What is the cause of reflection?

[A Dhingra]

what causes the reflection of light from a surface?? why can't light simply move through the shiny surface that reflects it...ya difference in penetrating strength... if we try to explain it at the microscopic level... why would the particles of the shiny surface make it bounce back making an angle of ( 90 + a)with its surface...where a is the angle of incidence... can the phenomenon of reflection be explained according to the particle theory ...if possible please explain ...

 

[lalbatros]

Yes it is explain on a microscopic basis.
Atoms interact with electric fields.
This is called the electric susceptibility and it is the direct cause for refraction and reflexion from dielectric media.
Electrons, when they are free, can react sharply to electric fields and this is the cause of conductivity and reflexion from conducting materials.

 

[Dale]

what causes the reflection of light from a surface?? why can't light simply move through the shiny surface that reflects it...ya difference in penetrating strength... if we try to explain it at the microscopic level... why would the particles of the shiny surface make it bounce back making an angle of ( 90 + a)with its surface...where a is the angle of incidence...

Reflection happens at the interface between two mediums with significantly different properties. For example, the conductivity of metal is very different from the conductivity of vacuum, so you will get reflection from a mirror in air. In this case, the high conductivity of the metal allows a current to flow in response to the e-field of the EM wave, and that current acts as the source of its own EM wave which is the reflected wave. In order for this to occur the surface must be smooth at the level of a wavelength of the EM wave, so for visible light this phenomenon is due to the collective behavior of many particles in the reflector. Most of the properties of solids (like metal) are not due to the behavior of the individual atoms, but rather due to their collective behavior.

can the phenomenon of reflection be explained according to the particle theory ...if possible please explain ...

Reflection from an individual particle is called scattering, and is a somewhat different beast. In the case of scattering, it is best understood in terms of the conservation of energy and momentum. If a photon has an energy level which does not correspond to any allowed internal transition in the scattering particle then it cannot be absorbed. This means that it must collide elastically conserving the total energy and momentum.

 

[GeorgeRaetz]

You may notice that shiny reflective metals such as gold, copper also have excellent electrical conductivity; that's not a coincidence. Electrons in metals are easily moved about by electric field and that includes the electric field of an incomming light wave. The electrons in the metal generate an equal but opposite electric field (they try to "short out" the incomming field), that opposite electric field propagates in the opposite direction; that's reflection.
Now, in the case of dielectrics, (glass for example) the electrons are not so free; they are bound to the molecules and so electrons tend to oscillate with the field but generally out of phase with it. That leads to a combination of partial reflection and transmission.

 

[cesiumfrog]

why would the particles of the shiny surface make it bounce back making an angle of ( 90 + a)with its surface...where a is the angle of incidence... can the phenomenon of reflection be explained according to the particle theory

Light scatters from each particle. These scattered light waves (from many different particles) are only in-phase at a certain angle, and interfere destructively in the other directions. You can even prove this using trigonometry.

 

[cesiumfrog]

You may notice that shiny reflective metals such as gold, copper also have excellent electrical conductivity; that's not a coincidence. Electrons in metals are easily moved about by electric field and that includes the electric field of an incomming light wave. The electrons in the metal generate an equal but opposite electric field (they try to "short out" the incomming field), that opposite electric field propagates in the opposite direction; that's reflection.
Now, in the case of dielectrics, (glass for example) the electrons are not so free; they are bound to the molecules and so electrons tend to oscillate with the field but generally out of phase with it. That leads to a combination of partial reflection and transmission.

Uh, what?

The reflection is not in the *opposite* direction.

And isn't the responsible displacement of electrons *always* nearly 180 degrees (off by a tiny fraction due to dissipation) out of phase with the incident electric field wave, regardless of the ratio of reflection to transmission? Regardless of whether they are locally bound as in a dielectric or comparatively free as in a conductor? As a consequence of simple harmonic motion?

 

[GeorgeRaetz]

Uh, what?

The reflection is not in the *opposite* direction.

And isn't the responsible displacement of electrons *always* nearly 180 degrees (off by a tiny fraction due to dissipation) out of phase with the incident electric field wave, regardless of the ratio of reflection to transmission? Regardless of whether they are locally bound as in a dielectric or comparatively free as in a conductor? As a consequence of simple harmonic motion?

