I've been taught before that mechanical waves, when they hit a boundary between a less dense (faster) medium and a more dense (slower) medium, the reflected wave is inverted because it acts like a fixed end reflection. And in a fixed end reflection, the end particle doesn't move and exerts a force on the other particles opposite the in direction of the wave, causing the reflected wave to invert.

But given that electromagnetic waves don't need a medium or any particles to propagate, I need some help understanding:

  1. What is the mechanism behind the reflection of electromagnetic waves? (As in why are they reflected)

  2. Why, when hitting a fast to slow medium boundary, are electromagnetic waves inverted?

Any help would be very much appreciated!

  • $\begingroup$ Why are electromagnetic waves inverted when reflected at a fast to slow medium boundary? This is not true. Assuming normal incidence, a reflection either inverts E and doesn't invert B, or it inverts B and doesn't invert E. The definition of inversion depends on whether you use E or B as the measure of amplitude. This kind of ambiguity turns up with mechanical waves as well. $\endgroup$ – user4552 Apr 29 '18 at 2:58

In a propagation of waves, we can see the interplay of two components: one represents force (pressure, voltage, electric field), the other represents motion (flow, current, magnetic field).

These two components, the yin and yang of waves, move a wave forward by continuously changing each other.

For example, when air pressure increases, it causes air molecules to move, which increases pressure downstream and causes other molecules to move, etc.

For a given type of a wave, in a given type of a medium, the ratio of these two components is determined by the medium and is called impedance, e.g., p/Q for acoustic waves, V/I for transmission line waves, E/H for electromagnetic waves).

Obviously, in a low impedance medium the same force will cause greater motion and vise versa.

As long as the impedance of a medium stays the same, a wave propagates at the same speed and in the same direction. However, when the impedance of the medium abruptly changes, say, due to a transition to a different medium, the same force won't be able to cause the same motion.

For instance, if the second medium has a higher impedance, only part of the wave would be able to proceed forward, forcing the remaining part to go back or reflect.

So, waves are reflected when they encounter discontinuity in the media or discontinuity of impedance, i.e., due to impedance mismatch.

The reflection can also be explained from the energy prospective. As we know, the energy (or power) transfer is at its maximum when the impedance of the load matches the impedance of the source. This means that, whenever the impedance of the medium changes, increases or decreases, not all energy of the wave will be able to go through, i.e., some of the energy will have to be reflected.

More precisely, the ratio of the force and motion components at the boundary will have to satisfy boundary conditions for both media.

Following this logic, we could derive a coefficient of reflection at the boundary, which is the ratio between the force representing the reflected wave and the force representing the incident wave:

Γ=(Z2-Z1)/(Z2+Z1), where Z1 and Z2 are impedances of the first medium and second medium, respectively.

Looking at this formula, we can see what happens with a wave at a boundary under different scenarios.

For instance, if Z1=Z2, Γ=0, so there is no reflection and the wave continues in its merry way.

If Z2 is very high, Γ=~1, i.e., the wave hits a wall or an open circuit and is completely reflected. The boundary condition here demands that nothing flows beyond the wall, and, therefore, the reflected wave must cancel the flow associated with the incident wave. To achieve that, the force at the wall has be twice as large as the force associated with the incident wave.

If Z2 is very low, Γ=~-1, i.e., the wave encounters no resistance or a short, the force must drop to zero. This means that the force associated with the reflected wave has to cancel the force associated with the incident wave or, we can say, that it will be inverted or out of phase with that force.

This last case could explain why electromagnetic waves reflect from metals with the inversion of the electric field. T

he EM wave impedance in free space (and in air) is about 377ohm. The wave impedance of a metal could be close to zero. This forces the electric field on the surface of the metal to be close to zero, which means the electric field associated with the reflected wave has to be out phase with the electric field of the incident wave, in order to cancel it.

The EM wave reflected by a metal is produced by an electric current flowing (mostly) on the surface of the metal and trying to kill the electric field associated with the incident wave in order to satisfy the boundary condition, E=0.

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your 1. question: when a photon interacts with an atom, three things can happen:

  1. elastic scattering, the photon keeps its energy and changes angle

  2. inelastic scattering, the photon gives part of it's energy to the atom and changes angle

  3. absorption, the photon gives all its energy to the atom and the valence electron goes to a higher energy level

In your case, reflection is elastic scattering. In a case of mirror, the angle is going to be opposite, and with glass it will be the same.

Your 2. question:

  1. The light wave inverts at a boundary with greater index of refraction

Why does a light wave invert at a boundary with greater index of refraction?

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