# Understanding reflection through polarization

I came across the following explanation:

The source of this so-called reflected light is not simply that the incident beam is reflected; our deeper understanding of this phenomenon tells us that the incident beam drives an oscillation of the charges in the material, which in turn generates the reflected beam. From the figure, it is clear that only oscillations normal to the paper can radiate in the direction of reflection, and consequently the reflected beam will be polarized normal to the plane of incidence. Now, my question is what determines the direction of reflection?

I know that the angle of incidence is equal to the angle of reflection. But given that it is the charges in the material that generate the reflected beam, why can't the direction of reflected beam be random? Why is its direction well defined? Should the above law of reflection be blindly accepted due to experimental evidence?

• In fact, the law of reflection, law of refraction and Brewster's angle and the can be derived from Maxwell's equations – K_inverse Oct 1 '18 at 9:14
• The material seems to be cut and pasted from a textbook, but there is no attribution. What is the source? It's also not very clear without more context. I think the argument partially reproduced here may be similar to one in the Feynman lectures. See physics.stackexchange.com/a/507534/4552 – user4552 Oct 10 '19 at 22:47

The angle of reflection is determined by requiring that the reflected wave matches the phase of the incident wave at all points of the boundary.

When you're working things out in full (usually under the heading of 'Fresnel equations' in EM textbooks), you are basically looking for solutions to Maxwell's equations which consist of

• an incident beam $$\mathbf E_\mathrm{in}(\mathbf r,t) = \mathbf E_{0,\mathrm{in}} e^{i(\mathbf k_\mathrm{in}\cdot \mathbf r-\omega t)}$$
• a reflected beam $$\mathbf E_\mathrm{re}(\mathbf r,t) = \mathbf E_{0,\mathrm{re}} e^{i(\mathbf k_\mathrm{re}\cdot \mathbf r-\omega t)}$$
• a transmitted beam $$\mathbf E_\mathrm{tr}(\mathbf r,t) = \mathbf E_{0,\mathrm{tr}} e^{i(\mathbf k_\mathrm{tr}\cdot \mathbf r-\omega t)}$$

where we implicitly assume that the physical fields are the real parts of the complex fields, and we require that the wavenumbers satisfy the dispersion relation $$\omega^2 = c^2 k_\mathrm{in}^2 = c^2 k_\mathrm{re}^2 = \frac{1}{n(\omega)^2} c^2 k_\mathrm{tr}^2$$ on both sides of the boundary.

Furthermore, if you put the $$x$$ axis along the boundary and the $$z$$ axis normal to it, then you can write the incident wavevector as $$\mathbf k_\mathrm{in} = k_\mathrm{in} \sin(\theta_\mathrm{in}) \hat{\mathbf{e}}_x - k_\mathrm{in} \cos(\theta_\mathrm{in}) \hat{\mathbf{e}}_z,$$ so that $$\mathbf k_\mathrm{in}\cdot \mathbf r = k_\mathrm{in} \sin(\theta_\mathrm{in}) x - k_\mathrm{in} \cos(\theta_\mathrm{in}) z,$$ and at the boundary itself $$\left.\mathbf k_\mathrm{in}\cdot \mathbf r \right|_\mathrm{boundary}= k_\mathrm{in} \sin(\theta_\mathrm{in}) x$$ since $$z=0$$ there.

Similarly, the reflected beam is expected to have a positive $$z$$ component, so that you can write it as $$\mathbf k_\mathrm{re} = k_\mathrm{re} \sin(\theta_\mathrm{re}) \hat{\mathbf{e}}_x + k_\mathrm{re} \cos(\theta_\mathrm{re}) \hat{\mathbf{e}}_z,$$ and at the boundary you again have $$\left.\mathbf k_\mathrm{re}\cdot \mathbf r \right|_\mathrm{boundary}= k_\mathrm{re} \sin(\theta_\mathrm{re}) x.$$

This is where the phase-matching comes in: the incident beam has a wave dependence as $$\mathbf E_\mathrm{in}(\mathbf r,t) \sim e^{i(k_\mathrm{in} \sin(\theta_\mathrm{in}) x - \omega t)}$$, while the reflected beam has the functional dependence $$\mathbf E_\mathrm{in}(\mathbf r,t) \sim e^{i(k_\mathrm{re} \sin(\theta_\mathrm{re}) x - \omega t)}$$. And here is the important thing: the coefficients of those two beams need to be matched using the Maxwell equations at the boundary, but that can only be done at a single point; if we want the two to match everywhere, the functional form needs to match: \begin{align} e^{i(k_\mathrm{in} \sin(\theta_\mathrm{in}) x - \omega t)} & \equiv e^{i(k_\mathrm{re} \sin(\theta_\mathrm{re}) x - \omega t)} \quad \forall x,t\\ \ \\ \text{or, in other words,} \quad k_\mathrm{in} \sin(\theta_\mathrm{in}) & = k_\mathrm{re} \sin(\theta_\mathrm{re}) . \end{align}

The rest then just falls in by itself: we know from the dispersion relation that $$k_\mathrm{in}^2 = \omega^2/c^2 = k_\mathrm{re}^2$$, which then requires that $$\sin(\theta_\mathrm{in}) = \sin(\theta_\mathrm{re})$$ and therefore that $$\theta_\mathrm{in} = \theta_\mathrm{re}$$.

