Seismic Wave Reflection Coefficient Calculation

This question asks us to find the reflection coefficient for a seismic wave hitting the boundary (interface) between two different geological layers. The wave is hitting the boundary straight on (normal incidence).

Understanding the Problem

We are given the properties of two layers:

• Layer 1: Density ($\rho_1$) = 3 g/cc, Velocity ($v_1$) = 5 km/s
• Layer 2: Density ($\rho_2$) = 2.5 g/cc, Velocity ($v_2$) = 4 km/s

The seismic wave travels from Layer 1 into Layer 2. We need to calculate the reflection coefficient ($R$), which tells us how much of the wave's amplitude is reflected back into Layer 1.

Formula for Reflection Coefficient

For a wave encountering an interface at normal incidence, the reflection coefficient ($R$) is determined by the acoustic impedances of the two layers. The formula is:

$$ R = \frac{Z_2 - Z_1}{Z_2 + Z_1} $$

Where:

• $Z_1$ is the acoustic impedance of the first layer.
• $Z_2$ is the acoustic impedance of the second layer.

Calculating Acoustic Impedance

Acoustic impedance ($Z$) is calculated by multiplying the material's density ($\rho$) by the wave velocity ($v$) in that material.

$$ Z = \rho \times v $$

Let's calculate the acoustic impedance for both layers. We need to ensure consistent units. Let's use CGS units (g/cc for density and cm/s for velocity).

• Convert $v_1$: $5 \text{ km/s} = 5 \times 10^3 \text{ m/s} = 5 \times 10^5 \text{ cm/s}$
• Convert $v_2$: $4 \text{ km/s} = 4 \times 10^3 \text{ m/s} = 4 \times 10^5 \text{ cm/s}$

Now, calculate $Z_1$ and $Z_2$:

• $Z_1 = \rho_1 \times v_1 = (3 \text{ g/cc}) \times (5 \times 10^5 \text{ cm/s}) = 15 \times 10^5 \text{ g/(cm}^2 \cdot \text{s)}$
• $Z_2 = \rho_2 \times v_2 = (2.5 \text{ g/cc}) \times (4 \times 10^5 \text{ cm/s}) = 10 \times 10^5 \text{ g/(cm}^2 \cdot \text{s)}$

Alternatively, we can note that the units in the reflection coefficient formula will cancel out. So, we can compute using the given units directly, just ensuring the multiplication is done correctly:

• $Z_1 = 3 \times 5 = 15$ (in arbitrary units, e.g., g·km/(cc·s))
• $Z_2 = 2.5 \times 4 = 10$ (in arbitrary units)

Performing the Calculation

Now, substitute the calculated acoustic impedances into the reflection coefficient formula:

$$ R = \frac{Z_2 - Z_1}{Z_2 + Z_1} $$

$$ R = \frac{10 - 15}{10 + 15} $$

$$ R = \frac{-5}{25} $$

$$ R = -\frac{1}{5} $$

$$ R = -0.2 $$

Conclusion

The reflection coefficient for the seismic wave at the interface between the two layers, under normal incidence, is -0.2. This negative value indicates a phase reversal upon reflection, meaning the reflected wave's polarity is inverted compared to the incident wave.