Spin Off Of - How We Can See Black?

Maybe is simplest to say it's reflected by materials which are unable to absorb it (by virtue of their 'stability' ) ?
Pretty much the same applies to throwing a ball at something....
Del

(Maybe Vermin will explain why he bounces?)

If you analyze Vermin from COV (Center of Vermin) coordinates, the world bounces.

Del, your answer is correct. But that opens up another question as to why there is absorption.

I don't remember 100%, but I think it has to do with the electronic structure of the atoms of the object that is hit by the light. The energy level of those outer electrons determines weather the energy is absorbed or not. I kind of remember that the photons are reflected when they hit the atom and then an outer electron is promoted to a higher energy level by means of the "photons energy" then is unstable in that energy level and falls back to its original energy level releasing the extra energy as a "bounced photon" of certain energy (wavelength) . Thus reflecting that wavelength.

So, are we headed for a black hole, or perhaps we'll bounce away from it as black-body radiation!!??

Don't forget that there is light scattering, too. Blackbox, dull, satin, glossy, and mirror finish all have different scattering even when the "color" and/or underlying chemistry is the SAME.

Also, especially on the white side, identical formulas, chemistry, AND surface scattering can result in different overall light scatter when the pigment refractive index, particle size, particle size distribution, particle separation, uniformity, porosity, shape, and particle density change (i.e. all clay, calcium carbonate, titanium dioxide products are NOT created equal). Not simple.

Wide or narrow illumination and view field, and angle of illumination and view are all important.

Add to that the term "color" depends on the illumination and viewer (are you color blind, using rose-tinted glasses, illuminating with fluorescent or daylight, on a store shelf for display or hiding from deer, at sunset or noon, trying to see color within a sealed box with no illumination, using night-vision glasses, determining the presence of government-issue postage stamps or not, etc.).

There are a variety of immuninations and instruments to measure specific characteristics of optical properties. The important thing with color is: decide what is important to your application (both illuminant and viewer), decide on a way to measure (angle of view, wavelength narrow or wide), whether there is an industry standard (turbidometer, or Laser light scattering Horiba, L*a*b*, CIE, etc), and go with it. Have fun!

Oops, that's the particle mode for absobtion and re-emission of photons. And if you stimilate that emmision you have amplificatiuon through stimulated emision of radiation. Anyhow on to the question originaly asked. The principles of Reflection (not scattering etc) use the wave modle for light. So as long as the material does not absorb or transmit the light wave, it is reflected. As mentioned about RF energy, it is not absorbed or reflected by most materuals except metal. But non-metals do refract RF energy just as translucebt/transparent materials do for light. As for scattering and such, you will need to go with the particle modle. For those who want to know a little more about the controversy regarding light may I suggest checking out The Photoelectric Effect on either your favirite search engine or that thing called a Library :) BBerried

Light may be scattered within materials that are not completely opaque by bodies within the material. If a material has particles within it (say, a layer of pigmented paint), the wavelength scattered is twice the wavelength of the particle size. This has nothing at all to do with photoelectric effect, and need not be related at all to the surface of the object. But, yes it has to do with the particle behavior of light.

I include scattering in the answer to the question because..

Incident light = Absorption + Transmittance + Scattering + Reflection

...if all the light is absorbed, transmited, or scattered, (and there is no emission) there will be none left to be reflected, and the object will be "black".

Black is the lack of reflected incident light from the object to the sensor. Pure Black can be perceived by difference with non-black, but cannot itself be seen. The purist answer is NO, BLACK CANNOT BE SEEN. The truth is, many items we are inclined to call "black" are in reality blue-black, deep grey, brown-black, etc. and can been seen by instrument or eye, providing there is incident light and suitable optical arrangement of the source, object, and sensor.

"...NO, BLACK CANNOT BE SEEN..."

After a fashion, it can. By contrast with whatever surrounds a black object (even a true black-body that absorbs ALL radiation) you can distinguish it's location and outline, although not any surface details.

The true black of an absence of light, such as in a deep cavern or mine shaft, is another matter. Not only can you see nothing, you can't even get much perception of where you, yourself, are in relation to anything except what you stand on.

well said!

If you can imagine looking through infrared binoculars, you might better understand how the brain and eye communicate in respect to the relationship of light and or energy. Going back to the original poser: Who said that black can be seen? Close your eyes. Let me spin a roll of duck tape around your head. Tell me what you see. Black? Or nothing?

It's kind of like PONG when you slide the little paddle back and forth really fast. Sometimes you bounce the little sucker back and sometimes you don't. Depends on the frequency of the little paddles.

Or another way is to think of a sifter. The big chunks don't pass through and the little ones go all over.

With the value of the dollar today, I'd rather have the pound or the euro.

Never figured out how to get that funny L !

Hi,

reflection is from smooth surfaces, scattering from rough surfaces.

Absorption is from damping of electromagnetic waves entering the material:

simple dielectric and/or magnetic losses (at the frequencies of "light"), also resonant losses if resonant molecules or atoms or color centers or quantum wells or any of these odd physical phenomena that have color as a side effect.

Also resonant losses of gas molecules (the masses of the gas atoms are bound by spring-like attractive forces), these resulting in infrared absorption of our atmosphere and UV absorption of the ozone layer and IR-opacity of carbon-dioxide and methane, the greenhouse effect.

Reflection can be from nonconducting or (electrically) conducting materials, this also thought at the frequency in question. In conducting material the electric field vector of the electromagnetic wave is short circuited in the direction of the surface plane, the magnetic changed according to the permeability.

Search fro the companies that produce reflectometers and ellipsometers they provide the suitable theoretical background, you can calculate from Maxwell's equations -if having the dielectric constant ε and the permeability µ - the reflectivity.

This is dependent on angle of incidence and on polarisation!

The index of refraction is n = √(ε/µ).

RHABE

smooth is a relative term. It's smoother The smaller the scale of surface roughness in relation to the wavelength of light.

And light can be scattered from within a material, not just at its surface (OK so it's scattered from interfaces within the material like bubble/material or pigment/matrix surfaces).

Anyway, not with our eyes closed we can't!