## 2 Materials and Methods
The terms \(U_{x1}\otimes U_{x2}^{*}\) and \(U_{x1}^{*}\otimes U_{x2}\) contain the information of in-plane displacement in the x-axis, and they are located at \(\left(\frac{2\xi}{\lambda d},0\right)\) and \(\left(-\frac{2\xi}{\lambda d},0\right)\), respectively. The terms \(U_{y1}\otimes U_{y2}^{*}\) and \(U_{y1}^{*}\otimes U_{y2}\) contain the information of in-plane displacement in the y-axis, and they are located at \(\left(0,\frac{2\eta}{\lambda d}\right)\), and \(\left(0,-\frac{2\eta}{\lambda d}\right)\) respectively. Figure 3: Schematic diagram of the relative positions of the four apertures in the exit pupil plane. Figure 2: Polarizing directions of the four imaging beams passing through polarizers. By properly selecting aperture diameter and distance, the five spectrums on the Fourier domain can be separated. By selecting the term. \(U_{x1}\otimes U_{x2}^{*}\), located at \(\left(\frac{2\varepsilon}{\lambda d},0\right)\) and performing the inverse Fourier transformation, \(u_{x1}u_{x2}^{*}\) can be obtained. In the same way, \(u_{y1}u_{y2}^{*}\) can also be obtained. The phase term can be calculated by the following relation:  
\[ \begin{split}\psi_{x}+\frac{4\pi}{\lambda d}\xi x=\arctan\frac{ \text{Im}\left(u_{x1}u_{x2}^{*}\right)}{\text{Re}\left(u_{x1}u_{x2}^{*}\right) }\\ \psi_{y}+\frac{4\pi}{\lambda d}\eta y=\arctan\frac{\text{Im}\left(u _{y1}u_{y2}^{*}\right)}{\text{Re}\left(u_{y1}u_{y2}^{*}\right)}\end{split} , \tag{4}\]  
where Im and Re denote the imaginary and real parts, respectively. \(\psi_{x}=\phi_{x1}(x,y)-\phi_{y2}(x,y)\), and \(\psi_{y}=\phi_{y1}(x,y)-\phi_{y2}(x,y)\) are the phase differences of the two pairs of mutually interfering beams. After displacement, the phase difference can be calculated using Equation (5):  
\[ \begin{split}\psi_{x}^{\prime}+\frac{4\pi}{\lambda d}\xi x= \arctan\frac{\text{Im}\left(u_{x1}u_{x2}^{*}\right)}{\text{Re}\left(u_{x1}u_{ x2}^{*}\right)}\\ \psi_{y}^{\prime}+\frac{4\pi}{\lambda d}\eta y=\arctan\frac{\text {Im}\left(u_{y1}u_{y2}^{*}\right)}{\text{Re}\left(u_{y1}u_{y2}^{*}\right)} \end{split} , \tag{5}\]  
Through subtraction, the displacement can be evaluated from the relative phase difference. The relationship between the phase change and the displacement is given by [21],  
\[ \begin{split}\Delta_{x}&=\psi_{x}-\psi_{x}^{\prime} =\overset{\rightarrow}{S_{x}\cdot\overset{\rightarrow}{d}}\\ \Delta_{y}&=\psi_{y}-\psi_{y}^{\prime}=\overset{ \rightarrow}{S_{y}\cdot\overset{\rightarrow}{d}}\end{split} , \tag{6}\]  
where \(\overset{\rightarrow}{d}=u\overset{\rightarrow}{i}+v\overset{\rightarrow}{j }+w\overset{\rightarrow}{k}\) represents the displacement vector of the surface. u and v are the in-plane displacements in the x and y direction, respectively; and w is the out-of-plane displacement. \(\overset{\rightarrow}{i}\overset{\rightarrow}{j}\), and \(\overset{\rightarrow}{k}\) are the unit vectors along the positive x-, y-, and z-axis, respectively. \(\overset{\rightarrow}{S_{x}}\) and \(\overset{\rightarrow}{S_{y}}\) are the sensitivity vectors, defined as functions of the illumination and observation directions. For the DSPI proposed in this study, the sensitivity vectors can be expressed by Equation (7):  
\[\begin{split}\overset{\rightarrow}{S_{x}}&=\left( \overset{\rightarrow}{K_{x}^{\uparrow}}-\overset{\rightarrow}{K_{i}} \right)-\left(\overset{\rightarrow}{K_{x}^{\uparrow}}-\overset{ \rightarrow}{K_{i}^{\downarrow}}\right)=\overset{\rightarrow}{K_{x}^{ \uparrow}}-\overset{\rightarrow}{K_{x}^{\uparrow}}=\overset{\rightarrow}{ \pi\frac{\sin\theta}{\lambda}}\overset{\rightarrow}{i}\\ \overset{\rightarrow}{S_{From Equations (6) and (7), in-plane displacement in two directions can be calculated by Equation (8):  
\[u=\frac{\Delta_{x}\lambda}{4\pi\sin\theta},v=\frac{\Delta_{y}\lambda}{4\pi\sin \theta}. \tag{8}\]