Question

Two-layer Earth: \(\rho_1=2000\ \text{kg m}^{-3},\ V_1=1800\ \text{m s}^{-1}\) (upper), \(\rho_2=3000\ \text{kg m}^{-3},\ V_2=2100\ \text{m s}^{-1}\) (lower). Compute the normal-incidence P-wave reflection coefficient at the interface. \([\,\text{round off to 3 decimals}\,]\)

Hint:

Normal incidence uses only impedances. For oblique incidence you'd need Zoeppritz (or Aki–Richards) equations with angles and Poisson's ratios.

Answer 1

Step 1: Acoustic impedance concept.
For a P-wave striking normally, the boundary behavior depends only on acoustic impedances \(Z_i=\rho_iV_i\) (units \(\text{kg m}^{-2}\text{s}^{-1}\)). Larger contrast \(\Rightarrow\) stronger reflection.

Step 2: Compute impedances with units.
\[ Z_1=2000\cdot 1800=3.6\times 10^6\ \text{kg m}^{-2}\text{s}^{-1}, \quad Z_2=3000\cdot 2100=6.3\times 10^6\ \text{kg m}^{-2}\text{s}^{-1}. \]

Step 3: Reflection coefficient at normal incidence.
\[ R=\frac{Z_2-Z_1}{Z_2+Z_1} =\frac{6.3-3.6}{6.3+3.6} =\frac{2.7}{9.9}=0.272727\ldots \Rightarrow \boxed{0.273}. \]

Step 4: Polarity and energy (optional insight).
\(R>0\) because the wave goes from lower to higher impedance; the reflected polarity is non-inverted at normal incidence. Fraction of energy reflected \(E_R=R^2\approx 0.074\) (about \(7.4%\)), remainder transmitted.

Final Answer:\ \(\boxed{0.273}\)