Question

A homogeneous rock layer Q of density \(2600 \, \text{kg/m}^3\) is lying below homogeneous rock layer P of density \(2400 \, \text{kg/m}^3\).
A compressional wave travels from P to Q. On reaching the interface, this wave is incident normally and gets reflected and refracted.
The velocity of the compressional wave is \(2.7 \, \text{km/s}\) in P and \(3.5 \, \text{km/s}\) in Q.
The ratio of reflection coefficient to transmission coefficient at the interface is:

Hint:

For wave reflection/transmission at boundaries, always calculate acoustic impedances first, then apply the formulae for \(R\) and \(T\).

Answer 1

Step 1: Acoustic impedance.
Acoustic impedance of a medium is: \[ Z = \rho \cdot v \] where \(\rho\) = density, \(v\) = velocity of compressional wave. For P: \[ Z_P = 2400 \times 2700 = 6.48 \times 10^6 \] For Q: \[ Z_Q = 2600 \times 3500 = 9.10 \times 10^6 \] Step 2: Reflection coefficient.
For normal incidence, \[ R = \frac{Z_Q - Z_P}{Z_Q + Z_P} \] \[ R = \frac{9.10 \times 10^6 - 6.48 \times 10^6}{9.10 \times 10^6 + 6.48 \times 10^6} = \frac{2.62}{15.58} \approx 0.168 \] Step 3: Transmission coefficient.
\[ T = \frac{2Z_Q}{Z_Q + Z_P} \] \[ T = \frac{2 \times 9.10}{15.58} = \frac{18.2}{15.58} \approx 1.168 \] Step 4: Ratio.
\[ \frac{R}{T} = \frac{0.168}{1.168} \approx 0.19 \] Final Answer:
\[ \boxed{0.19} \]