2
$\begingroup$

When a bunch of photons strike a surface then they get absorbed by the electrons on that surface which then starts oscillating and thus re - emit some photons but they do it in the same plane as that of the incident ray and the normal which is kind of strange if we think.

Why is this true ? There are infinite possible directions for electrons (with equal probabilities of being chosen I guess) to re - emit the photons , so why only a particular direction wins all the time (given the source is fixed) ?

$\endgroup$
2
  • 1
    $\begingroup$ Related : (1) Why one should follow Snell's law for shortest time?. (2) Snell's law in vector form. $\endgroup$
    – Frobenius
    Aug 8 at 11:41
  • $\begingroup$ Short and unsatisfying answer: the oscillating atoms in the medium don't oscillate randomly. They oscillate according to the field that excites them. Each atom can re-emit in some other direction, but the phase relationship among them remains fixed. This pattern of oscillations is such that the radiation generated by each atom in the collection interferes with all the others in such a way to generate the observed plane wave. $\endgroup$
    – garyp
    Aug 8 at 15:48
0
$\begingroup$

Let's first take it as read that the incident and reflected rays each travel in a straight line. You can prove this from Fermat's principle in one medium, Maxwell's equations in one medium or photonic $4$-momentum considerations, or just accept it as too obvious to need justifying.

When light strikes a point on a surface, it effectively strikes a tangential plane there. And in $\Bbb R^3$, given two lines and a plane not containing them that meet at a point, there's a second plane orthogonal to the first that contains both lines. We can work in this second plane hereafter.

So the real question is why angle of reflection equals angle of incidence. In particular, let's show why minimizing distance traversed (which is what Fermat's principle becomes in one medium) is equivalent to this angle equation. Suppose a ray incident from $A$ to a point $B$ on a reflecting surface subsequently traverses to $C$ (the points $A,\,C$ can be chosen arbitrarily on their respective rays, save for differing from $B$). Let $C$ have mirror image $C^\prime$ in the boundary, so $\vec{BC^\prime}$ is the mirror image of $\vec{BC}$. So$$\left|\vec{AB}\right|+\left|\vec{BC}\right|=\left|\vec{AB}\right|+\left|\vec{BC^\prime}\right|\ge\left|\vec{AC^\prime}\right|,$$with equality iff $B$ is where $\vec{AC^\prime}$ meets the boundary. You can do the angle-chasing to finish.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.