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Say I'm in a completely dark room, I'm facing a wall, and there's a tiny red spot right in front of me on the wall. I use a button to turn on the only light source in the room which is a tiny light bulb on the wall to my right.

I can now see the red spot right in front of me. Clearly, there's no path such that a photon leaves the light bulb, bounces on the red spot, and enters my eyes, as the angle of incidence when a photon from the lightbulb strikes the red spot is not $90°$. So the photon must bounce around the room before the photon strikes the red spot such that, with keeping angle of incidence equal to angle of reflection, it strikes my eyes.

Such a trajectory of the photon is hard to imagine -- if the red spot is truly directly in front of me, it seems to me that to get the right angle, many of the photons that do eventually enter my eye must bounce off of my face to get the right angle. Does that seem right?

Also, doesn't the photon lose some of its energy with each bounce? If $E = hf$, and it's losing it's energy with each bounce, doesn't that keep changing $f$? Doesn't that mean that not only the material, but the number of bounces of the photon is affecting the color we see? How many times can a photon bounce before it ceases to exist, or, $E = hf = 0$?

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  • $\begingroup$ This might be better deduced by removing the room/walls. If I'm a point object observing this spot in complete void with a point lamp to my right, I would I see the spot? I'm unsure. This could be tried in a open field on a dark night. It could be that the spot absorbs the light and then emits immediately after $\endgroup$ – Captain Morgan Aug 22 '18 at 1:29
  • $\begingroup$ Emits light in all directions? Wouldn't that violate the whole angle of incidence = angle of reflection concept? $\endgroup$ – StopReadingThisUsername Aug 22 '18 at 1:57
  • $\begingroup$ Putting aside your question about energy loss, which gets into one of the basic surprises of quantum electrodynamics --- is the first part of this question about "specular" versus "diffuse" reflection? $\endgroup$ – rob Aug 22 '18 at 2:39
  • $\begingroup$ @StopReadingThi... If the surface is rough on a scale larger than the photon's wavelength, the photon often bounces off bumps and dips on the surface in a way that would seem odd to a macroscopic observer who thought the surface was smooth. $\endgroup$ – Ricky Tensor Aug 22 '18 at 4:41
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Photons are scattered off the red spot in many directions even when the photons are coming from one angle. If you wore sunglasses, a dark shirt and pants and black face paint you would not see the red spot if it was in front of the light if there were no walls. Scattering is a form of reflection but angles are not equal.

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