# Theoretical Investigation for Angle Measurement Based on Femtosecond Maker Fringe
## 3 Simulation Results and Discussion
### The Investigation of Angle Measurement in the Optical Frequency Domain
It can be seen from Figure 12 that the peak wavelength \(\lambda_{\text{P}}\) almost monotonically increases or decreases as the incident angle changes. The slope of the angle dependence implies the measurement sensitivity. Similar to the intensity-based measurement, the averaged sensitivity \(S^{\prime}\) of optical frequency domain angular measurement can be calculated as Equation (14):  
\[S^{\prime}=\frac{\lambda_{\text{P}}(\theta_{\text{ie}})-\lambda_{\text{P}}( \theta_{\text{is}})}{\Delta\theta_{\text{i}}} \tag{14}\]  
where \(\theta_{\text{is}}\) and \(\theta_{\text{ie}}\) denote the start and end incident angle of the \(\lambda_{\text{P}}\)'s angular dependence curve, and \(\Delta\theta_{\text{i}}=\theta_{\text{ie}}-\theta_{\text{is}}\). Taking the p-s polarization of BBO as an example, we show \(\lambda_{\text{P}}(\theta_{\text{is}})-\lambda_{\text{P}}(\theta_{\text{ie}})\) and \(\Delta\theta_{\text{i}}\) in Figure 12a. The sensitivities \(S^{\prime}\) of other cases are shown in Table 3. It is clear that the p-s polarization's sensitivity is significantly higher than the p-p polarization and s-s polarization. Therefore, the p-s polarization is superior to the other two polarization combinations for the angular measurement in the optical frequency domain. The p-s polarization of BBO has the best measurement sensitivity 1.43 \(\times\) 10\({}^{-3}\) nm/arcsecond among the investigated materials, and only the KDP has a relatively higher sensitivity of \(-\)1.01 \(\times\) 10\({}^{-3}\) nm/arcsecond. Figure 9: The Maker fringe shift of p-s polarization of LiNbO\({}_{3}\). (**a**) The 2D plot; (**b**) 3D plot in the higher intensity area. The measurement sensitivity \(S^{\prime}\) is also closely related to the phase difference \(\psi\). This is because the peak shift is, in fact, the movement of the point where \(\psi\) = _constant_. As Equations (11) and (12), the phase difference \(\psi\) is related to the refractive index \(\Delta n\). Therefore, the peak shift is determined by the point of \(\Delta n\) = _constant_. Figure 13 shows the contour map of \(\Delta n\) of the p-s polarization. In this figure, the subplots (a), (b), (c), and (d) correspond to the crystal BBO, KDP, LiNbO\({}_{3}\), and Ag\({}_{3}\)AsS\({}_{3}\), respectively. In each subplot, the lines in different colors denote the different constants of \(\Delta n\), and the step of \(\Delta n\) between different lines is 0.001. It is easy to understand that the higher the slope of the line, the faster the wavelength changes as the angle changes in the line, i.e., the faster the peak shift. BBO has the highest line slope compared to the others, and therefore its measurement sensitivity is the best. We note that only the KDP has a negative linear slope, which means that as the  
Figure 10: The calculation results of peaks shift in the optical frequency domain. The figures (**a-d**) denote the nonlinear materials BBO, KDP, LiNbO\({}_{3}\), and Ag\({}_{3}\)AsS\({}_{3}\) in order, the subscript numbers 1, 2, 3 denote the polarization combinations p-s, p-p, and s-s, respectively. \begin{table}
\begin{tabular}{l c c c c} \hline \hline  & **BBO** & **KDP** & **LiNbO\({}_{3}\)** & **Ag\({}_{3}\)AsS\({}_{3}\)** \\ \hline p-s polarization & \(1.43\times 10^{-3}\) & \(-1.01\times 10^{-3}\) & \(2.05\times 10^{-4}\) & \(1.17\times 10^{-4}\) \\ p-p polarization & \(1.11\times 10^{-4}\) & \(4.53\times 10^{-5}\) & \(5.87\times 10^{-5}\) & \(1.90\times 10^{-5}\) \\ s-s polarization & \(9.80\times 10^{-5}\) & \(5.91\times 10^{-5}\) & \(1.98\times 10^{-5}\) \\ \hline \hline \end{tabular}  
* The unit of averaged measurement sensitivity is \({}^{\prime}\)nm/arcsecond\({}^{\prime}\). The \(\backslash^{\prime}\) means that the s-s polarization of KDP doesn’t have measurement sensitivity because of \(d_{\rm eff}=0\).  
\end{table} 
Table 3: The averaged sensitivity of angle measurement in the optical frequency domain. angle increases, the wavelength should decrease to keep \(\Delta n\) as a constant, and it causes its negative sensitivity.