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Name: Rachel
Status: student
Age: 17
Location: N/A
Country: N/A
Date: 8/28/2004


Question:
I was wondering exactly why is it that glass can have visible light pass through it, but no UV, and why brick for example does not allow either of them to pass through. I understand that it is due to a combination of the energy of the emr and the material/structure of the "thing" but i do not understand the relationship. Is it absorbed and then re-emitted? and if so, why are only some types absorbed for different materials, and others for other materials?


Replies:
Imagine these series', starting from a completely clear object:

scattering - clear glass cube, wavy glass block, pile of glass chips looking sparkly, pile of crystal sand looking almost white, bag of glass powder completely white, flat white brick.

absorption - clear glass, darkish transparent "smoky" glass, deep black glass looking glossy but jet-black.

frequency-selective absorption - clear glass, faintly yellowish clear glass, rich orangish clear glass (you can see everything through it, but blue things look black, just like "blue-blocker sunglasses").

The common impression of being "opaque" usually consists of combined strong scattering and some amount of absorption. With scattering but no absorption the object is "translucent". With a tiny amount of absorption the object is white. Actually 80%-98% white, the remainder black. Every real white paint is just such a very light gray. Adding more absorption causes darker grays until you reach flat black. If the absorption is frequency selective the opaque object has an opaque, "pigment-like" color. (i.e. "brick red")

Your example brick blocks UV by both scattering and absorbing it. Visible wavelengths, too. Some part of the light, UV or visible, will be bounced around inside by scattering and escape back out the front face before it is completely absorbed. This is why no flat black is ever completely black, and no red ever appears 0% blue.

If there is no scattering so there can be plenty of glass-like transmission, then I call the colors "dye-like" instead of "pigment-like". Imagine food-coloring in water for "dye-like" colors.

Only fluorescent colors absorb and re-emit light. Most of your familiar appearances involve transmission and/or reflection in varying degrees, but no re-emission.

Scattering is done by changes in index of refraction of the clear media the light is penetrating. Sudden changes reflect a small fraction of the light, and any change will bend the transmitted light if the light-ray approaches the surface of the changed zone at a slant.

Metal surfaces reflect most of the light strongly, and absorb the transmitted remainder in an extremely short distance. This is a different method of being opaque. If the surface is rough the light is reflected scattered instead of mirror-like. Sand-papered aluminum sometimes looks rather like white.

Any color can have a gloss if there is a smooth clear film over the surface. Any metal can be given a color by having frequency-selective absorption in a film over the surface.

Your UV-absorbing clear glass is just an example of frequency-selective absorption. You can think of UV as a "fourth color" in addition to the three our eyes see: red, green, and blue. It can be absorbed or not, independently of what is happening to the other colors. Although most near-UV-absorbing things are not perfectly selective and absorb a little blue as well, looking faintly yellowish. Most ordinary glass absorbs far-UV but not near-UV. It can easily look visually colorless. Always an open question with glass, how much UV it will absorb or transmit.

Jim Swenson



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