Can someone explain this statement on luminous transmittance?

  • #1
Rap
827
10
TL;DR Summary
Trying to understand the meaning of "luminous transmittance" as a function of wavelength for standard illuminant "C".
I have a table of wavelengths and percentage values. Can someone explain this statement:

"The following pages give percentage luminous transmittance at wavelengths 500 to 700 nm. for the standard Illuminant "C" adopted by the CIE."

What I need is the ratio of the energy coming out of the filter JF(w) to the energy going into the filter J(w) as a function of wavelength w, which has nothing to do with the response of the human eye and the luminous efficiency function V(w). If I have a wavelength of 450 nm followed by 78%, what does that 78% mean?
1707572354138.png
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Rap said:
TL;DR Summary: Trying to understand the meaning of "luminous transmittance" as a function of wavelength for standard illuminant "C".

Can someone explain this statement:
Hmm. If they quote the transmittance as a percentage at each wavelength then I don't see what more needs to be said about the filter. The percentages would apply to any wavelength and that would be sufficient for objective measurements. If you have just dived into an article then you may have missed their reason for the statement. If you are already across Colourimetry and colour reproduction etc. etc. then what `i'm saying may be bloomin' obvious. Look at this link (there are better sources, of course) for some ideas about colourimetry.

Now, the transmittance values are probably being quoted in the context of colourimetry and colours perceived by the (average) eye are subjective. The dark line with the dots on it in the middle of the coloured CIE chart represent the colour coordinates for several illuminants. I think that Colour Balance may be where they're heading. If you look at a coloured surface, illuminated by a particular source (illuminant C is one of several standard illuminants) and then under a different source, the subjective effect is to change all the colours of objects.

So basically, adding a filter to the illuminant will change where the colours of objects. In particular, they mention the non-spectral colours (purples etc) which are very much affected by the illuminant.

Hope this helps.
 
  • #3
Hi sophiecentaur - thank you for your response, but I still need a precise explanation of what that "78%" means when the table says "540 nm, 78%". If I shine a 540 nm monochromatic light beam of *radiant* intensity J thru the filter, will the emitted beam have a radiant intensity of 0.78 J ? By radiant intensity I mean physics energy, proportional to h-nu times the number of photons per second. If so, why talk about "luminous transmittance", and "CIE Illuminant C"? When I see luminous, lumens, etc, it means to me that it involves somehow the response of the human eye. For example, luminous intensity at w is radiant intensity at w times the luminous efficiency function V(w) which is the ratio of the eye's *perceived* intensity of a beam at wavelength w divided by the maximum perceived intensity of a beam with equal radiant intensity, which occurs at about 550 nm. I believe that the "total luminous transmittance" of a filter is the integral over wavelength of outgoing luminous intensity divided by incoming radiant intensity, but then there is no wavelength dependence. I don't understand what "luminous transmittance" vs wavelength means, nor why the "illuminant C" needs to be mentioned, despite some time spent trying to tie it down.
 
  • #4
Rap said:
If I shine a 540 nm monochromatic light beam of *radiant* intensity J thru the filter, will the emitted beam have a radiant intensity of 0.78 J
I think that's the right idea - except that it's not Energy (J or equivalent) but Power (W or equivalent). From what I have read recently, the the luminous transmittance is used for transparent materials and represents the total light transmitted through, less absorption and scattering; a very practical measure. (Try to forget about hν in this context. Photons do very little to help understanding in these classical problems; stick with Power or sometimes Power per unit Area.

You are right about how light flux / lumen defined. 'They' talk about "amount of light" rather than Power and are slppery about how to get from one to the other. As you say, there is an implied factor of the eye's spectral response to do a convertion and, to be precise, the spectrum of the illuminant has to be relevant too. Most of what I find is for consumers to choose the right light bulb!! Not much use here.

Before we continue, what is your application for all this? What level are we at?
 
  • #5
  • #6
Hi sophiecentaur - I am dealing with MacAdam ellipse raw instrument data, which is used to draw a "just noticeable difference" ellipse in the XYZ color space. Roughly speaking, it is an ellipse, inside of which any color is indistinguishable, to the human eye, from the color at the center of the ellipse (assuming constant luminance). I want to draw an ellipse in LMS space, which is the space which uses the response of the eye's cones, which is what should be used, but they didn't know what the cone response was back in 1942 when MacAdam analyzed his data. MacAdam compared two colors for his tests, which were generated by Wratten filters (which he listed), and we need the (radiant) transmittance of these filters as a function of wavelength in order to properly convert his raw data to LMS space. There are rough conversions from XYZ space to LMS space, but ultimately they are too inaccurate, we need the filter data. So, I have the Wratten filter data, but, as you see, I'm not sure what it means.

