- #1

Eoraptor

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1. Homework Statement

1. Homework Statement

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.

## Homework Equations

In general, the equations are:

$$ω_1+ω_2=ω_3$$

and

$$k_1 + k_2 = k_3$$

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:

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.

[CH-21]

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.

## The Attempt at a Solution

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

https://www.overleaf.com/2795897bqfyps

which results in

\begin{align}

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}}

\end{align}

For the second way, I got the following results:

\begin{align}

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}}

\end{align}

**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.

Thanks,

Eoraptor