## 2 Fourier-Transform Method  
The Fourier-transform method is commonly used to obtain a phase-shift image from a single interferogram including background fringes. In this section, taking the phase extraction from the interferogram shown in Figure 1a as an example, the procedure of the Fourier-transform method is shown and the advantages and disadvantages are shown. In the interferometer, a single optical wave is split into two waves that pass along different optical paths; one is called a reference wave which does not pass the object and the other called an object wave passing through the object with an additional phase-shift, \(\phi\). When their wave vectors are different but their amplitudes are provided by the same function, electric fields of both waves are expressed with the complex unit j as  
\[\mathbf{E}^{\intercal}(\mathbf{r}) =E_{0}(\mathbf{r})\exp\left[-\mathrm{j}\,\mathbf{k}^{\intercal}\cdot\mathbf{r} \right], \tag{1}\] \[\mathbf{E}^{\mathrm{O}}(\mathbf{r}) =E_{0}(\mathbf{r})\exp\left[-\mathrm{j}\left((\mathbf{k}^{\intercal}+ \delta\mathbf{k})\cdot\mathbf{r}+\phi(\mathbf{r})\right)\right], \tag{2}\]  
where superscripts 'r' and 'o' denotes the quantities related to the reference wave and those to the object wave, respectively, and the difference of the wave vector is denoted by \(\delta\mathbf{k}\). These waves are superposed on a screen and the interferogram, \(i(\mathbf{r})\), is observed as a power of the superposed electric field:  
\[ \begin{split} i(\mathbf{r})&=|\mathbf{E}^{\intercal}(\mathbf{r} )+\mathbf{E}^{\mathrm{O}}(\mathbf{r})|^{2}=2|E_{0}(\mathbf{r})|^{2}\left(1+\cos(\phi(\mathbf{r} )+\delta\mathbf{k}\cdot\mathbf{r})\right)\\ &=2i_{0}(\mathbf{r})+i_{\phi}(\mathbf{r})\exp\left(+\mathrm{j}\,\delta\bm {k}\cdot\mathbf{r}\right)+i_{\phi}^{*}(\mathbf{r})\exp\left(-\mathrm{j}\,\delta\mathbf{k} \cdot\mathbf{r}\right)\\ i_{0}(\mathbf{r})&=|\mathbf{E}_{0}(\mathbf{r})|^{2},\quad i_{ \phi}(\mathbf{r})=i_{0}(\mathbf{r})\exp\left(\mathrm{j}\,\phi(\mathbf{r})\right),\end{split}  \tag{3}\]  
where the symbol \"\(\mathbf{r}\)\" denotes a complex conjugate. To obtain the phase-shift \(\phi(\mathbf{r})\) from this equation, the first and the last term on the right-hand side in Equation (3) can be removed by using a 2-D Fourier transform. Applying the Fourier transform to Equation (3) and using a convolutional theorem, we obtain the following equation. \[I(\mathbf{k}) =\mathcal{F}\left\{i(\mathbf{r})\right\}=2I_{0}(\mathbf{k})+I_{\phi}(\mathbf{ k}-\delta\mathbf{k})+I_{\phi}^{*}(-\mathbf{k}-\delta\mathbf{k}), \tag{4}\] \[I_{0}(\mathbf{k}) =\mathcal{F}\left\{i_{0}(\mathbf{r})\right\},\quad I_{\phi}(\mathbf{k}) =\mathcal{F}\left\{i_{\phi}(\mathbf{r})\right\},\]  
where \(\mathcal{F}\left\{\right\}\) represents a 2-D Fourier transform operator. Figure 1b shows the Fourier transform of \(i(\mathbf{r})\) shown in Figure 1a. We can find three peaks in the Fourier transformed signal, \(I(\mathbf{k})\). These peaks correspond to three terms in the right-hand side of Equation (4); the peak appearing at the center in the frequency space, \(\mathbf{k}\), is a DC component the distribution of which is related to the profile of the waves \(i_{0}(\mathbf{r})\), and the other two peaks spreading around \(\mathbf{k}=\pm\delta\mathbf{k}\) include the information of \(\phi(\mathbf{r})\) related to the object. The two peaks appear point-symmetrically with respect to the origin of the frequency space. The distance between one of the two peaks and the origin represents the phase of the background fringes in the real space. This value, \(\delta\mathbf{k}\), will be needed in subsequent procedures, and it is determined as the maximum point around peak in \(\mathbf{k}\)-space. Next, as shown in Figure 1c, the DC component and one of the two peaks (this is referred to as a conjugate light component) is removed by filtering. To remove the DC component, we employed a method using a carrier peak isolation method [14], and to remove the conjugate component a half-plane filter [13] was employed.