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There are two sets of Fresnel coefficients depending on the polarisation of the incident light. At normal incidence these equations converge except there is a phase factor of $\pm1$ For the reflection coefficient

$r = \pm \frac{n_1 - n_2}{n_1 + n_2} $

For a normal incidence EM wave, how do you decide between these factors?

Edit - Additional Information

The reason I came to this question is that I was trying to derive the reflection equations for both electric fields and magnetic fields at an interface. At normal incidence, the boundary conditions for both the E and H fields are the same (as far as I can tell) but one must lag behind the other by $\pm \pi/2$. I can't figure out which one I should choose, and it seems to be an equivalent question to my original question.

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  • $\begingroup$ Use the Fresnel coefficient for the polarisation of the light for which you want to do the calculation. $\endgroup$
    – alanf
    Commented Sep 25, 2018 at 7:29
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    $\begingroup$ Isn’t it ambiguous at normal incidence? $\endgroup$
    – user668074
    Commented Sep 25, 2018 at 7:39

4 Answers 4

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Circular polarization flips its handiness on normal incident reflection, because the convention for handiness (and thus phases) are to "look" against the beam direction, not to look into a fixed lab direction.

The difference in sign for s and p at normal incidence is due to your degenerate choice of "scattering plane", since there are two ways to flip your reference direction.

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The minus sign is due to the different conventions for field direction for s and p waves. There is a conventional sign difference between the incoming and the outgoing wave for p-polarisation. You should use the plus sign for s and the minus sign for p to take this into account.

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  • $\begingroup$ Right, but don't these two conventions disagree with each other at normal incidence? The main example I can think of is standing waves. If $r=-1$ then the node is at the surface, if $r=1$ the antinode is at the surface. Am I incorrect? $\endgroup$
    – user668074
    Commented Sep 25, 2018 at 9:12
  • $\begingroup$ @user668074 It is the difference for reflection at an open end or at a fixed and of a rope. In optics, the phase depends on the optical density, $n$. $\endgroup$
    – user137289
    Commented Sep 25, 2018 at 9:18
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    $\begingroup$ @Pieter I agree, but isn't this built into the Fresnel equations through $n_1 - n_2$? So whether you go from a rare medium into a dense medium is already accounted for. $\endgroup$
    – user668074
    Commented Sep 25, 2018 at 9:23
  • $\begingroup$ How does this answer the question? $\endgroup$
    – Kyle Kanos
    Commented Sep 25, 2018 at 10:05
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Indeed at normal incidence, P and S polarizations are not well defined, because $\vec{k}\times\vec{n} = 0$ (where $\vec{n}$ is the normal of the plane). I would argue that in the extreme case of total reflection - such as in a mirror, you should use the plus sign. Here's why:

From wave theory, we know that the reflection coefficient of a wave propagating into a medium with different impedance is:

$$r = \frac{z_1 - z_2 }{z_1 + z_2}$$

See Derivation in David Morin's wave book (draft), sections 4.2.2 and 4.3. This is exactly the S polarized Fresnel coefficient ($r_s$) you get when you don't set $\mu_1 = \mu_2 = \mu_0$, see also this derivation of Fresnel coefficients. This is why it makes sense to take the plus sign as the normal incidence reflection coefficient.

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I too had this question and after staring on the diagram in my book and on the replies here I finally have an answer.

It boils down to what equation you use to describe the reflected wave. We work in complex representation, so your incident wave is: $$\tilde {\vec E}=\tilde {\vec E_0}e^{-i(\vec k \cdot \vec r-\omega \cdot t)}$$ I will now use the convention in my book to describe normal incidence (see attached image).

S-polarized incident wave is $$\tilde {\vec E_{0s}}=(0,E_y,0)$$ P-polarized incident wave $$\tilde {\vec E_{0p}}=(E_x,0,0)$$

S-polarized reflected wave:

$$\tilde {\vec E_{0sr}}=(0,E_{yr},0)$$

P-polarized reflected wave $$\tilde {\vec E_{0pr}}=(-E_{xr},0,0)$$

The sign of the vectors is + if they point along the x axis and - if against.
The books gives the coefficients: $r_s=\frac{n_1-n_2}{n_1+n_2}$ and $r_p=-\frac{n_1-n_2}{n_1+n_2}$.

If we multiply the complex amplitude by the corresponding coefficient, we get positive result in both cases. $$\tilde {\vec E_{0r}}=r \cdot \tilde {\vec E_{0}}$$ $$E_{yr}=\frac{n_1-n_2}{n_1+n_2}E_y$$ $$E_{xr}=-\frac{n_1-n_2}{n_1+n_2}\cdot (-E_x)=\frac{n_1-n_2}{n_1+n_2}\cdot E_x$$

In other words the choice of s and p polarization influences which direction is positive and which is negative for your reflected wave. There is many of these convention choices in physics. Just choose one and stick with it.

Image for derivation of Fresnel equations (Malý, P. (2013) Optika. Prague, Czech Republic: Charles University/Karolinum.)

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