Questions: Seismic Refraction Surveys and Interpretation

Question 1

A seismic refraction survey produces a travel-time curve with two straight-line segments. The first segment has slope 1/500 s/m, and the second has slope 1/2000 s/m. What do these two segments represent?

• A. The first segment is the reflected wave off the top of the lower layer; the second is the refracted wave returning through the upper layer
• B. The first segment is the direct wave traveling through the upper layer at V₁ = 500 m/s; the second is the head wave (refracted wave) that traveled along the faster lower-layer boundary at V₂ = 2000 m/s
• C. Both segments are direct waves — the first through soil, the second through bedrock — and no refraction is occurring
• D. The first segment is the P-wave arrival; the second is the S-wave arrival at a slower apparent velocity

Answer: (B)

In a refraction travel-time curve, the slope of each segment equals 1/V for that wave type. The first arrivals at short source-receiver offsets are direct waves through the upper layer — slope = 1/V₁ = 1/500, so V₁ = 500 m/s. At greater offsets, refracted head waves that traveled along the faster layer overtake the direct wave — their shallower slope 1/V₂ = 1/2000 gives V₂ = 2000 m/s. The velocity of each layer is read from slopes; layer depth is calculated from the intercept time where the refracted-wave line extrapolates back to zero offset.

Question 2

A geologist conducts a refraction survey searching for a buried soft-clay layer sandwiched between two harder, faster rock layers. Despite careful fieldwork, the survey reveals only two straight-line segments. Why might the clay layer be invisible?

• A. The clay layer is too thin to produce a detectable head wave at the geophone spacing used
• B. Refracted head waves only form when seismic velocity increases across a boundary; the clay has lower velocity than the overlying rock, so no head wave forms at its upper surface — it is a hidden low-velocity zone
• C. Soft materials like clay absorb seismic energy completely before it can return to the surface
• D. The clay layer would only be detected with S-waves, not P-waves

Answer: (B)

This is the fundamental limitation of seismic refraction: it requires velocity to increase with depth to generate head waves at each boundary. A head wave forms when the incident ray hits the boundary at the critical angle — which only exists when the lower layer is faster. If the clay (low velocity) is sandwiched between faster rocks, the lower boundary (clay-to-fast rock) does produce a head wave, but the upper boundary (fast rock-to-slow clay) does not. The clay layer goes undetected — it is a 'hidden layer' or low-velocity zone. Surveyors must consider this possibility when interpreting refraction data.

Question 3

In a seismic refraction survey, the velocity of each subsurface layer can be determined directly from the slope of the corresponding travel-time segment without knowing the layer depths.

• T. True
• F. False

Answer: True

Layer velocity is encoded in the slope of the travel-time segment: slope = 1/V, so V = 1/slope. This is independent of layer depth. The depth to each interface is then calculated separately using the intercept time — where the refracted-wave line, extrapolated back to zero source-receiver offset, crosses the time axis. This separation of velocity determination (from slope) and depth determination (from intercept) is what makes refraction surveys analytically tractable: you can extract velocities and depths sequentially using a layer-stripping approach.

Question 4

A seismic refraction survey can detect any subsurface boundary, regardless of whether velocity increases or decreases across it.

• T. True
• F. False

Answer: False

Refraction surveys require velocity to increase with depth across each boundary to generate detectable head waves. When a seismic wave hits a boundary at the critical angle, a head wave travels along the boundary in the faster lower medium and continuously radiates energy back upward into the slower upper layer. If the lower layer is slower than the upper layer, no critical angle exists — the wave is refracted downward away from the boundary rather than along it, and no head wave returns to the surface. Low-velocity zones sandwiched between faster layers are therefore invisible to refraction methods, which is the technique's most important limitation.

Question 5

Explain why seismic refraction surveys require that velocity increases with depth at each layer boundary, and describe what happens to seismic energy that encounters a boundary where the lower layer is slower.

Model answer: A refracted head wave forms only when a seismic ray hits the boundary at the critical angle (sin θ_c = V₁/V₂). This angle exists only when V₂ > V₁ — the lower layer must be faster. When the ray hits at this angle, it travels along the boundary at V₂, continuously shedding energy back upward at angle θ_c to reach the surface at progressively greater distances. If V₂ < V₁, Snell's law predicts no critical angle exists: the transmitted ray always bends away from vertical rather than along the boundary. Energy is transmitted into the lower layer (and partly reflected), but no wave propagates along the interface and returns to the surface. The boundary is therefore acoustically invisible to refraction methods. This is why a slow layer (e.g., clay, weathered rock) between two faster layers creates a 'hidden layer' that refraction surveys cannot resolve.

Understanding this limitation is essential for correct interpretation of refraction data. A two-segment travel-time curve that appears to show only two layers may actually be hiding intermediate slow layers. Surveys in areas with potential velocity inversions require complementary methods — seismic reflection, borehole measurements, or gravity surveys — to avoid missing critical subsurface features.