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What blocks electromagnetic waves in the double-slit experiment?

  1. Jun 17, 2011 #1
    In the double slit experiment, a wave is forced to go through a tiny slit. I do not understand what is stopping it from going through the barrier.

    After all, we can think of an electromagnetic wave as just the electric/magnetic fields of a sinusoidally oscillating particle, evaluated using the retarded position and time of that particle. Coulomb's Law and Biot-Savart say nothing of what is between the particle and the point in space its force is acting on--they work regardless, right?

    In other words, why doesn't the wave simply pass through the barrier? Why don't we see this:

    wave.png
     
  2. jcsd
  3. Jun 17, 2011 #2
    I don't think you know what the double-slit experiment is.
     
  4. Jun 17, 2011 #3
    I didn't feel like drawing the rest of it.
     
  5. Jun 17, 2011 #4
    I think I misunderstood your question. The thing stopping the wave from going through the barrier is the fact that the barrier is opaque. The reason your question is confusing, is because you added the idea of the double slit experiment to something completely different.

    What you apparently want to know, is why light doesn't pass through opaque walls. Well, it depends on the certain material, but mostly it's a matter of the energy of the light dissipating into the molecules in the wall.
     
  6. Jun 17, 2011 #5
    What causes that?

    I was under the impression that electric and magnetic fields continue indefinitely into space, regardless of any obstacles they might encounter--Coulomb's Law, for instance, asks only for the distance between two charged particles and says nothing about what is between them.

    Is this not the case, or is light an exception to this rule?
     
  7. Jun 17, 2011 #6
    In other words, is light different from an ordinary electromagnetic wave?
     
  8. Jun 17, 2011 #7

    sophiecentaur

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    No.
    An opaque screen is, by definition, opaque. There are a lot of them around - operating at all frequencies.
     
  9. Jun 17, 2011 #8
    An opaque screen blocks electromagnetic waves because it is opaque. Sorry, I'm looking for something more, maybe something about the nature of light that allows for this to happen.
     
  10. Jun 17, 2011 #9
    So, yea, light can be thought of as being an oscillating electric and magnetic field. The electric field in the light wave will create an electric force on the particles in a a material. This gives the particles energy, the EM wave gives the particle energy. In order to conserve energy, the energy of the light wave must decrease. The light wave gives the molecules in a wall energy (as you can image the light from the sun will make concrete hot). In certain materials, this energy dissipation can occur rather quickly (i.e. in opaque materials). There is certainly a bit more to it than that, and QED will also certainly have something to add. But, I hope that is an adequate enough explanation for the time being.
     
  11. Jun 17, 2011 #10
    Light is not different from other electromagnetic waves. Many books will use the two words light and electromagnetic waves interchangeably. Visible light is, as you probably know, just an electromagnetic wave at a specific range of frequencies.
     
  12. Jun 17, 2011 #11
    Yes, I think that answers my question. I'll need to spend time thinking about how an EM wave's energy can decrease...maybe by thinking about the 3rd law force pair between the particle that is affected by the field and the particle that creates it.

    Anyway, thanks for your responses all. I'll do some more thinking.
     
  13. Jun 17, 2011 #12

    sophiecentaur

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    That was a bit brief, wasn't it? Sorry.
    A screen can be opaque because it either absorbs or reflects the waves. It is the interaction of the em waves with the atoms in the screen that determine what happens. It is the complex dielectric constant (a macroscopic quantity) that counts - a combination of the resistivity and the permittivity. Low resistance materials (metals) will reflect rather than absorb. When the resistivity is higher, the energy is absorbed.
     
  14. Jun 17, 2011 #13

    Born2bwire

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    A few misconceptions that haven't been addressed. Classically, electromagnetic waves are not oscillating particles, they are the coupled electromagnetic fields. Coulomb's Law and Biot-Savart are not applicable here because they assume homogeneous background material and are valid only for electrostatics and magnetostatics respectively. Electromagnetic waves are time-varying by nature and thus cannot be modeled using these laws. These laws are simplifications of the overarching Maxwell's Equations. Maxwell's Equations will yield vector wave equations for the magnetic and electric fields. These vector wave equations, assuming a source free region or a charge neutral one, can be shown to be the same as a scalar Helmholtz wave equation. The boundary conditions imposed by the screens in our problem prevent the waves from penetrating and cause them to be reflected and/or diffracted where appropriate.
     
  15. Jun 18, 2011 #14

    sophiecentaur

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    (or absorbed?)
     
  16. Jun 18, 2011 #15
    What are these boundary conditions? My textbook derived the wave equations in a heuristic fashion using partial derivatives, so maybe I could understand it if I knew what these boundary conditions were.

    Perhaps the material stops inductance somehow? I was told that the wave was due to the changing electric flux inducing magnetic fields, and vice versa:

    [tex]-\frac{\delta B}{\delta x} = \mu_0 \epsilon_0 \frac{\delta E}{\delta t}[/tex]
    [tex]\frac{\delta E}{\delta x} = -\frac{\delta B}{\delta t}[/tex]

    If the partials were suddenly zero at the barrier, that might make sense.
     
  17. Jun 19, 2011 #16

    Born2bwire

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    The boundary conditions here pertain to a perfect electrical conductor. This is where the permittivity, \epsilon, has an infinite imaginary component. This corresponds to a material with infinite loss and thus inside the material the fields are zero. The resulting boundary conditions for the tangential electric fields (dirichlet) require that they be continuous and thus the tangential electric field on the surface is zero. The boundary conditions (neumann) for the normal electric field are not determined however because it will can be non-zero and is related to the induced charge on the surface.

    However, the normal magnetic (H) fields are continuous too, so the normal magnetic field is zero. The direchlet boundary conditions of the tangential E and normal H fields being zero on the surface are enough to solve the full set of differential equations as given by Maxwell's equations.
     
  18. Jun 19, 2011 #17
    Awesome! Thank you.
     
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