To keep the explanation simple I was envisioning an incoming EM wave normal to the conductor surface(E vector parallel to surface) but failed to mention that. Sorry about the confusion.

As for the second group of questions:
The electrons in a (perfect) conductor are so mobile they reduce the electric field to zero regardless of frequency. The simple harmonic oscillators used to explain dielectrics have a natural frequency that differs generally from the incomming EM wave. That introduces significant phase shifts.

 

[cesiumfrog]

Do you deny that the electrons in the surface of a conductor will undergo simple harmonic motion when a light wave is incident on the surface?

Are you saying that the momentarily displaced position of the electrons in the surface of a conductor is *in phase* with the electric field of the incident light wave?

The electrons in a (perfect) conductor are so mobile they reduce the electric field to zero regardless of frequency.

You don't think skin penetration plays any role in the interaction with light?

 

[A Dhingra]

In this case, the high conductivity of the metal allows a current to flow in response to the e-field of the EM wave, and that current acts as the source of its own EM wave which is the reflected wave.

i didn't get it... can the electric ( field ) component of the em wave affect the free electrons of a metal... to the extent that current is produced...
is it because both the fields are time varying and hence induce current...??
and this current reflects the source according t the lenz law... right?

If a photon has an energy level which does not correspond to any allowed internal transition in the scattering particle then it cannot be absorbed. This means that it must collide elastically conserving the total energy and momentum.

can you explain it in more elaborate manner...

have you meant that , if photon has larger amount of energy as compared to transitions within the atom then it collides... but how can that energy be ever more ... i mean the possible number of orbits to which transition can take place can even be at a distance of infinity...but to that we call the electron is out of the atom...right??

so possibly you mean that the energy of photon is less than the required amount for transition , so it will collide...

 

[A Dhingra]

The electrons in the metal generate an equal but opposite electric field (they try to "short out" the incomming field), that opposite electric field propagates in the opposite direction; that's reflection.

Now, in the case of dielectrics, (glass for example) the electrons are not so free; they are bound to the molecules and so electrons tend to oscillate with the field but generally out of phase with it. That leads to a combination of partial reflection and transmission.

one thing that is not clear is " out of phase "means what?
so glass allows light to pass through it (almost , to a good extent ...) and just make it reach the metal which reflects it ...in case of a mirror...

and also can you expalin this " The electrons in the metal generate an equal but opposite electric field "...

thanks for this explanation...

 

[Dale]

i didn't get it... can the electric ( field ) component of the em wave affect the free electrons of a metal... to the extent that current is produced...

Yes, an E-field produces a force on charges and in a conductor the charges are free to move and moving charges are a current.

have you meant that , if photon has larger amount of energy as compared to transitions within the atom then it collides... but how can that energy be ever more ... i mean the possible number of orbits to which transition can take place can even be at a distance of infinity...but to that we call the electron is out of the atom...right??

What you are describing here is called ionization, not scattering. With sufficiently high energy you can get ionization, pair creation, or creation of whole host of unstable particles. None of that is scattering. Scattering is when the photon and the particle collide elastically because the energy level of the photon doesn't match any allowable internal transition of the particle. This is not just limited to electron transitions in free atoms, but includes vibrational modes in molecules, etc.

 

[sophiecentaur]

To appreciate any of this you must first avoid using the dreaded Hydrogen Atom model when discussing em and solids (or, indeed, em and anything more than just one atom). It just serves to confuse.

A photon can interact with a structure as well as with an isolated atom. This form of interaction doesn't just happen for one frequency and can result in the waves being transmitted and absorbed and reflected. What actually happens, depends on the actual substance and its so called refractive index.

A dielectric substance like glass has a refractive index which is pretty much just equal to the ratio of speeds in the glass and in vacuum and there is very little absorption. (That's where you first come across the refractive index, in Science). The reflection at an interface is due to the 'mismatch' between the two media.

Once the substance has a significant 'resistive' property, the refractive index becomes a complex number and there will be loss / absorption. For very low resistances, as in metals, the refractive index is dominated by this and reflection can be regarded as being due to induced currents in the valence electrons - which can be 'made to oscillate' at almost any frequency up to Xrays, where they behave differently, and penetrate or are absorbed, rather than reflected (except at very oblique angles: see X ray telescopes).

