I've been trying to understand refraction, and even though I thought my background in mathematics would ensure my success, I've been mistaken.

Indeed, I've been reading the Wikipedia article, and it derives the phenomenon from the fact that the

Also, I've read a proof of Snell's law (https://www.brown.edu/research/labs/mittleman/sites/brown.edu.research.labs.mittleman/files/uploads/lecture12_1.pdf), which implicitly made the same assumption (ie. the wave retains its basic shape when entering the medium). Furthermore, the proof assumes that when entering the medium, wavelength and speed are multiplied by the same factor.

Now imagine a single photon enters the medium. What reason would this photon have to deviate from its straight path?

Or imagine the wave to be localised at a single point in space, ie. 1-dimensional. Then there is no wave front which could bend, but the light would still refract. Why?

I've found Why does light bend when it enters a medium? in the recommended questions, and there is a justified critique of a commonly given explanation, which hence I'd rather recommend to omit.

Thank you very much indeed for any answers.

  • $\begingroup$ Photons do not follow Snell's law of refraction. $\endgroup$
    – hyportnex
    Mar 7, 2022 at 18:11
  • 1
    $\begingroup$ The critique in the question you link to only appears to work because there's a finite set of uniformly distributed secondary sources, which gives a false impression of multiple alignment directions. If, instead of summing over this lattice, you integrate over the whole continuous interface, following the Huygens construction, you'll get a well-defined direction, reproducing the Snell's law. $\endgroup$
    – Ruslan
    Mar 7, 2022 at 18:57
  • $\begingroup$ @Ruslan That is reassuring, and I assume that the lack of uniformity in the distribution of the (discrete) photons that hit the surface means that in the limit, the continuous mathematical model is correct. $\endgroup$ Mar 8, 2022 at 9:09
  • $\begingroup$ You may also be interested in this question. $\endgroup$
    – Ruslan
    Mar 8, 2022 at 10:03

1 Answer 1


You raise several points, I'll try to discuss them one by one:

Why does the wave front remain a line? Actually the wave front is a two-dimensional surface not a line, but that's beside the point. The answer to this question is, that we assume a uniform material and a clean, even boundary on a scale large compared to the wavelength of the incident light and a large even wavefront. (Otherwise the law of refraction won't hold – you would have strong diffraction effects and be in a regime where ray optics don't work).

This in turn means, that there is the same field configuration and material configuration at each surface point, so by symmetry the wave front remains a straight line. (You can argue with the Huygens principle here).

If the intensity of the light is high and not entirely uniform over the wave front and the material has the right kind of non-linearity (changing the effective index of refraction with intensity) this may not hold in reality. Then there are lensing effects, where the wave fronts bend around local intensity maxima.

Assumption that speed and wavelength are multiplied by the same factor There is the basic relationship $c = \lambda f$ for the phase velocity of a wave (which comes from the observation, that within one period of the wave, the wave front must advance by one wave length by the very definition of the quantities). So for the the wave length and phase velocity to change by a different factor the frequency of the wave would have to change. This can't happen (up to non-linear effects which may induce higher harmonics of the base frequency), because the wave in the medium is excited by the incident wave, so the excitation there has the same frequency of the incoming wave.

Single photons A single photon does not have a definite path. Depending on the view you take, it either has an extended state functional, or (taking the Feynman integral construction for the propagator seriously) follows all possible paths at once. When measuring the is a probability distribution for the location of the photon, and its maximum is around the "classical path" taken by the rays a corresponding classical wave (because the contributions close to this path add their probability amplitudes are in phase and add up, while those from the non-classical paths cancel out). This is nicely described in Feynman's laymen introduction "QED: The strange theory of light and matter". Feynman also derives Snell's law using these arguments.

Light focussed to a single point This can't work on many levels. You can't focus light to a single point. At first there are practical problems due to the diffraction limit. Also, ray optics breaks down in such situations – so Snell's law can't be expected to hold – as it propagates such a collimated beam would diverge in all directions even in free space. There are also other limits: If you collimate the light more and more, the intensity of the light gets higher and higher – at some point your material will get non-linear and with even more intensity you will just ionize and destroy your material instead of observing refraction.

The bottom line is, that Snell's law is an approximation that holds very well in the geometry optics limit (that is, when your wave fronts are large compared to the wave length). You can derive it easily in the wave front picture under such circumstances. Your thought experiments don't get around this and at most will show the limits of validity of ray optics.

  • $\begingroup$ The wave front may, however, be very curved, even if it's projection to the plane perpendicular to the upward vector of the medium boundary is a line. Thus, it would be difficult to speak of the preservation of a property when referring only to the surface. $\endgroup$ Mar 8, 2022 at 9:25
  • $\begingroup$ I will accept this answer, because together with Mr. Ruslan's comment, it serves as an acceptable dispersion of my inquisitiveness. $\endgroup$ Mar 8, 2022 at 9:26

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