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Angle determination of Sum-Frequency Generation in BBO

  1. May 31, 2015 #1

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

    I have difficulties to get the correct angles for a third-order autocorrelator and I hope you can help me.
    In my setup are two BBO Crystalls, the first for SHG (Second-harmonic Generation) and the second for SFG (Sum-Frequency Generation).

    The Laser is a ultrashort-pulse Ti:Sapphire (λ_1 around 800 nm), the BBO Crystals are 50 µm big.

    The calculation for the first BBO (SHG) wasn't really easy (I didn't have a "Nonlinear Optics" class), but I needed only to search for one variable, Θ
    For my parameters, I got around θ= 28,65° (as turning angle for the crystal)
    The complete calculation is here: https://www.overleaf.com/2792476gxqntk

    The SFG is Type-II, so non-collinear and I need to determine $$θ_1, θ_2$$ (from the incoming rays) and Θ, the optical axis of the crystal. All I got are two Equations, which is rather difficult (for me) to solve.

    2. Relevant equations
    In general, the equations are:
    $$k_1 + k_2 = k_3$$

    In the more specific case of SFG in o-e-o Configuration, these are the equations:
    $$n_o(ω)\cdot sin(Θ_1) = n(θ+θ_2)\cdot sin(θ_2)$$
    $$n_o(ω) \cdot cos(Θ_1) + n(θ+θ_2) \cdot cos(θ_2)=2\cdot n(Θ,2ω)$$

    I have found two possible ways to calculate the angles:

    The first is similar to the calculation of non-collinear SHG like in this picture from "The fundamentals of photonics"(Saleh&Teich,2007) which I wanted to transfer to SFG.
    Here we have $$θ_1, θ_2$$ and the crystal axis $$θ$$

    Another attempt is, that there is $$θ_1, θ_2$$ and $$θ_3$$ as angle for each beam.

    3. The attempt at a solution

    For the first way (like in the fundamentals of photonics) I calculated the following:

    which results in
    n_1&=n_o(\omega) \\
    n_2&=n_e=n(\Theta+\Theta_2,2\omega)=\sqrt{\left( \dfrac{\cos^2(\Theta+\Theta_2)}{n_o^2(2\omega)}+\dfrac{\sin^2(\Theta+\Theta_2)}{n_e^2(2\omega)} \right) ^{-1}} \\
    n_3&=n_e=n(\Theta,3\omega)=\sqrt{\left( \dfrac{\cos^2(\Theta)}{n_o^2(3\omega)}+\dfrac{\sin^2(\Theta)}{n_e^2(3\omega)} \right) ^{-1}}

    For the second way, I got the following results:

    n_1&=n_o(\omega) \\
    n_2&=n_e=n(\Theta_2,2\omega)=\sqrt{\left( \dfrac{\cos^2(\Theta_2)}{n_o^2(2\omega)}+\dfrac{\sin^2(\Theta_2)}{n_e^2(2\omega)} \right) ^{-1}} \\
    n_3&=n_e=n(\Theta_3,3\omega)=\sqrt{\left( \dfrac{\cos^2(\Theta_3)}{n_o^2(3\omega)}+\dfrac{\sin^2(\Theta_3)}{n_e^2(3\omega)} \right) ^{-1}}

    My Question(s):

    Which is the correct way to embed these angles?

    And I always get only two equations but three variables, so the system is undetermined. What are some tricks to solve such a system? I have access to matlab and tried my luck with fsolve(…) or made 3 nested for loops, to test between -90° and +90° for each angle if the equations are 0.

    But I had no luck so far.

    I would really appreciate your help.


  2. jcsd
  3. May 31, 2015 #2
    I made a small mistake:
    These are the equations from the SHG exercise (from the book) and are not the equations for SFG

    The SFG equations (my first Version) would be:

    The equations under "3. The attempt at a solution" should be correct.
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