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Plane dipping reflector

The hyperbolic reflector in equation A-1 becomes a plane dipping reflector when $z_0=0$. In this case, equation A-5 simplifies to (Klokov and Fomel, 2012)

t_m(x_m) = \frac{2}{v} \frac{(x_m-x_0) \cos{\alpha} \sin{\beta}}{1-\gamma \sin{\alpha} \sin{\beta}}\;.
\end{displaymath} (21)

The dip of the image at a correct velocity is
\tan{\alpha_m} = \frac{v}{2} t_m'(x_m) = \frac{\cos{\alpha} \sin{\beta}}{1-\sin{\alpha} \sin{\beta}}\;,
\end{displaymath} (22)

which is equivalent to equation 8 in the main text. The dip of the reflector in this case is simply $\alpha_0=\beta$. It is easy to verify that when $\alpha=\beta$, $\alpha_m=\alpha$.