1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    with the xy-plane coinciding with the CCD plane.
    But I have several papers here saying that the intensity is actually proportional to
    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 ("*").
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook