Gloger.es@XEROX.ARPA (07/26/84)
Besides your rocks, notice that ice is almost colorless, but when shaved or crushed it turns milky white. I think what you're seeing must be that highly irregular surfaces reflect almost all colors of light in random directions directly from the surface, so that they appear white. This goes for objects of any color, from black to white to colorless transparent. To the extent the object material is transparent, it appears white as either its external surface or its internal structural faults are randomly reflective. If you take an object with an irregular surface and polish it, you're then able to see past the previously apparently white surface to the true color of the material, or, to the extent the object is transparent, you can then see past the white surface to the interior. If you take any material and reduce it to fine dust, it will tend to look white. (Have you ever seen flaming red dust?) So what color is the moon really? It looks off-white, but if you could polish a dust particle of the type which constitutes its surface, might it be green? By the way, is a yellow rose yellow? That is, does a yellow rose appear yellow because it reflects yellow light and absorbs other frequencies? Or does it not reflect yellow after all, does it perhaps instead reflect red and green which combination the eye senses as yellow? More generally, for objects in nature which appear in colors other than primary light colors, what number and kinds of objects actually reflect their apparent color in a single frequency band, which reflect multiple other colors that the eye combines to the apparent color, which do both? (I suppose I should go play like Isaac Newton with a prism. Short of that, there are presumably clues to be had from observing how many objects exist with apparent colors not contained in the rainbow spectrum, and thus their colors necessarily composed of multiple others - or are there such out-of-the-rainbow colors?) Paul Gloger <Gloger.es@XEROX.ARPA>
marcus@pyuxt.UUCP (M. G. Hand) (07/27/84)
A similar effect can be observed in fogs, smokes and colloids - the light is scattered producing a whitish appearance. There is one whole branch of spectroscopy that depends on this phenomenon (I cannnot remember what its called - Raman?) Its caused when the size of the particles (roughness of a stone) is close to a certain ratio wrt the light waves. Marcus Hand (pyuxt!marcus)
Gilman.ES@XEROX.ARPA (07/30/84)
But what about Xerox toner, which seems to be the size of flour particles, but is solid black?
bradford@Amsaa.ARPA (08/01/84)
From: Pete Bradford (CSD UK) <bradford@Amsaa.ARPA> In answer to a couple of queries that I have received on my comment that if you grind up a substance finely enough it will appear black, here is a simple explanation. Once the particle size becomes smaller than a value which is approximately the wavelength of light, radiation of that wave- length can no longer be affected by the said particles. no reflect- ion therefore takes place (no absorption either, presumably) and the 'powder' appears black. This is presumably the answer to Gilman's question about Xerox toner. PJB
MJackson.Wbst@XEROX.ARPA (08/02/84)
"Once the particle size becomes smaller than a value which is approximately the wavelength of light. . .This is presumably the answer to Gilman's question about Xerox toner." Wrong. Xerographic toner varies somewhat in size but typical values lie in the range of 5 to 15 microns. Visible light has wavelengths in the range of approximately 400 to 700 nanometers, or 0.4 - 0.7 microns. Xerographic toner is black because it absorbs the light incident upon it (mostly because of the carbon black in the [heat-fusable] polymer resin). The color of an object is the result of the way it interacts with incident light. One consideration is that to the extent that there is front surface reflection the spectral energy density of the illuminant is unchanged. (Try looking at a glossy print--unless you hold it so that direct reflections of light don't reach your eye the colors will be washed out). Light that isn't reflected at the front surface interacts with the bulk of the object; this may involve absorption and reradiation, absorption during transmission, reflection from internal surfaces, etc. This is how the SED of the light, and hence its perceived color, is changed. As to the original question, presumably rock dust scatters light at the particle surface, thus not changing the color. The random scattering gives a dull, white (color of the illuminant) appearance. By the way, when particles get small enough so that light no longer interacts with them they appear TRANSPARENT, not black (no absorption). That's why a solution is clear but a colloidal suspension is translucent or opaque. Mark
JGA%MIT-MC@sri-unix.UUCP (08/02/84)
From: John G. Aspinall <JGA @ MIT-MC>
Date: Wed, 1 Aug 84 8:33:15 EDT
From: Pete Bradford (CSD UK) <bradford at Amsaa.ARPA>
To: physics at sri-unix.arpa
Once the particle size becomes smaller than a value which
is approximately the wavelength of light, radiation of that wave-
length can no longer be affected by the said particles. no reflect-
ion therefore takes place (no absorption either, presumably) and
the 'powder' appears black.
Come on, if no reflection and no absorption take place, where does
the light go? Transmission? Be serious.