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Gaussian beam

  1. Jun 16, 2010 #1
    Can tell me the exact relationship between the intensity and the amplitude for a gaussian beam
    ?
    I know that I is proportional to |A|2 .. but i want the value of this proportianality constant
     
  2. jcsd
  3. Jun 16, 2010 #2

    Redbelly98

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    It's more properly called the irradiance (assuming you want the power per unit area), and for any electromagnetic wave (not just Gaussian beams) the relation is
    I = ½ c εo Eo2
    where Eo is the amplitude of the electric field.
     
  4. Jun 17, 2010 #3
    I(x) = [I0/σ(2π)0.5] exp[-x2/(2σ2)]

    where ∫I(x) dx = I0 and σ is the rms width.

    Bob S
     
    Last edited: Jun 17, 2010
  5. Jun 17, 2010 #4
    In engineering units you drop the "c" In the above intensity equation and you'll have Watts.
     
  6. Jun 18, 2010 #5

    Redbelly98

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    But that isn't in terms of the amplitude, as requested by the OP. You're also assuming the beam is Gaussian in just 1 transverse dimension. If the beam has a circularly symmetric intensity profile, things are different.

    Actually, dropping the c gives the energy density in J/m3.
    I'm using SI units. I'm not aware of any official "engineering units", unless you also mean the SI system?
     
  7. Jun 18, 2010 #6
    I believe this is the correct normalized Gaussian beam distribution in two dimensions:

    I(x,y) = [I0/(σxσy2π)] exp[-x2/(2σx2) - y2/(2σy2)]

    where ∫I(x,y) dx dy = I0, and σx and σy are the rms widths in the x and y dimensions.

    At any point x and y, the intensity is equal to the square of the amplitude.

    Bob S
     
  8. Jun 18, 2010 #7

    Redbelly98

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    I0 is perhaps a poor choice of variable name, in that it has different units than I(x,y). I.e., I0 is in Watts and I(x,y) is W/m2.

    Not in SI units, or Gaussian-cgs units either. Assuming "the amplitude" refers to the electric field (And I can't imagine what else it would mean.)

    (Electric field)2 has units of energy/m3 in Gaussian units, or [energy/(charge*distance)]2 in SI units. Neither is consistent with intensity = energy/(time*distance2), so I maintain that the relationship is proportional, but not equal.
     
  9. Jun 19, 2010 #8

    Integral

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    I had this image on hand, thought I would share it. This is a power cross section of a gaussian laser beam. Each color represents a power level. Other beam parameters are given to the left.

    4941far.jpg
     
  10. Jun 19, 2010 #9
    We consider a normalized bi-gaussian electromagnetic radiation beam travelling in the z direction.

    The Poynting vector is P(x,y) = E(x,y) x H(x,y) watts/m2 at every point in space, where E(x,y) is in volts per meter, and H(x,y) is in amps per meter.

    E(x,y)/H(x,y) = Z0 = 377 ohms in free space

    So the power density is w(x,y) = E2(x,y)/(2·Z0) watts/m2

    or E(x,y) = sqrt[2·Z0·w(x,y)] volts/m

    where the bi-gaussian distribution is w(x,y) = [W0/(σxσy2π)] exp[-x2/(2σx2) - y2/(2σy2)] watts/m2

    where the total beam power is W0 = ∫w(x,y) dx dy watts

    Bob S
     
    Last edited: Jun 19, 2010
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