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Yes, if the mirror was on the floor and facing upward. The transfer of momentum as photons recoil from it and are absorbed by the surroundings would make it heavier.


The problem with the suggestion of using polarization is that you now have the reflections off the polarizers to contend with. I think the short answer is "it depends on how 'black' you want it to be". "Truly black" = reflectance of 0. I am quite sure that is impossible - there will always be some probability of light scattering off a surface. All you can ...


What about heating up the mirror? I assume not all light is reflected but also absorbed. Energy is related to mass. Here is a summary of real experiments that show how kinetic energy in an object contributes to its gravitational mass: arxiv:gr-qc/9909014


Photons are elementary particles and they do have an invariant mass, when there are more than one of them: the addition of their four vectors. Example: pi0 into two gamma, those two gammas have the invariant mass of the pio0. A dark mirror in principle should show a smaller mass because the photons which hit and reflect , at the time of hitting have an ...


The only thing I can think of being true black would probably be a black hole. As light does not bounce off a black hole.


1) Light from the sun strikes a spherical raindrop across half of its entire surface. So every angle of incidence A from 0 to 90 is represented. 2) Some of the light will reflect (uninteresting), but some will enter the drop. Its path bends due to refraction, entering the drop at an angle B (found by Snell's Law) that is always less than A. 3) The light ...


Yes it does. Since the direction of the light beam changes with reflection also the direction of polarization. This is mostly because the observer is in a fixed coordinate system and the light beam changes its local coordinate system during reflection. For an idealized reflector and an observer which moves along with the light beam, the direction of the ...


Finally found out, what I was getting so confused about, here's the answer (credit to Dr Sebastian Steinlechner) with a relevant diagram. The incoming light is assumed to be unpolarised. We can, however, describe it as a combination of two orthogonal polarisations: one is polarised in the plane of incidence (the arrows in the picture), and the other is ...


Where is the light source? If the light source is above the midpoint of the mirror the mirror will weigh slightly more. However, if the light it below the mirror it's weight is going to drop. (This effect goes away if the mirror is absolutely perfect. Real world mirrors will absorb some photons, though.) There will be a considerably stronger (but still ...


Light from the source is emitted in all directions - including towards every point on the road surface. When the light hits the (wet, rough) surface of the road, it is scattered in a variety of angles. Some of the light rays will be scattered at just the right angle to enter your eye. Therefore reflected light from all different parts of the road between ...


You should read this page :http://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle The Huygens principle explain intuitively why the wave will spread after being "cutoff" by an obstacle, as the spherical sources at the edge will not interfere anymore with the adjacent ones (those being stopped by the obstacle) to form a plane wave. As shown in the ...


Have a look at the question How does a one-sided glass work?. The effect is due to the limited ability of the human eye to handle large contrast ranges and not to the effect of light intensity on the reflectivity of the glass. So without replacing the human eye with something else there is no way to achieve what you're asking.

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