In a recent paper [1] , there was a typo in (13), which should read as a scalar equation for the $ {z}$ -component of $\tilde {\mathbf {E}}\{\tilde {\mathbf {H}}\}$ in the form \begin{equation*} \tilde {E}_{ {z}} \{\tilde {H}_{ {z}}\} = F(\tilde {\xi }^{i}) \, \tilde {E}_{ {z}}^{i} \{\tilde {H}_{ {z}}^{i}\} + F(\tilde {\xi }^{r}) \tilde {E}_{ {z}}^{r} \{\tilde {H}_{ {z}}^{r}\} \end{equation*} from which the remaining field components may be obtained. The full vector equation for the electric field is then given by \begin{align*} \tilde {\mathbf {E}}=&F(\tilde {\xi }^{i}) \tilde {\mathbf {E}}^{i} + F(\tilde {\xi }^{r}) \tilde {\mathbf {E}}^{r} + \frac {e^{i \pi /4}}{\sqrt { 2 \pi \tilde {k} \tilde {\rho } \sin \tilde {\theta }_{0} }} e^{i \tilde {k} (\tilde {\rho } \sin \tilde {\theta }_{0} -\tilde { {z}} \cos \tilde {\theta }_{0})} \\[3pt]&\times \bigg \{ \cos \frac {\tilde {\phi }-\tilde {\phi }_{0}}{2} \left [{ \left ({ \tilde {\mathbf {E}}^{i}_{0} \cdot \hat {\tilde { {z}}} }\right) \hat {\tilde { {z}}} - \tilde {\mathbf {E}}^{i}_{0} }\right ] +\sin \frac {\tilde {\phi }-\tilde {\phi }_{0}}{2} \, \hat {\tilde { {z}}} \times \tilde {\mathbf {E}}^{i}_{0} \\[4pt]&+\,\cos \frac {\tilde {\phi }+\tilde {\phi }_{0}}{2} \left [{ \left ({ \tilde {\mathbf {E}}^{r}_{0} \cdot \hat {\tilde { {z}}} }\right) \hat {\tilde { {z}}} - \tilde {\mathbf {E}}^{r}_{0} }\right ] +\sin \frac {\tilde {\phi }+\tilde {\phi }_{0}}{2} \, \hat {\tilde { {z}}} \times \tilde {\mathbf {E}}^{r}_{0} \bigg \} \end{align*} where $\tilde {\mathbf {E}}^{i,r}_{0}$ denotes the incident (reflected) electric field amplitude at the edge of the half-plane in frame $\tilde {S}$ . In addition, $Z \tilde {B}_{h}$ in (24) should read as $\tilde {B}_{h}$ , yielding \begin{align*} \tilde {\bar {\bar {F}}}_{re}=&\dfrac {\tilde {Q}}{c} \\&\times \! \begin{bmatrix}\! 0 \!&\quad \! -\!\tilde {B}_{e} \sin \dfrac {\tilde {\phi }}{2} \!&\quad \! \tilde {B}_{e} \cos \dfrac {\tilde {\phi }}{2} \!&\quad \! 0 \tilde {B}_{e} \sin \dfrac {\tilde {\phi }}{2} \!&\quad \! 0 \!&\quad \! 0 \!&\quad \! -\tilde {B}_{h} \sin \dfrac {\tilde {\phi }}{2} -\tilde {B}_{e} \cos \dfrac {\tilde {\phi }}{2} \!&\quad \! 0 \!&\quad \! 0 \!&\quad \! \tilde {B}_{h} \cos \dfrac {\tilde {\phi }}{2} 0 \!&\quad \! \tilde {B}_{h} \sin \dfrac {\tilde {\phi }}{2} \!&\quad \! -\tilde {B}_{h} \cos \dfrac {\tilde {\phi }}{2} \!&\quad \! 0 \end{bmatrix}\!. \end{align*} None of the above-mentioned remarks affect the calculations and results presented in [1] .