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I was told that when a wave of visible light is reflected (or refracted) then the incident ray, the normal to the reflective surface (or interface of two optical media) and the reflected and refracted ray are coplanar, if we assume the rectilinear propagation of light (the geometric optics approximation). But I do not understand why this is the case. A classical explanation would be highly appreciated. Thanks.

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So you have 2 initial vectors: $\vec k$ is light's wave vector and $\hat n$ is the normal to the reflecting surface.

The final wave vector can be some combination of:

$$ \vec k' = a\vec k + b\hat n + c(\vec k \times \hat n)$$

where the prefactors can be combinations of numbers, and available scalars such as:

$$ 0, 1, k^2, \vec k\cdot \hat n, ||\vec v \times \hat n|| $$

For non-coplanar reflection, we need $c\ne 0$.

If we apply time reversal to the process, then:

$$ T(\vec k') = aT(\vec k) + bT(\hat n) + c(T(\vec k) \times T(\hat n))$$

$$ (-\vec k') = a(-\vec k) + b(+\hat n) + c((-\vec k) \times (+\hat n))$$

$$ -\vec k' = -a\vec k + b\hat n - c(\vec k \times \hat n)$$

So that $a$ and $c$ need to be time even, while $b$ is time odd. If we look at our list of pre-factors, the only time-odd one is $\vec k \cdot \hat n$, so we can write:

$$ \vec k' = a\vec k + b'(\vec v\cdot\hat n)\hat n + c(\vec k \times \hat n)$$

Now apply the parity operator:

$$ P(\vec k') = aP(\vec k) + b'(P(\vec k)\cdot P(\hat n))P(\hat n) + cP(P(\vec k) \times P(\hat n))$$

$$ (-\vec k') = a(-\vec k) - b'(\vec k\cdot \hat n)\hat n + c(-\vec k \times -\hat n)$$

$$ -\vec k' = -a\vec k - b'(\vec k\cdot \hat n)\hat n + c(\vec k \times \hat n)$$

For parity to be conserved, $c=0$, and:

$$ \vec k' = a\vec k + b'(\vec k\cdot\hat n)\hat n $$

At zero incidence (no reflection, or $\vec k' = \vec k$), this becomes:

$$ \vec k' = a\vec k = \vec k$$

so $a=1$.

So now:

$$ \vec k' = \vec k + b'(\vec k\cdot\hat n)\hat n $$

At normal incidence, $\hat n = -\vec k/k^2$, and $\vec k' = -\vec k$:

$$ -\vec k = \vec k + b'(-k)\hat n = \vec k + b'\vec k= \vec k(1+b') $$

$$ -1 = 1+b'$$

$$ b=-2$$

One could also argue that 2 normal reflections leave $|\vec k|$ unchanged, so that:

$$ k''= k'(1+b)=k(1+b)(1+b) = k$$

$$ b^2 +2b +1 =1 $$ $$ b(b+2)=0$$

which has roots $b=0$ (no reflections), $b=-2$ (2 reflections).

Finally:

$$ \vec k' = \vec k -2(\vec k\cdot\hat n)\hat n $$

is the only relation that conserves time-reversal symmetry, parity symmetry, and works at the 2 extreme cases.

Note: I could have just asked, "How would the light choose left or right if it were non-coplanar?", but that's not very clear.

You can include polarization, but with care. Circular polarization is a vector, but it is aligned with $\pm \hat k$, and hence does not add new information. Linear polarization is not a vector, it is a tensor alignment. Vertical polarization does not distinguish between up/down, and horizontal polarization does not distinguish between left/right, so it cannot make the choice either.

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  • $\begingroup$ It's always great to see the answer boil down to an argument based purely on fundamental symmetries. $\endgroup$ Commented Sep 30, 2020 at 20:19
  • $\begingroup$ @probably_someone Do you think dilation symmetry could constrain the the 1 and/or the -2? $\endgroup$
    – JEB
    Commented Sep 30, 2020 at 20:48
  • $\begingroup$ Thank you for your clear and thorough answer. I could understand your main argument, but I don't know how you chose the set of numbers that can be prefactors. Could you refer me to a beginner's guide to learning the basics of this topic (I don't know its name, is it quantum mechanics?) ? Thank you again. Your answer was so elegant that even someone like me who does not (yet) know the exact meaning of time reversal or parity could understand it. $\endgroup$
    – Meripadhai
    Commented Oct 13, 2020 at 16:55
  • $\begingroup$ it's anything to scale the magnitude of the vectors: any real number, maybe a physical constant...which would get into dimensional analysis, and then (pseudo)scalar products available in the problem. I am not sure whence I learned this...just seen it done enough that it stuck. $\endgroup$
    – JEB
    Commented Oct 14, 2020 at 13:51

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