I have ignored a lot of exceptions in this crude form of classification, of course, but what I have written is right in broad terms, I think. In many ways it is a lot better to start off in terms of classical behaviour and try to relate the quantum aspects once you have this well sorted.

 

[Grep]

one thing that is not clear is " out of phase "means what?

Kind of difficult to explain in a concise way. I suggest you read Richard P. Feynman's "QED: The Strange Theory of Light and Matter". It's a really great book, and I can't see how the basics of Quantum Electrodynamics could possibly be made simpler. He was a truly great teacher.

 

[sophiecentaur]

Don't you think it's demanding a bit much of someone who doesn't know what "out of phase" means, to tell them to read Feynman? Can't you suggest something more in line with the basic concept of phase?
You can be pretty sure that Feynman's target audience were expected to have a fair basis of classical Science knowledge. (I'm not talking of the 'Popular Science' aspect, where people go away from reading or listening to him and enjoyed the experience - that's a different matter)

 

[Grep]

Don't you think it's demanding a bit much of someone who doesn't know what "out of phase" means, to tell them to read Feynman? Can't you suggest something more in line with the basic concept of phase?
You can be pretty sure that Feynman's target audience were expected to have a fair basis of classical Science knowledge. (I'm not talking of the 'Popular Science' aspect, where people go away from reading or listening to him and enjoyed the experience - that's a different matter)

Perhaps you're right. I think at least the first part is simple enough most intelligent people would get the gist of it, but I could be wrong. He doesn't really use terms like "out of phase", for example. Just his "adding arrows" procedure which is pretty easy to understand visually.

And I don't see how things like partial reflection could be explained in any simpler way (doesn't mean there isn't a simpler way, but I haven't found it). Hence my recommendation. I think that perhaps, if the first part of the book can't be understood by someone, that they may not have the tools needed to understand it yet. It can only be made simple to a point, beyond which it's just nonsense.

But fair enough. However, after a bit of searching, I've failed to find a web page with a simple explanation of "out of phase". The wiki for "Phase" is full of mathematics, for example.

 

[sophiecentaur]

This link http://en.wikipedia.org/wiki/Phase_(waves)" should be a help, with or without the maths; the pictures are not bad in showing the gist of what a phase relationship means.
Try that.

 

[A Dhingra]

Yes, an E-field produces a force on charges and in a conductor the charges are free to move and moving charges are a current.


What you are describing here is called ionization, not scattering. With sufficiently high energy you can get ionization, pair creation, or creation of whole host of unstable particles. None of that is scattering. Scattering is when the photon and the particle collide elastically because the energy level of the photon doesn't match any allowable internal transition of the particle. This is not just limited to electron transitions in free atoms, but includes vibrational modes in molecules, etc.

thanks...
well one more thing ...you have mentioned vibrational modes...how do they affect
the reflection... rather collision...
is it that the amplitude of vibration affects the possibility of collision?

 

[A Dhingra]

This link http://en.wikipedia.org/wiki/Phase_(waves)" should be a help, with or without the maths; the pictures are not bad in showing the gist of what a phase relationship means.
Try that.

simply speaking ... out of phase means that the waves different phases ( or angle of phase shift is other than 0 degree)... is that right?

 

[Dale]

thanks...
well one more thing ...you have mentioned vibrational modes...how do they affect
the reflection... rather collision...
is it that the amplitude of vibration affects the possibility of collision?

If there is a vibrational mode of the right energy available then the photon will be absorbed instead of scattered. This is the principle behind infrared spectroscopy since the vibrational modes fall in the infrared energy range.

http://en.wikipedia.org/wiki/Infrared_spectroscopy

 

[sophiecentaur]

simply speaking ... out of phase means that the waves different phases ( or angle of phase shift is other than 0 degree)... is that right?

Simply speaking, waves with different phases have peaks and troughs at different times. You could 'see' this if you were to take a picture of the two waves - the peaks and troughs would appear in different places. Otherwise you could look at their variations in time (such as with an oscilloscope), in which case the peaks and troughs would be seen, also, to occur at different times.
Two signals (not necessarily waves) of the same frequency can also have a 'phase difference' between them. Two kids on swings of equal lengths may be going at the same rate but not coincide in time / phase.

The use of vectors (or, rather, Phasors) is a good way to show their phase relationship by displaying them as a 'frozen' circular motion. This method can show their amplitudes and phases and make it easy to show the result of adding the oscillations together. Wikkers is bound to have an entry on Phasors.

