# WHY does a mirror work?

1. Oct 5, 2004

### mwirth

Sure, everyone knows that the angle of incidence equals the angle of reflexion.
My question is rather aimed at the atomic level.
What is the interaction between electromagnetic radiation and matter? Why can a layer of metal (a mirror) reflect light? Does the interaction occur at the electron sheath? Can electron level changes be responsible? (I think not, how could the angle be preserved?)
Would an electric field achieve the same? (Never heard of it, never seen it.)

I asked a few fellow students this question, but noone could answer it. (No, it's not some kind of homework, we don't have physics any more. It's just my personal interest.)
I couldn't find anything useful on the web either, as all pages seem to focus on the macroscopic level.

I would appreciate any kind of answer, links to web pages, or whatever.

Greetings

Marcel Wirth
mawirth at student dot inf dot ethz dot ch

2. Oct 6, 2004

### Tide

Mirrors work generally because they are metallic and great conductors which means an abundance of mobile electrons. Since there are so many of them they respond VERY quickly to applied electromagnetic fields and they do it in such as way as to counteract the applied field. In other words, they generate their own electromagnetic fields equal and opposite to the applied field internal to the conductor and external to the conductor we see it as a reflected wave.

3. Oct 6, 2004

### gerben

in these great lectures, you can see Feynman explaining (among other things) how a mirror works:

http://www.vega.org.uk/series/lectures/feynman/index.php [Broken]

Last edited by a moderator: May 1, 2017
4. Oct 6, 2004

### aekanshchumber

I think that answering the question, Why it works? is difficult than, How it works?
I think, we are little bit pre-mature to answer questions why things work? this is biggest question in front of us is Why are we here?

5. Oct 6, 2004

### krab

Another important point is that the wavelength of visible light is large compared with the microscopic surface irregularities of a highly polished metal sheet. This tends to average out the irregularities

6. Oct 6, 2004

### aekanshchumber

Basically when a photon strikes the polished surface its energy is transfered to the atam with which it strikes obviously electron will gain this energy but it tends to loose this energy as soon as possible hence it eject a photon of same frequency, which appeared to be reflected light.

7. Oct 7, 2004

### mwirth

I'll have to read more about the interaction with mobile electrons, but now at least I know what to look for.

The Feynman lectures will take some time, because there are four of them, but I think it will be worth it.

8. Oct 7, 2004

### mathfeel

Mirror only works for certain range of frequency/wavelength. It doens't work at ultraviolet region. This has to do with the frequency (plasma frequency) in which free charge in conductor responds to oscillation (light). Below this frequency, the electron can respond fast enough that it generate its own e/m wave to damp out any propagation in metal. The wave in hence reflected (to conserve energy?). Light actually do penetrates metal a little bit, the so called "skin depth". Above this plasma frequency, the electron can not responds fast enough anymore. The wave is then partially propagated just like in dielectric. (This is the Drude model, which ignore a lot of feature, such as metal lattice, but it works extremely well against experiment).

It so happens that the plasma frequency for metal is in the ultra-violet region, so anything below that (visible) is reflected...For dielectric, the frequency is in the infrared region (correct me if I am wrong), which is (kind of) why glass is transparent.

9. Oct 7, 2004

### Ba

The best reflection you can have is complete internal reflection. How much difference does this have to your standard mirror?

10. Oct 8, 2004

You really ought to read Feynman's book QED. It will explain pretty much all that you want to know, in four 'simple' chapters!

11. Dec 19, 2004

### tozhan

would anyone mind giving me a quick lesson on how the photons conserve the angle of reflection... as far as i can tell it wasnt answered here. (maybe i missed it)

the only answer i can guess at is that maybe we have to take into acount the momentum of the whole object (mirror). when a photon approaches a mirror it is approaching the outer layer of atoms. these atoms have electrons orbiting (duh!). these electrons then absorb the energy of the photon and (as all matter does) re-emits it. the reason for the re-emission of photons (remember: packets of energy!) is that these electron cannot maintain this higher energy level and must discard this amount of energy, the smallest unit of enery known to exist, in the form of EM radiation. although the light will be re-emitted at a lower energy level (shorter wavlength EMR) as some of the energy will get transfered to other atoms via the momentary dncreased repulsion between the nucleus and the electrons new energy level. this newly obtained momentum will instantaneously be transfed around the matter i.e. heating the molecules in the form of increasing internal kinetic energy.

I know this summary is long winded but i need to show the holes in my knowledge ;)

Now... the only way i can see for angle to be preserved is the preservation of momentum of the whole system (photons and mirror). i know that in quantum theory photons have momentum (although with no mass this seems weird to me).

Is this theory correct of completly off the ball??

i know this is long but i felt i needed to write it, just recapping in my own mind...

any help would be great

cheers

Tom

12. Dec 19, 2004

### Integral

Staff Emeritus
The answer is in the reference, Fynmans QED. The short is answer is: They Don't.

13. Dec 19, 2004

### Gokul43201

Staff Emeritus
The result comes directly from matching boundary conditions on the incident and reflected waves.

For plane waves, let the incident E-field be :
$$\mathbf{E_1} = \mathbf{E_1^0}e^{i(\mathbf{k_1.r} - \omega t)}$$

And the reflected wave be :
$$\mathbf{E_2} = \mathbf{E_2^0}e^{i(\mathbf{k_2.r} - \omega t)}$$

For the two to have the same phase at the interface, we need :
$$(\mathbf{k_1.r})_{z=0}= (\mathbf{k_2.r})_{z=0}$$

But since $$|\mathbf{k_1}| = \omega \sqrt{\mu \epsilon}= |\mathbf{k_2}|$$

We must have $$cos(\frac{\pi}{2} - \theta _1) = cos(\frac{\pi}{2} - \theta _2)$$

Or, $$\theta _1 = \theta _2$$

To get Snell's Law for Refraction, just use the definition of the refractive index, given by
$$n = \sqrt{\frac{\mu \epsilon}{\mu _0 \epsilon _0}}$$