(Similarly, for the refracted beam, you're also required to have $$k_\mathrm{in} \sin(\theta_\mathrm{in}) = k_\mathrm{tr} \sin(\theta_\mathrm{tr})$$ through the same phase-matching argument, except that now the dispersion relation requires that $$k_\mathrm{tr} = n(\omega)k_\mathrm{in}$$, which then implies Snell's law, in the form $$\sin(\theta_\mathrm{in}) = n(\omega) \sin(\theta_\mathrm{tr})$$.)

Also, one important correction: your source's claim that

From the figure, it is clear that only oscillations normal to the paper can radiate in the direction of reflection, and consequently the reflected beam will be polarized normal to the plane of incidence.

is dead wrong. It is perfectly possible for the charge oscillations in the medium to contain a component in the plane of the page, and it is perfectly possible for the reflected light to contain a polarization component in that plane. There is indeed one and only one angle, known as Brewster's angle, at which the reflected beam is polarized normal to the plane of incidence, but this is a nontrivial consequence of the dynamics encoded in the Fresnel equations, and it is nowhere near obvious.

• "It is perfectly possible for the charge oscillations in the medium to contain a component in the plane of the page, and it is perfectly possible for the reflected light to contain a polarization component in that plane." But, in this case, the reflected light will only be partially polarized, right? – Gokulakrishnan Shankar Nov 21 '18 at 3:57
• Not at all. The angle of polarization might change, but the reflected light will be fully polarized. Whatever made you think otherwise? – Emilio Pisanty Nov 21 '18 at 4:30
• So, is it not necessary that the refracted ray and reflected ray be perpendicular to each other for the reflected ray to be fully polarized? – Gokulakrishnan Shankar Nov 21 '18 at 8:50
• Also, does the incident (unpolarized) ray cause the electrons in the medium to vibrate only along the plane perpendicular to the incident ray's direction, eventhough the incident ray contains vibrations in many planes? – Gokulakrishnan Shankar Nov 21 '18 at 9:00
• @GokulakrishnanShankar Basically, your question does not provide enough information about the polarization state of the incident light beam, so your questions here can't be fully answered. If the incident beam was fully polarized, then it will stay fully polarized. If the incident beam was completely unpolarized, and the incidence angle is away from Brewster's angle, then both the transmitted and reflected beams will be partially polarized. – Emilio Pisanty Nov 23 '18 at 12:09

You can draw lots of light paths from the source to some point on the reflecting surface and then to the receiver. Each is in itself valid. But the phase varies so rapidly that the contributions from neighbouring paths with slightly different lengths cancel out - except for the path which is an extremum, which is at $$\theta_i=\theta_{reflected}$$.

• Is this what Feynman was referring to with the principle of least action and the path integral? Thanks in advance Professor Barlow. – PhysicsDave Oct 3 '18 at 1:48
• Absolutely. He takes Fermat's principle, which applies to light, and generalises it to quantum mechanical wave functions. – RogerJBarlow Oct 3 '18 at 7:49
• It's interesting how light finds a path that has a path length equal to an integer multiple of its wavelength. The photon or wave function collapses where it 's path can have maximum E field at the impact point if I've understood in correctly. – PhysicsDave Oct 3 '18 at 13:04

Of course Emilio Pisanty's answer is correct, but let me add a few things.

First, when a photon interacts with an atom, it might elastically scatter off the atom, that is what we call reflection.

In the case of elastic scattering, the photon keeps its energy and phase, and changes angle. There are two types of reflection:

1. diffuse, when the photons' angle of reflection is random indeed

2. specular, when the photons' angle of reflection is the same for most of the photons, this is how a mirror image is created

You say that you know that the angle of incidence is equal to the angle of reflection. This is a common misconception. It is only true for specular reflection, that creates a mirror image. For diffuse reflection, the angle of incidence is not equal to the angle of reflection, but the angle of reflection can be random indeed, and that is why with diffuse reflection, the photons, that build up the classical EM wave, the photons' relative angle changes.

With specular reflection, the angle of incidence is indeed equal to the angle of reflection, and the relative angle of the photons' is kept. This is how specular reflection build a mirror image.

Now you are asking why the relative angle stays the same for specular reflection, and is random for diffuse reflection.

Well, the answer is QM. The surface of the material will decide. And the shape of the surface at the level of the size of the visible wavelengths. At 400-700nm zoom, you have to look at the surface.

If what you get is a picture of a random, non-smooth surface, then the reflection will be diffuse, and the angle of reflection will be random.

If the picture is smooth at a zoom of 400-700nm size, then this means that the visible wavelength photons "see" a smooth surface, and the angle of reflection will be the same as the angle of incidence, as you say, that is correct. This is why usually metals, which have a smooth surface at that level of size (scale), are reflective, and can build a mirror image.