 

[cesiumfrog]

Kind of difficult to explain in a concise way. I suggest you read Richard P. Feynman's "QED: The Strange Theory of Light and Matter". [..] He was a truly great teacher.

Don't you think it's demanding a bit much of someone who doesn't know what "out of phase" means, to tell them to read Feynman?

The use of vectors (or, rather, Phasors) is a good way to show[..]

For someone who is interested in how reflection works on the microscopic scale, but doesn't have a strong technical background yet, I think Feynman's little popular science book "QED" is indeed the best possible resource.

It captures the excitement of the science, and it conveys the true technical mechanism without misleadingly oversimplifying and without needing the technical jargon nor confusing math.

(Phasors, on the other hand, are such an abstract and esoteric topic among physics graduates nowadays, hardly the first concept I would recommend someone begin from..)

 

[cragar]

So are you guys saying that reflection has to do with light having an E and B field . And it gets affected by the electrons in the material. What I always had trouble with is that light doesn't have charge so why would this affect the photon. And when I shine a laser into a B field it doesn't alter its path. I am probably missing something fundamental here.

 

[Dale]

What I always had trouble with is that light doesn't have charge so why would this affect the photon.

There is more to EM than just the Lorentz force law. EM fields interact with the fields, not just with charges (see Maxwell's equations). Otherwise you couldn't have waves in vacuum.

 

[Born2bwire]

So are you guys saying that reflection has to do with light having an E and B field . And it gets affected by the electrons in the material. What I always had trouble with is that light doesn't have charge so why would this affect the photon. And when I shine a laser into a B field it doesn't alter its path. I am probably missing something fundamental here.

The problem is that you, and this seems to be common, are mixing in both classical and quantum ideas. Thinking about charges and currents and using that as an explanation for the classical field reflection is perfectly fine. The incident fields induce currents on the surface and interior of the material via the Lorentz force. These induced currents create secondary fields which cancel out the transmitted fields and produce the reflected fields. Thus, the total field is an incident and reflected field without any transmitted wave (assuming total reflection). In terms of photons however, we need to look at the energy modes of the material and the band structure all of which are determined using quantum mechanics. It then can become a problem of scattering (in the case of reflection) and/or of transmission into one of the supported modes of the problem (where the diffraction can be solved for example via conservation of momentum). One can also use the idea of a path integral to get the same results (whose classical analogue is Fermat's principle).

 

[sophiecentaur]

For someone who is interested in how reflection works on the microscopic scale, but doesn't have a strong technical background yet, I think Feynman's little popular science book "QED" is indeed the best possible resource.

It captures the excitement of the science, and it conveys the true technical mechanism without misleadingly oversimplifying and without needing the technical jargon nor confusing math.

(Phasors, on the other hand, are such an abstract and esoteric topic among physics graduates nowadays, hardly the first concept I would recommend someone begin from..)

You have apoint there, of course, but reflection, described as a wave phenomenon, requires a certain 'base level' of understanding. The idea that one should need to look at a book with Quantum Electrodynamics in the title suggests that it's even harder than it is.

It seems to be the fashion to leap into fundamental particles as an initial explanation of everything when a few very basic classical ideas can relate easily to concrete reality. I would say that any Physics graduate who struggled with the idea of expressing the amplitude and phase of a wave as a vector (just polar coordinates, surely?) would certainly be struggling with 'true understanding' of anything on the topic of fundamental particles.

Phasors, "esoteric"? I have talked to may 'technician grade' workers who can deal more than adequately with the phasor in power engineering. Look at the threads on power factor on this forum. They're taken as read.

 

[cragar]

thanks dalespam and born2bwire for your answers.

 

[A Dhingra]

i think by now i have a rough idea as to what causes reflection...
to just make sure what i have got is correct i am giving this reply...
for a mirror , an em wave moves through the surface of glass ( by scattering if the energy is less than that required for transition...)and at the surface of the metal ...the wave induces current ( here the incoming wave is absorbed in the conductor...) ..and thus gives out another em wave of same property as the existing one... making the angle of incidence equal to the angle of reflection...

is it correct by far?
if yes then only query left is that how do the angles become equal... for 0 degree angle of incidence the em wave produced will move opposite to the entering direction...
what happens for other angles...