# Why does a mirror work?

Discussion in 'Physics & Math' started by Omnignost, Apr 23, 2004.

1. ### OmnignostRegistered Senior Member

Messages:
88
I have tried to find a cause-effect answer to why a mirror works. I understand that a photon interacting with a metal surface will cause a collective excitation of the conduction electrons called a surface plasmon but I haven't found any explanation why the reflection law is obeyed when the photon is reemitted. Please dont' answer any crap about the conservation of momentum. That is a calulation rule, not an explanation. Too often confused in physics education. Any ideas?

3. ### John ConnellanValued Senior Member

Messages:
3,636
The cause for the angle of reflection is due to the rule that light always follows the shortest path from a to b. Now if u add the constraint that the light must also hit the mirror (as it must in this example) then u will find it is the angle of reflection that is observed experimentally (because this is the shortest path). I really have to run now, I will find u a link 2moro about it. That crap was just off the top of my head. Gotta go, c u!!!

5. ### OmnignostRegistered Senior Member

Messages:
88
Than you John, that's all very well it just that it is not an explanation, it is just a rule for calculation. I want a microscopic explantion why the reemitted photon chooses an angle that is in accordance with the reflection law=shortest path.

7. ### Quantum QuackLife's a tease...Valued Senior Member

Messages:
23,277
I've been trying to get an answer to this one for ages... good question.

Why the angle of the dangle....so to speak.

8. ### ballisticRegistered Member

Messages:
12
Am I being thick or is it not just a matter of illusion? If different people stand looking at the same mirror, they all see a different reflection, so it is only the observer who observes the shortest route.

9. ### John ConnellanValued Senior Member

Messages:
3,636
Put it this way: why does the angle of (reflection off the ground) of a tennis ball equal its angle of incidence?

A photon has momentum so when it strikes the mirror the component of momentum perpendicular to the mirror surface will be reversed but not the parallel component. Hence the angle of incidence = angle of reflection (just like a tennis ball).

10. ### OmnignostRegistered Senior Member

Messages:
88
Hmm yes. This is exactly the stupid (sorry about that one) answer you get in the physics books. It of course correct but it is not a microscopic explanation. It is a tool for calculation that leads you to the right answer but I want to understand how the sloshing conduction band electrons manage to emit a photon in the right direction. I won't accept that there isn't such an explanation. It must have something to do with how the conduction electrons are excited.

11. ### John ConnellanValued Senior Member

Messages:
3,636
The electrons are excited but remember the photons momentum and direction (since it IS a vector quantity). Now, why would the electron emit the photon out in the same direction it came in if the angle of incidence was less than 90<sup>o</sup>? Think about it. What is so "stupid" about this explanation?

12. ### oxymoronRegistered Senior Member

Messages:
454
Huygen's principle! Are we forgetting this???

You are having trouble grasping this idea because you are thinking of just one impeding photon reflecting off a molecule. Rather, think of a beam composed of many wavefronts impinging on your smooth surface. This way it is much easier to visualise your question.

Follow a wavefront as it hits the surface. As it iteracts with the surface molecules it energizes each one that the front reaches and releases photons (the scattering effect, I wont bore you with the details). The radiating photons must be regarded as a spherical wavelet in the incident medium.

Do you understand so far? Keep in mind Huygen's principle. Think of a row of molecules (draw it on paper if you want) draw a row of dots. Then as the beam strikes each dot one after the other it releases a spherical wavefront from each scatterer - draw circles around each dot BUT the first will be bigger than the second in the line and so forth (this makes sense because the dot further to the left of your drawing was hit first). So you should have a line of dots (molecules) each with a circle around it (with the circles getting smaller). Now if you have drawn it correctly the circles will overlap. Draw a line through the points where they overlap. Voila! The line will be your Huygens' wavefront!

This wavefront is a result of each molecule re-emitting a spherical wavefront and interfering constructively in only one direction. Note how the circles in your drawing will not overlap as nicely in the other direction! That is, there is ONLY ONE WELL-DEFINED REFLECTED BEAM off any optically dense medium. (of course this is not true if the incident radiation had shorter wavelength than the separation of the molecules in the medium - hence why it must be optically dense).

So now we know geometrically/molecularly how the relfected beam is generated. Capiche!?

13. ### oxymoronRegistered Senior Member

Messages:
454
By the way, I can explain this mathematically if you want me to.

14. ### OmnignostRegistered Senior Member

Messages:
88
Yes! Thank you. This is exactly what I was aiming for. I don't know what Huygens principle is but I understand what you mean. This is a truly beautiful phenomenon. I find it almost magical that the shortest path, reflection law and conservation of momentum automatically follows. And the refraction law too. Feel free to add the maths if you feel like it.

15. ### John ConnellanValued Senior Member

Messages:
3,636
Light effects can usually be explained in terms of particles or waves. I must admit, Huygens principle is a more beautiful explanation alright

16. ### ballisticRegistered Member

Messages:
12
errr, I'm lost ok. perhaps I'm in the wrong place.

