Black Mirrors

My theory is that the backing surface, black or white is some distance back from the front glass as in a mirror. The front glass should reflect to some extent or else it would appear invisible. If the white is reflecting due to refraction you will get some minute blurring due the the presence of two images, very close to one another. The degree of this over the black surface would be much less although the intensity of light returning would be much less.

As far as a plain black surface, versus a plain white surface polished to mirror finish. I'm not sure. Possibly the wavelength of light returning off a white surface is sctattered more easily than that off of a black surface? Is it because white is a combination of all color wavelengths where black is more along the lines of a singular wavelength of color therefore being clearer? Couldn't tell you for sure.

-T

Yours is a very interesting question. We tend to think of reflection as something related with a beam (a wave, if you wish) bouncing on an ideal surface. This can be true in a few cases, if we neglect distances that are a fraction of the wavelength: metallic reflection, surface reflection and total reflection on ideally transparent media. Polished surfaces reflect the light in two different ways. Let's think of polished stone, granite for example. Some parts of the surface act as transparent media, reflecting some light as a primary reflection. But part of the light goes thru the surface, and may be absorbed or reflected by the particles inside. This secondary reflection reaches the surface at different angles, because the reflecting particles' surfaces are not aligned with the main surface (for example, the mica laminas inside are randomly oriented), or they are too small and diffuse the light instead of reflecting it. This leads to a "diffuse" reflection.

"Black" particles will absorb much more light than they reflect. The macro effect you can see depends on the ratio between primary and secondary (diffuse) reflections. For a "black" specimen, it is high, because the secondary reflection is low, so the secondary reflection doesn't fade the image you expect to see, while for a white sample it does.

Please, notice that the terms "primary" and "secondary" are mine, and not necessarily main stream.

Guest, you ask "Why the reflected immages by the black "mirror" are remarkably clearer than by the white one?"

Is it true the image from the black reflective surface is clearer? I guess I have never seen a distinct difference. To be honest, never occured to me to look.

Is this an observation of yours, or are there study data available for us to review?

Also, I assume you mean a single reflective surface (as a painted piece of steel, polished to be very reflective), as opposed to a double reflex (piece of glass over reflective) surface. Is this correct?

Hi Doorman;

Yes it's true, you can easily find both, black and white objects with such a finish and see for yourself.

This is an observation from my younger years that I never bother to resolve.

Your assumtion is correct they are single reflective surfaces; no glass or any other coating on top.

Yahlasit

This is just me, coming in from another angle. You're comparing with your eyes; your pupils dilate to adjust for light. When you look at the image with a black outline, your eyes will open and let in more light, like the aperture on a camera. When you look at the same image outlined in white, your eyes will close some. Your eye is adjusting to the total light entering, and it may affect your opinion of the image.

In this scenario, the image on the white surface would be sharper since, just like the aperature of a lens, the depth of field is increased as the aperature is reduced. This is caused by reducing the "circles of confusion."

Our eye is senses the ratio of the darkest to lightest to perceive details. In reflections from a white surface this ratio is less than from the black surface.

That is exactly what I thought at first, but now I not so sure since I put both, black and white mirrors one besides the other, closed one eye (to avoid the odd posibility of adjusting my pupils independently) and still saw better reflections on the black one.

Thanks for answering, we've got to figure this out with absolute certainty, hopefully someone versated on optics is just waiting to give us all a wisdom pearl.

Yahlasit

Since the gamma ratio is better for black reflections than for the white the eye perceives the black reflection better.

What is all too often forgotten is that all we sense are from our biological sensors, in this case, our eyes. As with all sensors, there is a limited range of operability, and all sorts of factors exist that may interfere with the level or clarity of detection (resolution). Call it noise, interference,signal too weak, stimuli too strong, or what have you...the fact is that our sensors (eyes) are better suited for detecting images with a bit more contrast, and less brightness (white noise, in this case) in images. The filters / auto-adjust functions of our eyes as described by mikeh are great, but have their limitations. Much rather have my relatively low speed eyes than those of some insects that are well aware of the 60 on/off cycles of flourescent lamps every second... Just signed up, my first post...please don't be too harsh!

