# Prism-Based Spatial Heterodyne Spectrometer with a Fixed Fringe Localization Plane
## 2 Principles and Designs
### Performance Parameters  
Now, we calculate the theoretical spectral resolving power of the system. Assuming that the incident light with a wavenumber of \(\sigma_{1}\) produces \(2N\) fringes, the maximum \(OPD\) on the interferogram can be expressed as \(N\lambda_{1}\). If the half-width of the light source is \(d_{0}\), due to the small angle between the wave fronts of the two arms and the FLP, the half-width \(d\) of the wave fronts of the two arms after passing through the refracting surfaces of the prisms can be approximated as  
\[d(\sigma_{0})=\frac{d_{0}}{\cos 30^{\circ}}\times\sin\Bigl{(}90^{\circ}-\sin^{ -1}(\sin(30^{\circ})n(\sigma_{0}))\Bigr{)} \tag{5}\]  
Thus, the maximum \(OPD\) can also be expressed as  
\[OPD_{\rm max}=N\lambda_{1}=2d(\sigma_{0})\sin\beta_{1} \tag{6}\]  
where \(\beta_{1}\) is the angle between each wave front and the FLP when the wavelength of the incident light is \(\lambda_{1}\). Since \(\beta_{1}\) is small, substituting Equation (3) into Equation (6), we can approximately get  
\[N=2d(\sigma_{0})\sigma_{1}\Bigl{(}\sin^{-1}(0.5n(\sigma_{0}))-\sin^{-1}(0.5n( \sigma_{1}))\Bigr{)} \tag{7}\]  
Assuming that the number of fringes increases by one when the wavenumber of the incident light changes to \(\sigma_{2}\), the same way we obtain  
\[N+0.5=2d(\sigma_{0})\sigma_{2}\Bigl{(}\sin^{-1}(0.5n(\sigma_{0}))-\sin^{-1}(0. 5n(\sigma_{2}))\Bigr{)} \tag{8}\]  
The difference between these two wavenumbers is the maximum spectral resolution. When defining the function,  
\[f(\sigma)=\sigma\sin^{-1}(0.5n(\sigma)) \tag{9}\]  
Combining Equation (7) and Equation (8), we can get  
\[0.25=d(\sigma_{0})\Bigl{(}(\sigma_{2}-\sigma_{1})\sin^{-1}(0.5n(\sigma_{0}))- (f(\sigma_{2})-f(\sigma_{1}))\Bigr{)} \tag{10}\]  
Since the difference between \(\sigma_{1}\), \(\sigma_{2}\) and \(\sigma_{0}\) is small at a given tuning, the following approximation can be made:  
\[f(\sigma_{2})-f(\sigma_{1})=f^{\prime}(\sigma_{0})(\sigma_{2}-\sigma_{1}) \tag{11}\]where \(f^{\prime}(\sigma)\) denotes the derivative function of \(f(\sigma)\). Substituting Equation (11) into Equation (10), we can get the maximum spectral resolution:  
\[\sigma_{2}-\sigma_{1}=\frac{1}{4d(\sigma_{0})\Big{(}\sin^{-1}(0.5n(\sigma_{0})) -f^{\prime}(\sigma_{0})\Big{)}} \tag{12}\]  
The theoretical spectral resolving power of the system at wavenumber \(\sigma\) can be expressed by the following equation:  
\[R=\frac{\sigma}{\delta\sigma}=4\sigma d(\sigma)\Big{(}\sin^{-1}(0.5n(\sigma)) -f^{\prime}(\sigma)\Big{)} \tag{13}\]  
The variation of the theoretical spectral resolving power of the spectrometer is shown in Figure 4. The theoretical spectral resolving power in the spectral range (10,000 cm\({}^{-1}\)-25,641 cm\({}^{-1}\)) is approximately 1381-27,595; that is, the highest spectral resolution is approximately 0.014 nm/390 nm (0.93 cm\({}^{-1}\)/25,641 cm\({}^{-1}\)). In some applications, the required spectral resolving power differs in different spectral ranges, allowing the selection of the appropriate material and apex angle for the prisms accordingly. Next, consider the tuning angles of the system. According to the Nyquist-Shannon theorem, spectral information can be recovered if a detector uniformly samples an interferogram at a frequency that exceeds the fringe frequency by at least a factor of two. The spectral range of an arbitrary tuning is limited by the maximum fringe frequency of the interferogram, which is determined by the pixel pitch of the detector.