I don't see how the spectrum of the illuminant is important in the sense that any illuminant can be used, as long as you know its intensity as a function of wavelength. Results at one wavelength don't depend on the intensity or transmittance at some other wavelength. When you integrate over wavelength, then, yes, you need the whole picture.

I will stick with energy, power, etc., I was just trying to make it very clear what I meant by "intensity" since there are too many sources that don't make that clear. I know what you mean - I search for the definition of luminous transmittance, and they talk about "amount of light" and you don't know whether its radiance or luminance. Wikipedia is no help. I might edit the "transmittance" article there as soon as I think I know what I am talking about. :)
 
  • #7
Hi Rap
It's good to have some context to this; thanks.
I still have a problem with converting between luminous intensity, Lumens or luminous flux and Watts (of EM wave power). This may not be relevant in this context but it would be nice to relate one unit to another.
Rap said:
Results at one wavelength don't depend on the intensity or transmittance at some other wavelength.
Objectively that's got to be true but when we consider JND's, that involves the values of three retinal sensor spectral sensitivities. Wouldn't any full set of subjective tests have to involve reference to a White (illuminant) so you'd be able to find what actual level of luminosity (or, rather, Power) was coming from each of the coloured sources. I never read that paper but I imagine the work would have had to start with Power Flow in a given wavelength interval, rather than luminosity or else, wouldn;t they be pulling themselves up by their own bootstraps?
Whenever I read "lumen" I also read "perceived", which seems to be missing something objective out.

All this is taking me back more than half way to the time the McAdam work was done, in the 30's. I did my colourimetical stuff. in the late 60s but 'everyone' still harks back to the CIE publications. It's surely about time, now we have such good TV displays, that the whole thing was re-visited. Perhaps that would go against commercial interests.

I await responses to the other thread.
 
  • #8
Yes, MacAdam does start out by shining a beam of light from "Illuminant A" through two filters producing two colors, which are later mixed and matched. The "white" Illuminant A is just a special tungsten lamp which emits very nearly black body radiation corresponding to 2848 K, at a particular radiant intensity. There is a dial which the observer can adjust until there is a perceived match. Then the fun begins, translating the raw data (dial readings) into points in XYZ space. The design of the instrument is kinda genius actually, it automatically maintains constant luminosity no matter how the colors are mixed. MacAdam only gives the XYZ coordinates of Illuminant A shining through each of the filter he used. This is all that's needed to convert the raw data to XYZ coordinates. He specifies the Wratten filters he used, but doesn't give much information on the filter transmittance vs wavelength which he used to arrive at those XYZ coordinates. If I want convert the raw data to LMS coordinates, I need that information.
 
  • Like
Likes sophiecentaur
  • #9
Rap said:
it automatically maintains constant luminosity
My problem is about how they actually measure luminosity. Any sensor will basically measure Power received. Luminosity involves perception. Somewhere there must be a conversion which gives luminosity from the measured power and they must use that in a feedback loop. I haven't found how to do that. Sorry to burden you with a further problem but it basically affects you too (and you need an answer while I am just ribbernecking).
Illuminant A is very 'red' for TV use but the filament is cooler than illuminant C colour temperature, which is basically daylight but halogen filaments wouldn't last at that temperature.
 
  • #10
Yes, "white" is a very flexible term. I used to work at a lamp factory, and they had a filter whose transmittance closely matched V(λ), the luminous efficiency function for the human eye. Then they just integrated the radiant power out of the filter over λ to get lumens.

Also, I found some old Kodak publications listing Wratten filter transmittance vs λ in percent, and they very clearly state that it is (radiant intensity out)/(radiant intensity in). I guess you could call that "spectral radiant transmittance". These transmittances are similar to the ones in the reference I showed you, in the sense that they don't invariably go to zero outside the 400-700 wavelength interval, like V(λ) does. So I am coming to believe the example of 78% does mean spectral radiant transmittance. Maybe the "luminous transmittance" is a common but unfortunate misuse of terminology, and the reference to Illuminant C is just an irrelevant side note on the illuminant used to make the measurements.
 

Similar threads

Replies
16
Views
710
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Other Physics Topics
Replies
4
Views
14K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
8K
Replies
8
Views
2K
Replies
2
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
10
Views
7K
Back
Top