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in this animation from the Fresnel equation wiki we can see a wave bouncing back in the "negative" compared to the "positive" incoming wave. What physics model govern such behavior in transverse waves?

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    $\begingroup$ Wave equation: hyperphysics.phy-astr.gsu.edu/hbase/Waves/waveq.html $\endgroup$ – K_inverse Dec 3 '18 at 1:32
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    $\begingroup$ what does "S" polarized mean? $\endgroup$ – JEB Dec 3 '18 at 5:23
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    $\begingroup$ One more clarification needed: if the wave is incident perpendicular to the surface, there is no distinction between S and P polarization. Why are you asking about S polarization specifically? $\endgroup$ – S. McGrew Dec 3 '18 at 15:03
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    $\begingroup$ Please read [edmundoptics.eu/resources/faqs/optics/polarizers/…, where the s- and p- polarizations are defined. $\endgroup$ – S. McGrew Dec 3 '18 at 20:48
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    $\begingroup$ Normally the terms S- and P- polarization are used for oblique incidence. When a wave is incident perpendicular to the surface, there is no way to distinguish between S- and P- polarization. In the Edmund Optics explanation, "the page" is in the plane of the incident, reflected, and transmitted beams. In the case of perpendicular incidence, there is no such plane. $\endgroup$ – S. McGrew Dec 3 '18 at 21:09
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You would like a conceptual answer, so here it is. Imagine that, at the surface of the medium, the incident light wave stimulates movement of electrons in response to the changing E field in the incident wave. The movement produces an electromagnetic wave that propagates symmetrically in both directions but which has opposite phase to the incident wave: it moves both in the same direction as the incident wave (forward) and in the opposite direction to the incident wave (backward).

If the stimulated wave is equal in magnitude and opposite in phase to the incident wave, then the transmitted wave will be perfectly cancelled by the portion of the stimulated wave moving in the "forward" direction. The backward moving portion of the stimulated wave is the reflected wave.

In the case of partial reflection illustrated by the animation you referenced, the stimulated wave has a reduced amplitude with respect to the incident wave. Accordingly, the transmitted incident wave is reduced, but not to zero. The reflected wave (the backward portion of the stimulated wave) is also reduced because it is the stimulated wave, which is lower amplitude than the incident wave.

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  • $\begingroup$ -thanks for the answer, if you have time, could you please add an expanation to such behavior if we were talking about 2 pieces of ropes (of different density) attached together ? $\endgroup$ – Manu de Hanoi Dec 4 '18 at 1:37
  • $\begingroup$ Furthermore, please notice that the incident wave in the animation isnt symmetrical around 0 (it's just a positive pulse), so a 180 phase shift (if it meant anything on a single pulse) wont make anything negative $\endgroup$ – Manu de Hanoi Dec 4 '18 at 1:41
  • $\begingroup$ The stimulated electromagnetic wave is symmetrical in the sense that it consists of mirror-image waves moving in the forward and backward directions. The incident wave is not symmetrical; it moves in only one direction. Its shape is irrelevant. If the incident wave is decomposed into its Fourier components, we can refer to the phase of each component. All of the components are transmitted &/or reflected independently unless the interaction between the incident wave and the reflecting/transmitting medium is nonlinear. $\endgroup$ – S. McGrew Dec 4 '18 at 1:57
  • $\begingroup$ A conceptual explanation of partial reflection at the junction between ropes of different mass is basically the same. We can't talk about electrons, but the underlying concept is that at the junction a wave gets launched in both directions. The portion traveling in the same direction as the incident wave cancels part of the transmitted wave, and the portion traveling in the opposite direction serves as the reflected wave. To work out the magnitudes of the various waves, you need to invoke energy conservation. A textbook explanation probably will be better than the way I would explain. $\endgroup$ – S. McGrew Dec 4 '18 at 2:17
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    $\begingroup$ You're right that it's related to Huygens principle, but I think it might be confusing to think about it that way. Instead, think of the stimulated wave as being launched by forcibly jiggling a point on an infinitely long rope. It's pretty obvious that it would launch waves in both directions, and that they would be mirror images of each other. Of course, "mirror image" has a slightly different interpretation if the refractive index (for light) or the mass density (for a rope) is different in the two directions away from the jiggled point. $\endgroup$ – S. McGrew Dec 4 '18 at 3:44
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The physics model governing electromagnetic waves is of course, Maxwell's equations. A plane incident on an interface between two dielectric material is governed by Maxwell's equations in linear media:

Gauss's Law replaces the electric field, $\vec E$, with the displacement field, $\vec D$, which with no free charges, becomes:

$$ \nabla\cdot\vec D = 0 $$

where:

$$ \vec D = \epsilon \vec E $$

Ampere's Circuital law (with no free currents) becomes:

$$ \nabla \times \vec H = \frac 1 c \frac{\partial \vec D}{\partial t} $$

where:

$$ \vec H = \frac 1 {\mu} \vec B $$

The other (source-free) equation remain the same

$$ \nabla \cdot \vec B = 0$$ $$ \nabla \times \vec E = -\frac 1 c \frac{\partial \vec B}{\partial t}$$

When applying these equations to a plane wave at an interface between 2 media, you need to distinguish 2 polarization states. Now all polarization is perpendicular to the direction of travel, but that space is spanned by two orthogonal directions. In one formalism, they are called S and P , where P has the electric field vector(s) coplanar with all the directions of propagation and S has it orthogonal to all directions of propagation.

So: applying the linear-media equations to the plane waves at the interface, the following conditions determine the scattering:

(1) The tangential component of $\vec E$ is continuous.

(2) The normal component of $\vec D$ is continuous.

(3) The normal component of $\vec B$ is continuous.

(4) The tangential component of $\vec H$ is continuous.

Now in the situation you described (normal incidence), the normal components of all the fields are zero, so you just apply the tangential conditions.

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  • $\begingroup$ -Thanks for your answer, although the animation comes from a wiki page about optics, my question isnt restrained to EM waves. Furthermore I believe EM waves arent shaped like the animation shown. Lastly my question focuses on the reversal of the waveform upon reflection, your answer doesnt seem to address that $\endgroup$ – Manu de Hanoi Dec 3 '18 at 19:45

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