= Solution
A transmitted P-wave <seismic head wave> occurs when its refraction angle reaches $90^\circ$, requiring $\sin i_c=V_{P,\rm incident}/V_{P,\rm transmitted}\leq1$. At the limestone–sandstone boundary the ratio is $2500/2000>1$: \b[there is no P-wave <seismic critical angle> for incidence from limestone]. A slower lower layer cannot carry that transmitted head wave.
At the sandstone–salt boundary,
$$
\sin i_c=\frac{2000}{4000}=\frac12,\qquad \boxed{i_c=30.00^\circ}.
$$
The head wave travels along the faster salt boundary and radiates energy back into the slower overlying medium. Reversing the direction of incidence would change which interfaces admit such a head wave.
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