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CCD measures intensity prop. to z-component of C-Poynting vector OR |E-field|^2

  1. Aug 22, 2011 #1
    intuitively, i would say that in a CCD, the photoelectrons are being "created" only by the electric field of the incident light, so that the intensity is proportional to
    [itex]|E_x|^2+|E_y|^2[/itex],
    with the xy-plane coinciding with the CCD plane.
    But I have several papers here saying that the intensity is actually proportional to
    [itex](\textbf{E}\times\textbf{B}^*)_z[/itex].
    could someone shed some light on why this should be the case? thanks!
     
  2. jcsd
  3. Aug 22, 2011 #2
    They both basically mean the same thing. The energy flow due to electrodynamic fields in general is described by the Poynting vector S:

    S = E x H

    For air, the magnetic field H and the magnetic field B are related according to B = μ0 H, leading to:

    S =(1/μ0) E x B

    The intensity in a certain direction is typically defined as the time-average of the Poynting vector dotted by the direction. For sinusoidal waves, the time-average is found by complex conjugating the magnetic field and multiplying by a half:

    I = (<S>t)z =(1/2 μ0) (E x B*)z

    If the wave is a transverse plane wave (not always the case), then the electric field and magnetic field are related according to: B= sqrt(μ0ε0) z x E which leads to:

    I =(1/2Z0) |E|2

    For a transverse wave, there is no z-component to the electric field, only x and y components, so we have finally:

    I =(1/2Z0)(|Ex|2 +|Ey|2)

    So your last expression is more general, and your first expression only applies to transverse plane waves.
     
  4. Aug 26, 2011 #3
    thank you for your clear explanation.

    forgive my ignorance, but could you point out an example of a non-transversal EM wave?

    i'm trying to calculate the intensity on the CCD resulting from a focused light beam. the two formulas give slightly different values, the more general one has a complex component, do you think one should simply take the real part or the modulus?
     
  5. Aug 29, 2011 #4
    If you are close enough to light sources, the fields are not necessarily transverse. This is called the near field. Also, light in a waveguide (like a fiber optic) is not necessarily transverse. If you are using the expression I =(1/2 μ0) (E x B*)z, then the result should be automatically real-valued only. That is the point of the complex conjugation operation ("*").
     
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