1
$\begingroup$

A wave that meets the boundary of two substances with different propagating velocities is partially reflected.

Does anybody know the reason behind reflection?

$\endgroup$
3
$\begingroup$

The question is general, so one can start with the simplest waves, the ones that are easiest to picture: transverse waves on a string.

First one needs to consider reflection at the ends of strings. There are two cases: where the end is fixed, or where the end is free. Those are the extreme ends of boundary conditions: where the string on the right is infinitely heavy or where the string on the right is massless.

It is often is a bit of math to solve differential equations with boundary conditions, but at these types of ends it can be done with picture-type arguments, and the principle of superposition. In both of these cases, one can add a mirror wave, a wave traveling in the opposite direction, like in the animation below:

animation of a pulse reflected at a fixed and at a free end

The wave is on the blue string. In the first case, the string end is fixed at the black dot. One could imagine the string continuing behind this point (pictured in red), and a wave coming from the other side. When the distance is the same and the phase is inverted, the dot remains stationary in the superposition of the two waves traveling through eachother. So this satisfies the wave equation with the boundary condition of a fixed end.

Similar for a free end, but now the reflected wave has the same phase.

Now to the question. Imagine ropes with different masses connected at the dot. Then part of the wave continues, but there will also be reflection. The phase depends on whether the wave velocity to the right is higher or lower, the amplitude depends on how large the difference is. This is very similar to water waves (wave speed changes at a change of the depth of the water), or sound, or light. Or even the quantum mechanical problem of a particle transmitted at a potential step.

See also this page (with animations): http://www.acs.psu.edu/drussell/Demos/reflect/reflect.html

More mathematically: http://www.people.fas.harvard.edu/~djmorin/waves/transverse.pdf

In 4.2 there, also the boundary condition is explained. It is obvious that the displacement of the string must be continuous at the boundary. But also the slope must be continuous.

$\endgroup$
2
$\begingroup$

I believe it is due to surface charge carriers which are free enough to absorb and re-emit the light at the same frequency, and as they receive vibrational energy in one direction the re-emission is directional, like an antenna. So metals (conductors) tend to be reflective, plastics (insulators) generally less common.

$\endgroup$
  • $\begingroup$ Sound is also reflected $\endgroup$ – veronika Dec 23 '16 at 22:21
  • $\begingroup$ Well said. Yes, water waves / sound waves are different from light fundamentally in that there is a propagation medium. When this medium is constrained $\endgroup$ – JMLCarter Dec 23 '16 at 22:31
  • $\begingroup$ Well said. Yes, water waves / sound waves are different from light fundamentally in that there is a propagation medium. There can also be a surface tension effect, but let's consider the weight of an infintesimal column of water in a wave exerts a pressure, an surrounding columns of water. The momentum of the column can be transfered to neighbouring columns. If there is a fixed boundary not another column of water in one direction, some momentum can no longer be transfered forward. It goes backwards, pressure is omni-directional (as is surface tension along the tangent of the surface). $\endgroup$ – JMLCarter Dec 23 '16 at 22:41
  • $\begingroup$ Fresenel Huygens resolves the direction of reflection. $\endgroup$ – JMLCarter Dec 23 '16 at 22:42
  • $\begingroup$ There is interaction with a boundary involved whether or not there is a propagation medium (light or water), and as long as that boundary has a different capability to store potential energy of the wave on each side (no capability for water in the wall, some in the water; no capability for light in free space, some in the conductor). $\endgroup$ – JMLCarter Dec 23 '16 at 22:49

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.