## 3 Analysis of the Influence of the Diffraction Effect
\begin{table}
\begin{tabular}{c c c c} \hline \hline  & **r** & **Peak** & **Percentage** \\ \hline
1 st Maximum & 0 & 1.0 & 83.9\% \\
1 st Zero & \(1.22f\lambda/(2a)\) & 0.0 & - \\
2 nd Maximum & \(1.635f\lambda/(2a)\) & 0.0175 & 7.1\% \\
2 nd Zero & \(2.233f\lambda/(2a)\) & 0.0 & - \\
3 rd Maximum & \(2.679f\lambda/(2a)\) & 0.0042 & 2.8\% \\
3 rd Zero & \(3.238f\lambda/(2a)\) & 0.0 & - \\
4 th Maximum & \(3.699f\lambda/(2a)\) & 0.0018 & 1.5\% \\
4 th Zero & \(4.24f\lambda/(2a)\) & 0.0 & - \\
5 th Maximum & \(4.72f\lambda/(2a)\) & 0.00078 & 1.0\% \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Intensity distribution of each ring of the diffraction pattern. For spectrometers, at least 80% of the Airy disk energy is typically required to fall within the size range of the sensor pixel in the optical design process. Taking 84% of the diffraction pattern energy falling within just one pixel as the benchmark, by covering 95-99% of the signal power of the spectrum as the criterion, it can be observed from Table 1 that up to the fourth dark ring of the diffraction pattern, the energy occupied represents 95.3% of the whole diffraction pattern energy. Limited by the size of the pixel, we can approximate that the energy outside the fourth dark ring has little effect on other pixels and can be regarded as negligible noise. Then, the effective coverage of light with any wavelength \(\lambda\) on the image plane is:  
\[r\leq\frac{4.24f\lambda}{2a} \tag{13}\]  
Obviously, the \(r\) value is larger than \(\frac{1}{2}\) of the pixel size. Without loss of generality, when we analyze with the \(i\)th pixel of the sensor, then the distance of the boundary of the pixel along the y-axis from the origin is:  
\[\left\{ \begin{array}{l}y_{\min}^{i}=(i-\frac{M}{2}+1)P\\ y_{\max}^{i}=y_{\min}^{i}+P\end{array} \right. \tag{14}\]  
According to the previous analysis, the spectral energy that the pixel can receive is the sum of all spectral energies in the wavelength range determined by Equation (13). As shown in Figure 4, the red and black curves represent two boundary wavelengths, and the corresponding fourth dark ring of the diffraction pattern is at the boundary position of pixel \(i\). At this point, any wavelength spectrum located in the two wavelength ranges will contribute to the energy of pixel \(i\). In Figure 4, let the wavelengths corresponding to the black and red curve be \(\lambda_{I}^{i}\) and \(\lambda_{r}^{i}\), respectively, and the positions of the centers of the two curves on the sensor image plane are denoted by \(y_{I}^{i}\) and \(y_{r}^{i}\), respectively. Then we can obtain:  
\[\left\{ \begin{array}{l}y_{I}^{i}=y_{\min}^{i}-\frac{4.24f\lambda_{I}^{i}}{2 }\\ y_{r}^{i}=y_{\max}^{i}+\frac{4.24f\lambda_{r}^{i}}{2a}\end{array} \right. \tag{15}\]  
Figure 3: Intensity distribution of Fraunhofer diffraction of circular aperture. Combining Equations (11), (14) and (15), we can obtain wavelengths \(\lambda_{l}^{i}\) and \(\lambda_{r}^{i}\), corresponding to the black and red curves, and all spectra in this wavelength range contribute to the energy of the \(i\)th pixel. The wavenumber range corresponding to this wavelength range is \([v_{r}^{i},v_{l}^{i}]\), where \(v_{r}^{i}=1/\lambda_{r}^{i}\), \(v_{l}^{i}=1/\lambda_{l}^{i}\). Obviously, this spectral range is larger than that expressed in Equation (6), which is mainly due to the optical diffraction effect. Next, we analyze the influence of the diffraction effect on the bandpass sampling interferogram and recovered spectrum. For light with wavelength \(\lambda\), the distribution of its diffraction pattern at the image plane is shown in Figure 5. Taking the center of the Airy disk as the center of the circle, given any \(r\), the diffraction pattern intensity value at \(r\) is \(S(r)\).