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Nonlinear optics: second order polarization calculation

  1. Mar 10, 2014 #1
    This is a problem from Boyd Nonlinear Optics chptr 1 problem 2.

    1. The problem statement, all variables and given/known data

    Numerical estimate of nonlinear optical quantities. A laser beam of frequency ω carrying 1 W of power is focused to a spot size of 30μm diameter in a crystal having a refractive index of n =2 and a second order susceptibility of [itex]\chi^{(2)}=4\times 10^{-11} m/V[/itex]. Calculate numerically the amplitude P(2ω) of the component of the nonlinear polarization oscillating at frequency 2ω.


    2. Relevant equations

    [tex]P(2\omega)=\epsilon_0^{(2)}E^2[/tex]
    [tex]I=\frac{cn\epsilon_0}{2}E_0^2[/tex]

    3. The attempt at a solution

    I solved for [itex]E_0[/itex] assuming a uniform distribution across the spot, with I=P/A and got
    [tex] E_0=\sqrt{\frac{2P}{Acn\epsilon_0}}[/tex]
    and put that into the equation I gave for [itex]P(2\omega)[/itex]. The value I got was [itex]1.89\times 10^{-11}[/itex], which is almost exactly 4 times the value given in the text of [itex]4.7\times 10^{-11}[/itex]

    I feel like it's possible that the value given in the text accidentally uses the diameter of the spot to calculate the area, which would give them a factor of 1/4 that I don't have, but I also thought that maybe my problem lies in my assumption that the spot is uniform. Maybe I actually need to integrate numerically assuming a gaussian beam profile-which would kind of make sense considering the problem title (numerical estimate of nonlinear optical quantities).
     
  2. jcsd
  3. Mar 11, 2014 #2

    TSny

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    Gold Member

    Hello, dnp33.

    Here's a possibility. Note that below (1.2.14b), Boyd (3rd edition) gives the expression for intensity as ##I = \frac{1}{2}n_0\epsilon_0c\varepsilon^2## where ##\varepsilon## is the field amplitude as defined by (1.2.12). But this definition of field amplitude differs by a factor of 2 from the field amplitude ##E## defined by (1.2.1). Note ##\varepsilon = 2E##.

    Very confusing. Would have been nice if Boyd had given an explicit numerical example to help sort out the notation.
     
  4. Sep 24, 2015 #3
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