How long does it take for a photon to be reflected? Starting with the photon being absorbed by some atom to the point it's reemitted?
And what's the same point with pressure waves, like sound?
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$\begingroup$ How long reflections/transmissions take (modulo the frequency) is usually called the "phase shift" of scattering, and it depends on the details of the scattering potential. $\endgroup$– Ron MaimonCommented Apr 13, 2012 at 7:26
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2 Answers
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If we're talking reflection, then it's best not to think of the photon being absorbed and then re-emitted by an atom. Rather, think of the photon as having very many virtual interactions with the various electrons in the reflecting surface, the complex phases of which all get added up in a coherent way. This will be described by the evolution of the photon's wave function.
Since we are dealing with wave phenomenon, which are not localized in time, it is best to ask the question what happens to a wave packet being reflected from the surface. The reflected wave packet will be delayed and stretched out. The delay of the packets peak, or group delay, is a reasonable thing to identify with the time it takes to be reflected. This group delay will in turn be related to the skin depth of the reflecting surface. For a metal say at optical frequencies, the skin depth will be on the order of several nanometers, and so the group delay will be on the order of 10-17 seconds. This reflection time will be much smaller then the uncertainty in the photon's time of arrival.
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$\begingroup$ So if I'd emit a short, let's say 10^-18s, burst of visible light to, let's say a silver plate, there could be a moment where there are no photons outside the reflector? $\endgroup$– bot47Commented Jun 7, 2011 at 20:07
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1$\begingroup$ For such a short pulse, you would also have to consider dispersion, the stretching of the pulse. The reflected pulse would be delayed 10^-17 seconds but also stretched by about the same amount, so not all of the photons will be inside the reflector at any time. $\endgroup$– user1631Commented Jun 7, 2011 at 20:23
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$\begingroup$ ""where there are no photons outside the reflector? "" You fall back to photons another time! Interpreting linear optics in the photon way is pure masochism! (Its possible, but You need to know quantum theory thoroughly) Learn wave optics. $\endgroup$– GeorgCommented Jun 7, 2011 at 20:52
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Georg is right that it doesn't make much sense to consider this phenomenon in a photon context. However, using wave optics, you can calculate that there is a phase shift upon reflection of a plane wave from a surface. That means, roughly, that the wave is delayed by a certain amount upon reflection.
However, you can't really think about this as a 'signal' being delayed - because plane waves are ideal and infinite in time and space. If you interrupt them or modulate them to carry information, they aren't monochromatic plane waves anymore and you need to consider dispersion as user1631 explains in their answer.
Use the amplitude form of the Fresnel equations:
$$ r_s = \frac {n_1 \cos\theta - n_2 \sqrt{1-\left(\frac{n_1}{n_2}\sin\theta\right)^2}} {n_1 \cos\theta + n_2 \sqrt{1-\left(\frac{n_1}{n_2}\sin\theta\right)^2}} $$
and
$$ r_p = \frac {n_1 \sqrt{1-\left(\frac{n_1}{n_2}\sin\theta\right)^2} - n_2 \cos\theta} {n_1 \sqrt{1-\left(\frac{n_1}{n_2}\sin\theta\right)^2} + n_2 \cos\theta} $$
These will give you a complex reflection coefficient as a function of the angle of incidence $\theta$. Your wave is traveling in a medium of refractive index $n_1$ and reflects off the interface with a medium of refractive index $n_2$. The first equation applies to $s$-polarized light, and the second to $p$ polarization. The phase shift is the argument (angle in the complex plane) of the complex reflection coefficient.
What do these equations actually mean? Try putting in different refractive indices.
For two dielectrics ($n$ is real), where the wave travels in the less dense of the two ($n_2 > n_1$), the reflection coefficients are also real. $r_s$ is negative, meaning the wave is flipped on reflection. $p$-polarization is a bit more complicated, because it has a Brewster angle at which nothing is reflected, and the reflection coefficient changes sign there; so above the Brewster angle, the wave is not flipped anymore. I don't really consider this a time delay (shifting a plane wave by 180 degrees is indistinguishable from flipping it.)
However, if the situation is reversed ($n_2 < n_1$), the reflection coefficients are positive to start out with - there is no flipping or phase shifting. The $p$-polarized wave starts to be flipped once it gets past the Brewster angle. However, when the angle gets to be above the critical angle for total internal reflection, the reflection coefficients become complex and you do get a phase shift as a function of the angle of incidence.
Finally, in the case where one of the media has a complex refractive index (for example, a metal), you always get a complex reflection coefficient and therefore a phase shift.
