= Solution
Decomposing a plane <displacement> into components parallel and perpendicular to its <wavevector> gives compressional and shear speeds
$$
\boxed{c_p=\sqrt{(\lambda+2\mu)/\rho},\qquad c_s=\sqrt{\mu/\rho},\qquad
k_p=\omega/c_p,\quad k_s=\omega/c_s.}
$$
The incident <polarization> is longitudinal. Its reflected <longitudinal wave> reverses the vertical <wavevector> component while retaining its horizontal component. The converted <shear wave> has <wavevector> $k_s(\sin\beta,-\cos\beta)$ perpendicular to <polarization> $(\cos\beta,\sin\beta)$.
Every term must have the same <phase> along the rigid boundary. Thus
$$
\boxed{k_s\sin\beta=k_p\cos\alpha,\qquad
\sin\beta=\frac{c_s}{c_p}\cos\alpha.}
$$
Choose $\cos\beta>0$ so the reflected <shear wave> propagates downward into $y<0$. At $y=0$, both <displacement> components must vanish for every boundary <phase>:
$$
(1+R_1)\cos\alpha+R_2\cos\beta=0,\qquad
(1-R_1)\sin\alpha+R_2\sin\beta=0.
$$
Solving this two-by-two system yields
$$
\boxed{R_1=\frac{\sin(\alpha-\beta)}{\sin(\alpha+\beta)},\qquad
R_2=-\frac{\sin2\alpha}{\sin(\alpha+\beta)}.}
$$
For a stable isotropic material $0<c_s/c_p<1$. An incident wave directed toward the boundary has $0<\alpha<\pi$; choosing $\beta\in(-\pi/2,\pi/2)$ by the sine relation gives real reflected waves throughout this range and no zero denominator. In the usual convention restricting incident horizontal propagation to positive $x$, this is $0<\alpha\leq\pi/2$. Grazing endpoints have no incoming normal energy flux and are limiting cases. There is no <critical angle> for incident compressional-to-shear conversion here, because the converted wave is slower; the opposite incident-wave type has different critical-angle behaviour. At <normal incidence> $\alpha=\pi/2$, $\beta=0$, $R_1=1$ and $R_2=0$, cancelling the incident <displacement> with the reversed longitudinal <polarization>. This is <compressional-wave reflection from a rigid elastic boundary>.
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