# B Polarisation Brewster's angle

1. Feb 18, 2017

### Glenn G

I get the idea that at the Brewster's angle dipoles aligned at the surface can not emit p-oscillations in the reflected direction as the dipole is aligned parallel to this direction. What I don't get is that if the dipoles are aligned as shown what is emitting the s oscillations that make up the reflected beam?
Thanks ,
G.

Last edited by a moderator: Feb 18, 2017
2. Feb 18, 2017

### kuruman

Your figure only shows surface p-dipoles. There are also oscillating surface s-dipoles. Don't forget, the incident beam is unpolarized and has a mix of p and s oscillations as you show in your figure.

3. Feb 18, 2017

### Glenn G

That makes sense. Thank you.

4. Feb 18, 2017

### sophiecentaur

it may be better to say that the waves have p and s components. Your description could suggest that p and s are already there.

5. Feb 19, 2017

### Glenn G

What's still troubling is that if the dipoles that emit p components can do so in all directions except parallel to the dipole itself why do we just see a strong refracted beam in the direction we do as given by snells law.

Huygens construction gives us a means of seeing why waves change direction when refracted, snells law gives details of the actual angles involved and I love this absorbed and re-emitted by dipoles concept as to what is actually going on but don't get why we see such a strong refracted beam at the angle we observe. I also assume that once in the water there must be lots of absorption re-emission events, so why does it maintain its forward direction.
G.

6. Feb 19, 2017

### sophiecentaur

Not clear what you mean by this except that Huygen (in your following sentence) actually accounts for it. In the ideal case (and for a large enough reflector) the result of the Huygens construction is to give a ray just in the forward direction. It's hard to do the construction accurately with paper and pencil but if you look at what happens in a line that's not in the refracted ray direction, all the secondary wavelets do not emerge in phase so there's no constructive interference. Google Huygens construction (Images) and you are bound to find a picture that will satisfy you. I'm loth to suggest one because there will be so many.
This is a very dodgy approach, The energy levels are all wrong for it - plus, if it really did work like that, the "re-emitted" photons would have random phase relationships so you would get no coherent wavefront out of the process (Huygens would not apply). You have to look at it in terms of interaction with the bulk material - with trillions of atoms involved. (Just try to ignore the photon thing here - it isn't appropriate and (despite what they seem to tell you in School and later) it is no more 'fundamental' than the wave approach.

7. Feb 19, 2017

### Glenn G

Thanks Sophiecentaur, so I should think of a continuous wavetrain coming in and interacting with many atomic surface dipoles such that due to interference only adds up to constructive interference in the direction that we observe the refracted beam travelling in? G.

8. Feb 19, 2017

### sophiecentaur

If you like. It's not a bad model to work with in your mind if you don't like the macroscopic Wave approach. You could also regard it as a problem in Diffraction. (Most optics boils down to it in the end - the fringes etc/ are only there when the dimensions of the problem call for it.