Question

A homogeneous rock layer \( Q \) of density 2600 kg/m³ is lying below homogeneous rock layer P of density 2400 kg/m³. A compressional wave travels from P to Q. On reaching the interface of P and Q, this wave is incident normally and gets reflected and refracted. The velocity of the compressional wave is 2.7 km/s in the rock layer P and 3.5 km/s in layer Q.
The ratio of reflection coefficient to the transmission coefficient at the interface is .......... (rounded off to two decimal places).

Hint:

Reflection and transmission coefficients depend on the acoustic impedance of the two layers. Make sure to use the correct formula and units when calculating these values.

Answer 1

The reflection and transmission coefficients can be calculated using the following formulas:
\[ R = \frac{Z_2 - Z_1}{Z_2 + Z_1}, \quad T = \frac{2 Z_2}{Z_2 + Z_1} \] where: - \( R \) is the reflection coefficient, - \( T \) is the transmission coefficient, - \( Z_1 \) and \( Z_2 \) are the acoustic impedances of layers P and Q, respectively.
The acoustic impedance is given by:
\[ Z = \rho \times V \] where: - \( \rho \) is the density of the material, and - \( V \) is the velocity of the compressional wave in the material.
For layer P:
\[ Z_1 = 2400 \times 2.7 = 6480 \, {kg/m}^2{s} \] For layer Q:
\[ Z_2 = 2600 \times 3.5 = 9100 \, {kg/m}^2{s} \] Now, calculate the reflection and transmission coefficients:
\[ R = \frac{9100 - 6480}{9100 + 6480} = 0.18, \quad T = \frac{2 \times 9100}{9100 + 6480} = 0.82 \] Thus, the ratio of reflection coefficient to transmission coefficient is:
\[ \frac{R}{T} = \frac{0.18}{0.82} = 0.22 \] Therefore, the ratio of reflection coefficient to transmission coefficient is between 0.18 and 0.23.