> writes
>
>What the RGB->HLV conversion will do is give you an approximation of a
>single wavelength that will approximate the perception of the color of
>the original scene.
>

Whether or not that colour in the original scene was produced by a
single wavelength or not.
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's pissed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)

Kennedy McEwen, Nov 26, 2006
10. ### SkywiseGuest

"Yeppers" <> wrote in news:1164506705.856175.226690

> Hi all,
>
> Is there a simple straightforward set of equations
> for converting normal visible light
> * red
> * green
> * blue
> values from my digital camear valuesto
> * wavelength
> * intensity
> ?
>
> Second question: how about if I put a IR-pass visible-block filter
> such as the Hoya R72 on my digital camera? Can I compute
> infrared wavelength and intensity from remaining RGB data?
>
> Thanks!

In addition to the answer posted by others, I'd like to direct
you to the following page,

http://www.efg2.com/Lab/Graphics/Colors/Chromaticity.htm

You should read up on concepts such as "color space" and "color
gamut".

Many years ago I spent considerable time on this problem. I
eventually emperically derived an approximation curve by
comparing the spectrum of the sun off a grating to views on
the monitor. The result was used for a chart of laser lines.

You can see it here,

http://www.skywise711.com/lasers/reference.html

Look for the "Visible Laser Spectrum Chart".

Brian
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
Seismic FAQ: http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
Quake "predictions": http://www.skywise711.com/quakes/EQDB/index.html
Sed quis custodiet ipsos Custodes?

Skywise, Nov 27, 2006
11. ### IoannisGuest

"Skywise" <> wrote in message
news:...
[snip]

> In addition to the answer posted by others, I'd like to direct
> you to the following page,
>
> http://www.efg2.com/Lab/Graphics/Colors/Chromaticity.htm
>
> You should read up on concepts such as "color space" and "color
> gamut".
>
> Many years ago I spent considerable time on this problem. I
> eventually emperically derived an approximation curve by
> comparing the spectrum of the sun off a grating to views on
> the monitor. The result was used for a chart of laser lines.
>
> You can see it here,
>
> http://www.skywise711.com/lasers/reference.html
>
> Look for the "Visible Laser Spectrum Chart".

Brian,

The HeNe spectrum is outstanding. I'd be interested to know what kind of setup
did you use to achieve such resolution and photo accuracy. Can you perhaps
replicate it here using some ASCII art?

Thanks!

> Brian

--
Ioannis
-------
The best way to predict reality, is to know exactly what you DON'T want.

Ioannis, Nov 27, 2006
12. ### SkywiseGuest

"Ioannis" <> wrote in
news:1164589689.499664@athprx04:

<Snipola>
> Brian,
>
> The HeNe spectrum is outstanding. I'd be interested to know what kind of
> setup did you use to achieve such resolution and photo accuracy. Can you
> perhaps replicate it here using some ASCII art?
>
> Thanks!

top down view (not to scale)

_______ grating
|\
_[ ]_\
|_____| \
camera \
\
\
\
\
() laser

Thanks for the compliment. That was really fun to do.

First, I have several HeNe lasers. I took one of my larger tubes
out of it's casing and stood it vertically on a wooden chair.

Yep. I'm high tech.

I have this old beat up reflective diffraction grating that I
bought at a military surplus many years ago.

I set up my digital camera to look at the grating at the diffracted
image of the HeNe bore. The camera was set as close to the grating
as possible without physically blocking the incoming light.

I zoomed the camera all the way out and started snapping pictures
at the highest resolution. I think I needed 3 or 4 images to cover
the whole spectrum. I had to do several test shots to get the
exposure settings right. I wanted to 'go deep' without saturating
the brighter lines and over exposing the background noise caused
by the poor condition of the grating.

Then I stitched it all together in software.

Figuring out all the emission lines was done by referencing some
of Lasers.

Brian
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
Seismic FAQ: http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
Quake "predictions": http://www.skywise711.com/quakes/EQDB/index.html
Sed quis custodiet ipsos Custodes?

Skywise, Nov 27, 2006
13. ### IoannisGuest

"Skywise" <> wrote in message
news:...
[snip]
> top down view (not to scale)
>
> _______ grating
> |\
> _[ ]_\
> |_____| \
> camera \
> \
> \
> \
> \
> () laser
>
>
>
> Thanks for the compliment. That was really fun to do.
>
> First, I have several HeNe lasers. I took one of my larger tubes
> out of it's casing and stood it vertically on a wooden chair.
>
> Yep. I'm high tech.
>
> I have this old beat up reflective diffraction grating that I
> bought at a military surplus many years ago.
>
> I set up my digital camera to look at the grating at the diffracted
> image of the HeNe bore. The camera was set as close to the grating
> as possible without physically blocking the incoming light.
>
> I zoomed the camera all the way out and started snapping pictures
> at the highest resolution. I think I needed 3 or 4 images to cover
> the whole spectrum. I had to do several test shots to get the
> exposure settings right. I wanted to 'go deep' without saturating
> the brighter lines and over exposing the background noise caused
> by the poor condition of the grating.
>
> Then I stitched it all together in software.

Your diagram looks like it doesn't use any slit. How did you collimate the
laser beam into nice vertical lines without a slit?

> Figuring out all the emission lines was done by referencing some
> of Lasers.
>
> Brian

--
Ioannis
-------
The best way to predict reality, is to know exactly what you DON'T want.

Ioannis, Nov 27, 2006
14. ### Don Stauffer in MinnesotaGuest

Yeppers wrote:
> Hi all,
>
> Is there a simple straightforward set of equations
> for converting normal visible light
> * red
> * green
> * blue
> values from my digital camear valuesto
> * wavelength
> * intensity
> ?

No, there is no simple set of equations. The problem is that the
filter responses for the red, green, and blue filters may not have
easily mathematically describable passbands. The response curve for
each is likely to be a very complex equation, either a power series
with many coefficients or some other type of curve fitting equation.

>
> Second question: how about if I put a IR-pass visible-block filter
> such as the Hoya R72 on my digital camera? Can I compute
> infrared wavelength and intensity from remaining RGB data?
>
> Thanks!

Same problem. The IR bandpass filter is unlikely to have a simple
equation, although some of them can be modeled by a sin(x)/x or a
normal curve equation with sigma representing the passband. So this
problem is likely to be easier than the original one. However, you
need to solve the original one before you can do the second.

Don Stauffer in Minnesota, Nov 27, 2006
15. ### SkywiseGuest

"Ioannis" <> wrote in news:1164625065.599242@athprx03:

> "Skywise" <> wrote in message
> news:...
> [snip]
>> top down view (not to scale)
>>
>> _______ grating
>> |\
>> _[ ]_\
>> |_____| \
>> camera \
>> \
>> \
>> \
>> \
>> () laser
>>

<Snipola>

> Your diagram looks like it doesn't use any slit. How did you collimate the
> laser beam into nice vertical lines without a slit?

Didn't need one since the laser bore was already a narrow line.
The further from the camera the laser is, the narrower the 'line'
of the bore appears.

The distance I used was about 5 feet.

Brian
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
Seismic FAQ: http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
Quake "predictions": http://www.skywise711.com/quakes/EQDB/index.html
Sed quis custodiet ipsos Custodes?

Skywise, Nov 27, 2006
16. ### TomasMarny

Joined:
Sep 19, 2012
Messages:
1

TomasMarny, Sep 19, 2012