I am an A-level student and I am studying waves these days. It is said that the direction of the oscillation of particles of transverse waves is perpendicular to the direction of the energy transferred. How can the energy be transferred if there is no horizontal movement of particles? Then how can the waves spread out such as water waves? Why is it hard to explain polarisation if considering light as longitudinal waves? When polarising the lights with a mixture of planes twice, why will the second light filter allow some lights with only plane passing through if the transmission axis of the filter is not the right angle with the polarisation plane?(why doesn't it block all lights passing through)? Why is the light polarised when it deflects?

  • $\begingroup$ As about why's there is a polarization on reflection, check brewster's angle $\endgroup$ – Agnius Vasiliauskas Feb 11 at 9:21
  • $\begingroup$ the water particles forming the apparently transversal surface water (gravity) waves move in a circular pattern (see en.wikipedia.org/wiki/Airy_wave_theory and upload.wikimedia.org/wikipedia/commons/a/ab/… ), the wave's gross motion is transversal not that of the particles that form the wave. Also, almost all EM propagating modes in dielectric and/or metallic waveguides have longitudinal components, only in free space is guaranteed a pure TEM mode operation. $\endgroup$ – hyportnex Feb 11 at 13:11
  • $\begingroup$ I think you have 6 questions here--should limit it to 1 or 2. Preferably one. $\endgroup$ – user45664 Feb 11 at 17:06

I'm hearing a couple of distinct questions being asked. I'll try my best to address them individually:

Q: "How can energy [in a wave] be transferred if there is no horizontal movement of particles?"

A: While the particles themselves do not move in the horizontal direction (more precisely, in the direction of wave propagation), the shape of the wave does. One way to see how this transfers energy is to take the example of a wave traveling down a rope. When the wave is passing a given point on the rope, that piece of the rope is moving in the transverse direction (i.e. away from its initial resting position). As this first piece of rope is displaced, the action of that piece of rope being displaced exerts a tug on the next bit of rope which causes it to move in the transverse direction as well (and so on and so forth all the way down the rope). You can imagine a water wave transferring energy similarly, except now the 'next bit of rope' corresponds to an ever expanding concentric circle about the initial disturbance that caused the wave.

Q: "Why can't light be a longitudinal wave?"

A: Light waves being transverse is an experimental fact. This is something which is also baked into Maxwell's Equations, the classical theoretical framework underlying optical phenomena. (More specifically, in the absence of charge, electric fields must be divergenceless, a requirement which longitudinal waves fail to satisfy).

Q: "Why don't two polarizers rotated relative to each other block out all the light passing through them?"

A: Let's make two assumptions here:

  1. The light incident on the first polarizer is not polarized in the $y$-direction of the first polarizer (the meaning of this will become clear momentarily).
  2. The angle between the two polarizers is not $90^\circ$.

In order to determine how a polarizer will affect incident light, we can imagine a polarizer as an $xy$-plane and draw the incident light's electric field vector $\mathbf{E_0}$ on the plane. The effect of the polarizer is to only allow the $x$ component (relative to the polarizer) of the electric field to pass through, so if the field is initially $$ \mathbf{E_0} = E_0 \hat{i} \cos \theta + E_0 \hat{j} \sin \theta, $$ then the electric field which is transmitted will be $$ \mathbf{E_1} = E_0 \hat{i} \cos \theta, $$ which is entirely in the $x$-direction (again, $x$ is simply the name we have given to the direction on the polarizer which allows light to pass through).

When considering the effect of the second polarizer on this transmitted light (that is, $\mathbf{E_1}$), we now consider the axes $x'y'$ of the new polarizer as defining the directions which permit/deny the transmission of polarized light, and repeat the process above. Thus, as long as the second polarizer is not rotated $90^\circ$ relative to the first, then there will be a component of the light which passes through the second polarizer even if the rotation of the second polarizer relative to the polarization of the light incident on the first polarizer was $90^\circ$.

tl;dr: this is because the first polarizer changes the direction of polarization of the light.

Q: "Why is reflected (initially unpolarized) light polarized when deflected from a surface?"

A: We can think of the process by which the reflected light ray is generated as being the absorption and re-emission of the original incident light ray by many electric dipoles which constitute the material of the surface. A fundamental fact about dipoles is that they radiate parallel to the direction of their oscillation. Depending on the angle of incidence of the original ray, some (or at Brewster's angle, many) of these electric dipoles in the surface will begin oscillating in the direction parallel to the surface of the material and perpendicular to that of the reflected ray. It is this subset of the dipoles in the material which result in the partial (or total at Brewster's angle) polarization of the reflected ray.

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