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Why does reflection of EM waves not also cause transmission?

  1. Nov 8, 2014 #1
    I understand that reflection of EM waves is the electrons on the surface of an object being caused to oscillate, and transmission is the electrons all the way through the object being caused to oscillate.

    Why, in reflection, when the electrons on the surface of an object are caused to oscillate, are the electrons all the way through not caused to oscillate also?
     
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  3. Nov 8, 2014 #2

    jedishrfu

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    Here's a discussion on Wikipedia about it

    http://en.m.wikipedia.org/wiki/Reflection_(physics)

    In it they mention the classical interpretation of electron oscillation acting as a dipole antenna radiating in all directions resulting in a reflection back so I guess the oscillations disepate as they penetrate further into the solid.
     
  4. Nov 8, 2014 #3
    One way to answer this is to cite what is called the "skin effect". A field outside a conductor causes currents to flow within the conductor, but the currents are confined to a "skin" or layer that gets thinner and thinner as the frequency increases. The current in deeper layers is negligible.

    A more detailed picture is that the outer layers produce their own field, which partially cancels out the original applied field on the inside. So inner layers are effectively exposed to a weaker field. These inner layers in turn partially cancel out this (weaker) field that has managed to penetrate to that depth, so that layers that are deeper still will experience an even weaker field. Finally you have almost no field by the time you get to the other side of the metal sheet.

    The reflected wave on the outside of the conductor is the combined effect of all the fields produced by all the currents in different layers in the conductor.
     
  5. Nov 8, 2014 #4

    Simon Bridge

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    Since this is generally phrased in terms of reflection from an "object", then it should be pointed out that, at a surface, you do get reflection and transmission of EM waves in varying amounts depending on the material properties of the object.

    Further, we no longer understand reflection and transmission in terms of "oscillating electrons".
    There are a series of models of varying complexity covering the different situations where you need to know about reflection and transmission.
    i.e. see: http://vega.org.uk/video/subseries/8 (2nd lecture).
     
  6. Nov 8, 2014 #5
    I guess my question is then:

    What is the difference in structure between an object which only reflects, and an object which both reflects and transmits?

    Consider light shining upon both a brick wall and a glass window. What is the difference in structure between a brick wall and a window, which causes only reflection in the brick wall, and both reflection and transmission in the glass window?
     
    Last edited: Nov 8, 2014
  7. Nov 8, 2014 #6

    Simon Bridge

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    See the video series linked in my post. But I'm not sure you are after that kind of fundamental detail.

    Lets lay some groundwork: the comparison between a brick and a sheet of glass is problematical because there is so much that is different between the two it is hard to draw out the principles.

    Compare, instead, a clear, polished, sheet of plate glass like for a window with the same glass after it has been sand-blasted. It's the same material chemically but quite different reflection and transmission. What do you think has had that effect?

    Compare a thin sheet of glass with a thick sheet - less transmission through the thick sheet.
    At this bulk-level description, light may be reflected, transmitted, or attenuated in a material.

    What you probably want to ask about is more like the comparison between a sheet of polished glass and a sheet of polished metal.
    The metal would have to be very thin, compared with the glass, to let any light through.

    The main difference is between the way the electrons are distributed in the material.
    Then things like material density and overall shape add in extra effects.

    To know what to tell you, we really need to know what you need the information for.
    Otherwise we could be writing out details for days without actually helping you.
    But the above should get you thinking properly.
     
  8. Nov 8, 2014 #7
    What kind of electron distribution enables transmission of light?
    I want the information for my own understanding. I do not attend any school or university.
     
  9. Nov 8, 2014 #8

    Simon Bridge

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    Generally you need the electrons to be scattered through the material and well attached to their atoms - for something like a metal, there are a lot of electrons close to the surface. It's more involved than that though.

    ... well I don't think anyone can help you.
    It is a big and involved subject and there is no way to tell what would count as "for your own understanding" unless you tell us.
    At the non-technical level, transmission and reflection are sort-of material properties derived from the relative refractive indexes on each side of the classical surface. It's just how light responds to regeons of different speeds.
     
  10. Nov 8, 2014 #9
    In a brick wall the electrons are well scattered through the material, and well attached to their atoms, but light is still not transmitted through. Why?
    What are you talking about?
     
    Last edited: Nov 9, 2014
  11. Nov 9, 2014 #10

    Simon Bridge

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    This is not quite correct - the brick is not a uniform substance so bits of it have different properties, which have different electronic distributions. It also has a rough surface like the frosted glass, and lots of holes, cracks, and other imperfections all the way through, making extra interfaces.
    The brick is, in short, very complicated.

    Another way life gets more complicated than the general rule is that most materials are selective about how they reflect and transmit different wavelengths - this is to do with the fine details of the electronic configuration and also helps determine the color of the object. An object opaque to visible light may be transparent to other wavelengths.

    Please follow the links I gave you. Please remember that this is a big big subject which we cannot do justice to here.
     
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