Rookie mistake, I meant to credit mike k not mikeh... Alittle more on the operating range of our eyes, one that aggrevates me to no end with my BlackBerry...a 1W bluie LED will appear brighter than a 1W red LED...and whatever the actual power is on this thing...it's too bright for me!

Are you sure that the surface roughness is the same on both "mirrors"?

Roughness will generate non-specular reflection = stray-light. This is distributed ideally into any angle of the half-sphere above a flat surface.

In reality there are surface marks from machining and scratches that result in deviations from ideal behaviour. That is measured in BRDF= bidirectional reflection distribution function.

Specular reflection divided by the part of the stray-light that enters your eyes pupil is signal to noise ratio.

Your black surface seems to be better with respect to this!

RHABE

"What about the black surface? it is supposed to adsorb all visible light, but it reflects immages apparently better."

This assumes that the 'black' in question absorbs all visible wavelengths of light completely... I would love to get my hands on some material with that property... If the material you are using were to completely absorb all visible wavelengths then there would be no reflection at all (and it would probably get really hot).

When you look at a mirror the light reflecting to your eyes comes from multiple directions and consequently reflects in multiple directions. Your eyes will pick up these mis-aligned lightwaves as well, which causes 'bluriness'. I am near certain that the improved clarity on a black mirror is largely due to a reduction in this visible 'noise' because more wavelengths are absorbed instead of reflecting back vs. a white mirror.

Good question. Most materials only reflect efficiently at very small angles, and are much less efficient at angles close to normal (perpendicular to the surface). Even a reasonably smooth piece of wood will reflect a crude image if the angle is small enough. I assume that you are talking about reflections at angles close to normal.

You don't mention what these plates are made of, but let's assume they are not metal (I can't think of any metal that is white or black). Polished metals, such as gold, silver, and aluminum reflect 85% to 99% of the incident light. They also keep their high efficiency at almost any angle. Other polishable non-metalic material such as glass, plastics, ceramics, and stone reflect a fraction of that under the best (small angle) conditions.

So when you look at the black reflector, the light energy in the reflection is just a few percent of the incident light. The rest passes through the reflecting surface, and is absorbed in the sub-surface layers, never to be seen again. This produces a dim but clear image. When you look at the white reflector, again just a few percent of the light is reflected. Assuming that the surfaces are equally well polished, the reflection from the white surface is just as dim but clear as from the black surface.

But for the white reflector, the rest of the light passes through the reflecting surface, and rather than being absorbed, it is scattered back in all directions, including into your eyes. So you are looking at a dim reflection against a white background, and the contrast is greatly reduced.

The eye-brain system has a number of contrast increasing features, some of them in the form of negative feedback loops in the retina itself, others in the optic nerve, and still more in the visual cortex. In the case of the reflection from the black surface, these features get the job done, because there is little or no stray light in the reflection. For the white surface however, the amount of stray (scattered) light may be greater than the amount in the reflected image, and your neural circuts don't work well. The white reflector has a much worse signal to noise ratio. So while both reflections are equally sharp, the veiling glare from the white surface overwhelms the image. If you look at a refelction at very small angles from both surfaces, the percentage of the reflected light will be greater, and the difference between the images will be much less.

I agree with you (as well as other posters) this is a great topic for consideration.

I find it interesting that my thought of reflective surface, black and white, is a polished automobile hood, trunk lid, bonnet... something along that line. No one else is considering this... of course, I am a little off the norm sometimes.

Are the elements of your discussion and explanation still valid for a painted and polished surface? I would think yes, but don't know. How about varied types of wax toppings for the finish paint?

This IS an interesting topic. Good Job!

Are the elements of your discussion and explanation still valid for a painted and polished surface?

Yes, because the paint is not really opaque: it consists of a pigment and a carrier. The carrier is at least partly transparent. Otherwise for a 'color' paint you would not be able to see the light reflected from the pigment, and for a 'black' paint you wouldn't notice the absense of light reflected from the pigment. My understanding is the waxes and other gloss coats are supposed to be nearly clear also.