Can Perfect White Light Consist of Only Out-of-Phase Photons Resulting in Black?

[Edi]

As with two same length sound waves 180degrees out of phase travelng the same direction, would out of phase photons in the same weay would not be percievable? (do they even exist at all?)
For perfect white, where every frequency is taken (of course, it would take pretty much infinite amount of energy, but still, for the sake of the thought), it would all consist of out of phase (180) photons and would not be visible.. Black. Is this true?

Does generating a by 180 out of phase waves really creates a wave at all?

 

[Simon Bridge]

You seem to be mixing models here ... the wave-model for light (as EM waves) and the particle model for light (as photons).

A surface is color black if (in context) all visible wavelengths are absorbed.
A surface registers as white if it scatters visible light. It's also white if it is very bright.
White and black, as colors, are about the experience we have.
Of course you can define what you like...

A "black" body, for example, is a bit more complicated than that.

In the wave model - summing over every wavelength, as per your question, may look something like:$$f(x,t)=\int_0^\infty \sin\left (\frac{2\pi (x-ct)}{\lambda}\right )d\lambda$$... you can look this up - it's fun.

In the photon model, the picture is somewhat different...
http://vega.org.uk/video/subseries/8

Does generating a by 180 out of phase waves really creates a wave at all?

... i.e. what is the difference between two waves destructively interfering to give a zero amplitude everywhere and not having a wave in the first place?

The difference is usually that something is doing work to produce those waves so they can interfere.

 

[Darwin123]

As with two same length sound waves 180degrees out of phase travelng the same direction, would out of phase photons in the same weay would not be percievable? (do they even exist at all?)
For perfect white, where every frequency is taken (of course, it would take pretty much infinite amount of energy, but still, for the sake of the thought), it would all consist of out of phase (180) photons and would not be visible.. Black. Is this true?

Does generating a by 180 out of phase waves really creates a wave at all?

Reflected waves don't travel in the same direction as incident waves. Incident and reflected may start at the same interface, but they move in different directions.

No. White is not black.

The color of an object is determined by the direction of reflection as well as the efficiency of reflection. Reflection can be diffusive or specular. My personal belief is that diffusive reflection isn't mentioned early enough in the coursework. However, this post will partly address that issue.

The type of reflection first discussed in physics classes is specular reflection. An image is preserved by specular reflection. The law of reflection is valid for specular reflection. This is the reflection observed in smooth and homogenous surfaces.

The more common type of reflection is diffusive reflection. Diffusive reflection does not preserve an image. The direction of the light wave is randomized by diffusive reflection. This is the type of reflection seen from rough or inhomogeneous surfaces.

Many material surfaces have a combination of specular and diffusive reflection. A surface where both types of reflection are simultaneously present is called glossy. The theory of glossy surfaces is really interesting but complicated.

The word "color" usually refers to the spectrum of the diffusive reflection, not the specular reflection. Usually, the material under the surface affects the diffusive reflection more than the specular reflection. So when we say that a ball is red, we are usually talking about the diffusive reflection of red light from the ball.

Mirrors are examples of objects that have much more specular reflection than diffusive reflection. A mirror is not referred to as "white". A mirror has a smooth and homogenous material that specularly reflects light. In a way, a mirror is closer to "black" than to "white". A smooth black paint will often have specular reflection. Next time you see a shiny black car, look at the reflection of yourself in the paint.

The law of reflection is satisfied by a specular surface. A beam of light is reflected in one specific direction. If your eye isn't turned in the precise direction of the reflected beam, then you won't see it. This is the same if you were looking into a matte black finish.

Other surfaces reflect in a diffusive manner. Small irregularities near the surface take parts of a beam of light and scatter them in different directions. The effect is as though each spot on the surface is a separate light source. The law of reflection is either not satisfied on a macroscopic scale on a diffusive surface.

A surface that reflects all visible frequencies in a diffusive manner is white. Images are not preserved by diffusive reflection, even though the light energy may be preserved. Chalk is generally white.

The rules that you learned about "phase difference" don't apply to diffusive reflectance. Diffusive reflection "randomizes" the phase of the reflected wave. A perfect conductor or dielectric with a perfectly smooth surface causes specular reflection with a phase shift of 180 degrees.

Your comments about "white being black" imply that diffusive reflection causes a 180 degree phase shift. This is not true. However, specular reflection for small incident angles has a 180 degree phase shift. The word "shiny" often refers to strong, specular reflection. So there is a partial truth if you say that "shiny" is "black".

Most of what you have learned applies to specular, not diffusive, reflection. For instance, the thin film calculations that you learn are appropriate to specular reflection, not diffusive reflection. The etalon calculations refer to specular, not diffusive, reflection. However, the lens calculations that you learn really apply only to diffusive reflection, not specular reflection.

Matte black describes a surface that reflects a small amount of light in a diffusive manner. Generally, it is a material with a rough surface that absorbs all visible frequencies of light. I think that is what you mean by black. Most of the light is absorbed by a matte black surface. However, there is a little bit of diffusive reflection from a matte black surface. Ideally, there is no specular reflection from a matte black surface.

"Glossy black" is a shade of black from a smooth surface. As the name implies, there is always a specular reflection superimposed on diffusive reflection. Most of the light is absorbed, but there is a little bit of specular reflection from a glossy black surface. Ideally, there is no diffusive reflection from a glossy black surface.

Sometimes, it is hard to distinguish a glossy black surface from a mirror. They both reflect specularly. The glossy black surface reflects less light than a mirror. However, under strong illumination they look very similar. If the incident light is strong, then the reflected light is strong whether or not it is glossy black or mirror-like. So the difference is more psychology than physics.

There is a separate theory of color that involves as much psychology as physics. However, diffusive reflection is very important because this is where most images come from. When you analyze lenses later, the assumption will be implicitly made that the objects reflect in a diffusive manner.

What drove me a little nuts was that no teacher or textbook told me that the objects in the lens diagrams were diffusive reflectors. I had managed to learn about how specular reflection works. However, the lens diagrams do not show specular reflection. I wasn't even told there was such a thing as diffusive reflection!