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Author Topic: color and light  (Read 12099 times)

Offline gsmollin

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Re: color and light
« Reply #25 on: 09/09/2004 17:51:34 »
Actually, I was posting an answer to my question, but it never happened. Here is what I meant to say:

R+G+B=white, so these are the additive primary colors. This is true if they are truly mixed, say using prisms, mirrors, or lenses. It is also physiologically true if they are mixed by showing dots too small to be resolved by the eye at a distance. The dot pitch on your computer monitor, for instance is about 0.26 mm. With a magnifying glass you can see the red, green, and blue dots. The use of additive mixing in CRT screens is a natural for the way a CRT works. When the electron beam is off, the tube looks black. The surface of the tube has a black coating on it to make it look black. The RGB guns light up the RGB phosphors, and we see colors.

The opposite is true of the photo, painting, or half-tone-printed page. The paper is white, so to see colors, we must subtract the colors we don't want to see. This is obvious to anyone who has colored or painted. The problem I was having was with half-tone printing. It is a mosaic of CMY dots and other shapes. (Many times special colors are used, however, so you must be sure you are looking at a CMY-half-tone-printed image) But these dots reflect light individually, so their contributions are added- giving us additive color mixing. That was my point of confusion. The apparent answer to this question is that the subtractive primary dots are combined by our visual process, and appear to us as if they had been mixed together like paints.

I imagine the visual process works something like this:
The light from the C, M, and Y dots all falls on 1 cone cell, since the dots are too close together to be resolved individually. The cone cell responds as if the C, M, and Y inks had been mixed together, since it sees them as one. The result is subtractive color mixing. The same visual process is happening for additive color mixing.

OBTW, if also found out that "k" is used for black, since "b" was already used for blue. Duh.
 

Offline McQueen

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Re: color and light
« Reply #26 on: 10/09/2004 07:28:38 »
An interesting corollary to the theory which I have put forward  is that all light is due to absorption and re-emission of photons of  a particular colour. The proof of this is very simple , just look in a mirror !!  A mirror as we all know is made up of a glass  sheet generally  of from 3 5 mm thickness at the back of which is a silvered layer. Now , it is common knowledge that light travels through substances like glass by being absorbed and re-emitted by electrons in the glass , therefore in order to see the colours of an image reflected in a mirror , the photons of those particular colours must first journey through the glass being in turn absorbed and then emitted by electrons in the atoms making up the glass , till they reach the silvered surface where they are absorbed and re-emitted one final time before making the reverse journey back through the glass from atom to atom and from there to the retina. The question of why the silvered layer reflects light back in the direction from which it has come is due to the properties of metals but it seems that this property must be shared at least partially by all substances , although it is obvious that with metals all light is emitted and absorbed while other substances selectively absorb and emit light.  Thus it is plain to see that all colours are due to a process of absorption and emission of photons of the requisite wave length.  Now supposing we have an object which is pure blue under white light what is happening ? We have to assume that the object is made up of atoms whose electrons are particularly susceptible to photons with the energy and wave length  of blue light (around 550 nm ) and practically ignore all other wave-lengths and energies. We know that something very much like this must happen from our study of spectroscopy , where certain substances only absorb and emit light of certain colours , for instance sodium will only emit and absorb yellow light , ignoring all other wave lengths and energies. No-one would suggest that sodium is absorbing all the photons of other colours and reflecting only the yellow light , in fact it is simply ignoring the other photons and selectively absorbing and emitting yellow photons ! Returning to the case in question of the blue object , we must assume that when white light shines on this surface , the electrons of the atoms making up this object absorb and re-emit only the blue light so that the  object looks blue. This is an intrinsic property of the  object , thus it retains this property even when there is no light. However if light of a different frequency , (as for instance ultraviolet light ) falls on the surface , it activates different electrons with the result that photons of a different wave-length are absorbed and emitted , while the blue photons which were predominant under white light are suppressed , and the object changes colour. The most interesting part of this discussion so far as I am concerned lies in coming to an understanding of just how fast and how continuously electrons emit and absorb photons under the influence of radiation even by  white light . The emission and absorption of photons by electrons under the influence of radiation  is seen to be a practically   continuous and unremitting process. It is difficult to imagine while looking into a mirror or through a window ( this for those who  still have reservations  about my ideas )  the number and speed at which these emissions and absorptions of photons  must  take place in order to form the continuous  images which we perceive.
 

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Re: color and light
« Reply #26 on: 10/09/2004 07:28:38 »

 

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