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Geometric optics vs Electromagnetic waves

  1. Sep 1, 2006 #1
    Light which are electromagnetic waves have an electric field component and magnetic field component. The electric field can be phase shifted but the magnetic field never does.

    In geometric optics, light is modeled as a straight line and its rules upon reflection and refraction are according to the electric field only. So its as if in this model light was only a travelling electromagnetic wave which is false.
  2. jcsd
  3. Sep 1, 2006 #2
    The electric and magnetic fields are related, because of the fundamental laws of electromagnetism.
    In vacuum or in isotropic media, without attenuation, the two fields are perpendicular to each other and perpendicular to the wavevector which is also the direction of propagation.
    The electric and magnetic fields can both be derived from the "potential vector".
    As you know, only one 4-vector, the potential vector, is all that is needed to describe electromagnetism. Electric and magnetic fields are simply more familiar objetcs in classical physics.
    In classical physics, the potential 4-vector may look like a pure mathematical object (except for the electrostatic component).
    In quantum physics it receives a more important status, due to the Aharonov-Bohm Effect, at least.

    The relation between magnetic and electric fields is not only related to their mutual directions.
    The amplitude of the electric field (E) and magnetic field (H) are related by:

    E = Z H
    where Z in the charactistic impedance of the medium
    the vaccum characteristic impedance is Zv = 377 Ohms​

    The Fresnel laws for refraction and reflection are determined by the conditions on the boundary between the two media.

    What do you mean by these statements:

    Last edited: Sep 1, 2006
  4. Sep 1, 2006 #3


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    Surely, if the phase (with respect to a fixed axis) of the electric field of a plane wave changes, the magnetic field follows.

    I thought people talked about the electric field only because it's far easier to get measurements from.
    Last edited: Sep 1, 2006
  5. Sep 1, 2006 #4


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    In geometric optics we simply disregard the wave-nature of light. It's a good approximation whenever the wavelength (or relevant wavelengths if you will) are very small compared to the dimensions of the problem. It's certainly easy to work with.
  6. Sep 1, 2006 #5

    The fresnel laws, for reflexion and refraction, are hard to explain without the electromagnetic theory of light. Without it remains, at best, an empirical law.

    So, I would say that even in geometric optics we cannot have a closed/complete theory without reference to the electromagnetic wave nature of light.

  7. Sep 1, 2006 #6
    What I mean is that in geometric optics, when light reflects, its phase can shift or not according to the same rule as when an electric field is incident on a medium.
    But when a magnetic field is incident on a medium, it can reflect differently to the electrifield.
    Light consists of both mangetic and electric fields. So why only consider the electric field when determining how light reflects in geometric optics?
    i.e. If light was only considered as a mangetic wave in geometric optics than it would reflect according to different rules compared to what is in the books today.
  8. Sep 2, 2006 #7

    This never happens. Magnetic waves do not exist. Electric wave do not exist.
    Only electromagnetic waves do exist.

    The wave phenomenon itself is an oscillation where the variations of magnetic field create and induced electric field, and conversly the variation of the electric field induce a magnetic field. Both fields are not separable.

    In addition the conditions at the boundary are rather similar and can be derived from one another. Continuity of the tangential components.

    Note that different situation may occur depending on the polarisation of the wave with respect to the incidence plane.

  9. Sep 2, 2006 #8

    I take your point about the EM wave. But when it reflects, the E component can be phase shifted whereas the M component cannot.

    When it comes to geometric optics we treat light as only one wave and it reflects according to the properties of the E component only. Why is that? Why not make it reflect according to the B component?
  10. Sep 2, 2006 #9

    I don't understand what you mean.

    Assume one of the three rays (incident, reflected, refracted) has a certain space-time dependence.
    Assume the following for the electric field:

    E = Eo exp(i( kr - wt + phase ))​

    then, the space-time dependence for the magnetic field is determined by the Maxwell's equations. Therefore, the phase of the magnetic field will mimic the phase of the electric field.

    I trust that if you explain me more about what you mean by
    I could give you some interpretation.

  11. Sep 2, 2006 #10
    I think what pivoxa15 means is that when we do optical analysis, we only bother with the details of the electric field, and rarely the magnetic field.

    As far as I know, the only reason we do this is due to convenience and choice. Eg. we have known materials which act as polarisers on the electric component of the EM field, although if we wanted to, we could say we were considering the magnetic field.
  12. Sep 2, 2006 #11
    Checking my notes again, it could be that I was wrong. the phase of E and M fields are equal. If one is phase shifted by 180degrees than so does the other one. Although their magnitudes change by different amounts depending on the material. If that is the case than sorry about the confusion I have created.
  13. Sep 3, 2006 #12

    There is no need to be sorry.
    I was very happy to go back to my preferred book (Landau).
    The boundary conditions to be applied made it clear that geometric optics needs waves to be explained.

    For example, the law for refraction occur because the phases of the three waves on the interface must be compatible. The intensity of the different "rays" are further consequences of the EM wave nature of light.

  14. Sep 3, 2006 #13

    What are the three waves? I thought light was modelled as 2 types of waves (that are always together) in EM theory and 1 in geometric optics.

    This brings about another question. if the E wave induce the M wave and viceversa than does that mean only one of E or M can exist at one time? Although when they reach their destination, they map out a trajectory as if they had travelled together.
    Last edited: Sep 3, 2006
  15. Sep 3, 2006 #14
    By three waves I'm fairly sure lalbatros means the incident wave, the transmitted wave, and the reflected wave. There is no reason for anything to match between these waves until one applies boundary conditions (or thinks about it for a sec)
  16. Sep 3, 2006 #15

    Claude Bile

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    For completeness, we only need to find either E or B. Once we know one, we can then figure out the other (in principle) using Maxwell's equations.

    There is no absolute convention as far as I know as to people using E or B (or D or H for that matter). I guess E is popular because the resultant phenomena such as radiating dipoles, polarisation etc are more inutitively explained, whereas for magnetic-related phenomena such as the magneto-optic effect, the B field is the obvious choice.

    Last edited: Sep 3, 2006
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