17. ### Mr. ChipsBannedBanned

Messages:
954
Reflected light, no big deal. Did you know that if you have a mirror made in the shape of a saddle, two 90 degree opposed hyperbolas, it is non-reversing? Probably all of the mirrors you have seen to date show a reversed image.

18. ### wesmorrisNerd Overlord - we(s):1 of NValued Senior Member

Messages:
9,845
I'd think it's because the material is incapable of absorbing the energy that the photon is attempting to transfer to it.

19. ### OmnignostRegistered Senior Member

Messages:
88
No, wesmorris that is not the reason. A metal has a high density of electronic states and will match all photons in the visible range, and more too. It is just that the excited state is short lived.
Oxymoron, is it possible to expand the explanation to include the change of polarization of the light after reflection?

20. ### oxymoronRegistered Senior Member

Messages:
454
Omnigost; absolutely! However I have not the time at the moment to explain. But I will get around to it in a couple of hours. Good question though!

21. ### oxymoronRegistered Senior Member

Messages:
454
During the day our light comes from the sun. But before it reaches us it is scattered by molecules in the air. It makes sense that without an atmosphere the sky would be black and we would only see the light that is shining directly at us from the sun. With the atmosphere the red end of the spectrum passes straight through whereas the blue end is scattered. The scattering causes beams of light to hit us from many directions making the enture sky look blue.

After this brief introduction I will add that many people will say that light from the sun is unpolarized. Well this is a falacy. There is actually no such thing as unpolarized light, however, I will tell you that the light is randomly polarized - this is a big difference. The closest we can get to completely unpolarized light is perhaps two orthogonal incoherent P-state waves. But we wont go into this...

When we deal with polarization of light, especially that of relfection (and scattering) we talk about linearly polarized plane waves. This means a 'flat' (for lack of a better term) wave similar to a plucked string.

So we have a linearly polarized plane wave (light from the sun) which hits a molecule on the surface of a mirror. The incoming wave of light has a electric field vector (E) and a Poynting vector (S) - the electric field vector points in the direction of a normal to the wave plane and the Poynting vector points in the direction of energy transfer (for idealism - where the light is pointing).

Anyway, the point is, the incoming light strikes the atom which then starts to vibrate because it just got hit with a heap of energy. The molecule will oscillate in a special way according to its dipole moment. This is such that the vibrations induced into the molecule are parallel to the electric field vector and perpendicular to the Poynting vector. NOTE: The Molecule will not re-radiate in the direction of its dipole axis.

To summarize:

A linearly polarized plane wave consists of an electric field vector. The wave interacts with the molecules, and in particular, drives the electrons near the surface normal to the surface. The electrons reradiate the energy in the form of a reflected wave. (This process I discribed in the above post).

At this point something peculiar happens. You might all be shouting "Ooh I know this is Brewtser's angle!". Well that is a result of this phenomenon.

What happens is that the relfected wave's flux density is much lower than the incident wave. Picture this: The incoming wave makes an angle &theta; with the dipole axis of the molecules in the metal. The dipole axis need not be parallel with the surface! The reflected wave will have a small angle with the dipole axis too. (forget about the angle to the normal for now).

When this small angle is zero then the incoming randomly polarized wave made up of those two orthogonal P-states I told you about earlier, then only the component parallel to the surface is reflected. This is quite hard to visualise I know but I am trying to stress the fact that when a wave (randomly polarized -sunlight) is incident on a (dielectric) surface - the wave consists of two orthogonal P-state components - the component parallel to the surface is only reflected - the reflection occurs at the Brewster's angle. So from randomly polarized (unpolarized if you will) light we have linearly polarized light, polarized parallel with the surface it just relflected off!!

Since Brewster's Angle (BA) + Transmission angle (TA) = 90 degrees then from Snell's Law we have:

n_i sin &theta;_BA = n_t sin &theta;_TA

it follows that tan &theta;_BA = n_t / n_i

So for air whose n is 1 and if the transmission medium is glass whose n is 1.5 the polarization angle is &theta;_BA = arctan(1.5/1) = 56 degrees.

So to finish off: sunlight consists of a bunch of randomly polarized light. It hits the surface of a piece of glass whose n is 1.5. The component of the randomly polarized sunlight which is parallel to the surface of the glass is completely relfected (the rest is partially transmitted). However the angle of this relfected component is calculated by Brewster's angle (and the whole dipole axis thingy). Brewster's angle is a consequence of the dipole axis and Snell's law which gives the formula above.

22. ### OmnignostRegistered Senior Member

Messages:
88
You seem to be very much into optics. I find this very interesting so I am going to go un bugging you. There is one little detail though, since I am a chemist I have to complain about your mentioning metals and molecules. Metals are not molecular in the usual sense. There has to be differences between the reflection off a metal surface (No molecules, no dipoles but conduction electrons) and a dielectric like glass (molecules, dipoles, non-conducting) where you have reflection AND transmission.
Since we are getting into the nature of polarized light. How does the spin states of photons relate to photons? Photons are spin 1 particles so they have three spin states. There are only two orthogonal polarization directions or...do we have polarization along the Poynting vector too?

23. ### oxymoronRegistered Senior Member

Messages:
454
I'm getting there!