## Appendix B  
We focus on the scattering properties of a PIM slab with thickness of \(d\) in air, as shown in Figure 1. Consider monochromatic acoustic waves with frequency \(\omega\) and incident angle \(\theta\). As discussed in the previous section, the dispersion relations of acoustic laser mode and CPA mode can be expressed as \(\left(\frac{1-\eta}{1+\eta}\right)^{2}=e^{j2k_{x}^{\prime}d}\) and \(\left(\frac{1-\eta}{1+\eta}\right)^{2}=e^{-j2k_{x}^{\prime}d}\), respectively.  
In general, these two modes appear in different situations. However, when \(\left(\frac{1-\eta}{1+\eta}\right)^{2}=e^{i2k_{x}{{}^{\prime}}d}=e^{-j2k_{x}{{}^{ \prime}}d}\), i.e., when \(\frac{e^{jk_{x}{{}^{\prime}}d}}{e^{-jk_{x}{{}^{\prime}}d}}=\pm 1\), the CPA mode and the laser mode can be satisfied simultaneously.  
If \(\frac{e^{jk_{x}{{}^{\prime}}d}}{e^{-jk_{x}{{}^{\prime}}d}}=1\) holds, we can get \(\eta=0\). This indicates that the wave vector in the x-direction is zero, i.e., the wave cannot propagate forward in the air, so this solution can be ignored. And if \(\frac{e^{jk_{x}{{}^{\prime}}d}}{e^{-jk_{x}{{}^{\prime}}d}}=-1\) holds, we can get \(\eta=-i\) and \(e^{ik_{x}^{\prime}d}=\pm i\).  
Case 1: \(e^{ik_{x}^{\prime}d}=i\)  
For \(e^{ik_{x}^{\prime}d}=i\), the result \(k_{x}{{}^{\prime}}d=(2m+\frac{1}{2})\pi\) can be derived, where, m is an arbitrary integer related to the mode order. And since \(\eta=\frac{\rho k_{x}}{\rho k_{x}{{}^{\prime}}}=\frac{-i\rho\rho k_{x}}{\rho k _{x}{{}^{\prime}}}=\frac{-i\rho k_{x}}{k_{x}{{}^{\prime}}}=-i\), we can get \(\frac{ek_{x}}{k_{x}{{}^{\prime}}}=1\). Thus, the problem is transformed into solving the following set of equations,  
\[\left\{\begin{array}{l}\frac{ek_{x}}{k_{x}{{}^{\prime}}}=1\\ k_{x}{{}^{\prime}}d=(2m+\frac{1}{2})\pi\end{array}\right., \tag{12}\]  
where \(k_{y}=k_{0}\sin\theta\) is the wave vector along the \(y\) directions, and \(k_{x}=\sqrt{{k_{0}}^{2}-{k_{y}}^{2}}\), \({k_{x}}^{\prime}=\sqrt{{n^{\prime}}^{2}{k_{0}}^{2}-{k_{y}}^{2}}\) are the \(x\) directional wave vector in the air and PIM media, respectively. \(n=\pm\sqrt{\rho\beta_{s}/\rho_{0}\beta_{0}}=\pm\sqrt{\rho_{r}\beta_{r}}\) is the refractive index. The solution of Equation (12) is as follows:  
\[\left\{\begin{array}{l}\rho_{r}=\sqrt{\frac{\frac{\pi^{2}}{k_{0}^{2}d^{2}} (2m+\frac{1}{2})^{2}}{(1-\sin^{2}\theta)}}=\frac{\pi}{k_{0}d}\times(2m+\frac{1 }{2})/\cos\theta\\ \beta_{r}=\left[\frac{\pi^{2}}{k_{0}^{2}d^{2}}(2m+\frac{1}{2})^{2}+\sin^{2} \theta\right]/[\frac{\pi}{k_{0}d}\times(2m+\frac{1}{2})/\cos\theta]\end{array} \right.. \tag{13}\]  
Case 2: \(e^{ik_{x}{{}^{\prime}}d}=-i\)  
Similarly, for \(e^{ik_{x}{{}^{\prime}}d}=-i\), the result \(k_{x}{{}^{\prime}}d=(2m-\frac{1}{2})\pi\) can be derived. Combined with \(\eta=\frac{\rho k_{x}}{\rho k_{x}{{}^{\prime}}}=\frac{-i\rho_{r}\rho k_{x}}{ \rho\psi k_{x}{{}^{\prime}}}=\frac{-i\rho_{r}k_{x}}{k_{x}{{}^{\prime}}}=-i\), i.e., \(\frac{ek_{x}}{k_{x}{{}^{\prime}}}=1\). The problem is transformed into solving the following set of equations,  
\[\left\{\begin{array}{l}\frac{\rho k_{x}}{k_{x}{{}^{\prime}}}=1\\ k_{x}{{}^{\prime}}d=(2m-\frac{1}{2})\pi\end{array}\right. \tag{14}\]  
The parameters are set in the same way as in case 1. The solution of Equation (14) is  
\[\left\{\begin{array}{l}\rho_{r}=\sqrt{\frac{\pi^{2}}{k_{0}^{2}d^{2}}(2m+ \frac{1}{2})^{2}}{(1-\sin^{2}\theta)}=\frac{\pi}{k_{0}d}\times(2m-\frac{1}{2} )/\cos\theta\\ \beta_{r}=\left[\frac{\pi^{2}}{k_{0}^{2}d^{2}}(2m-\frac{1}{2})^{2}+\sin^{2} \theta\right]/[\frac{\pi}{k_{0}d}\times(2m-\frac{1}{2})/\cos\theta]\end{array} \right.. \tag{15}\]