## 2 Formulation of the Problem
Using the boundary conditions for tangential electromagnetic fields at \(z=-L_{m}\) for \(m=1,2,3,4\), and \(5\), we obtain  
\[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]0 \[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]1 (20) \[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]2 \[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]3 (21) \[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]4 \[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]5 (22) \[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]6 \[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]7 (23) \[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]8  
where \(\Gamma_{m}=\left(\alpha^{2}-k_{rm}^{2}\right)^{1/2}\) with \({\rm Re}\Gamma_{m}>0\) for \(m=1,2,3,4,5\), and  
\[\Psi_{(+)}(x,\alpha)=\Phi_{+}(x,a)-A\,\frac{e^{-ikx\sin\theta_{0}}}{\alpha-k \cos\theta_{0}},\ A=\frac{1}{(2\pi)^{1/2}i}, \tag{10}\]9  
\[\Phi_{1}(x,\alpha)=\sum_{m=1}^{4}\Phi_{m}^{\prime}(x,\alpha)+\Phi_{1}^{0}(x, \alpha)+\Phi_{-}(x,\alpha). \tag{11}\]0  
The transformed wave equations for \(|x|<b\) are represented by (16) and (19)-(23). Considering the radiation condition, the solution to (14) can be written as:  
\[\Phi(x,\alpha)=\left\{ \begin{array}{l}-\gamma^{-1}\Psi^{\prime}_{(+)}(b, \alpha)e^{-\gamma(x-b)}\ {\rm for}\ x>b,\\ \gamma^{-1}\Psi^{\prime}_{(+)}(-b,\alpha)e^{\gamma(x+b)}\ {\rm for}\ x<-b, \end{array} \right. \tag{26}\]  
with the prime denoting the \(x-\) derivative, the expression has been derived by utilizing (9) and applying the boundary conditions for tangential electromagnetic fields across \(x=\pm b\). Equation (26) denotes the representation of the scattered field for \(|x|>b\). Due to the medium discontinuities in \(|x|<b\), the transformed wave equations contain unknown inhomogeneous terms \(f_{m}(x)\) and \(g_{m}(x)\) for \(m=1,2,3,4,5\). For convenience, we expand these functions into the Fourier cosine series as  
\[\left. \begin{array}{l}f_{m}(x)\\ g_{m}(x)\end{array} \right\}=\frac{1}{b}\sum\limits_{n=1}^{n}\
u_{n}\left\{ \begin{array}{l}f_{mm}\\ g_{mm}\end{array} \right\}\cos\frac{n\pi}{2b}(x+b),\ \left\{ \begin{array}{l}v_{n}=1,\ n \geq 1\\ v_{0}=1/2\end{array} \right. \tag{27}\]  
for \(|x|<b\). Then, by following a procedure similar to that in our previous paper [46], we arrive at the solutions of (16) and (19)-(23), resulting in  
\[\Phi_{-}(x,\alpha)=\frac{e^{-i\alpha L_{1}}}{b}\sum\limits_{n=1}^{\infty}\
u_{ n}\frac{C_{1n}^{-}(\alpha)}{\alpha^{2}+\Gamma_{1n}^{2}}\cos\frac{n\pi}{2b}(x+b), \tag{28}\]  
\[ \begin{array}{l}\Phi_{1}^{0}(x,\alpha)+\Psi_{(+)}(x,\alpha)=\Psi^{\prime}_{(+ )}(b,\alpha)\frac{\cosh\gamma(x+b)}{\gamma\sinh 2\gamma b}-\Psi^{\prime}_{(+)}(-b,\alpha)\frac{\cosh \gamma(x-b)}{\gamma\sinh 2\gamma b}\\ -\frac{1}{b}\sum\limits_{n=0}^{\infty}n\frac{C_{5n}(\alpha)}{\alpha^{2}+\Gamma _{2n}^{2}}\cos\frac{n\pi}{2b}(x+b)\end{array}