A geologist uses ultrasound pulses to investigate the depth of a rock layer buried beneath the ground. The ultrasound is emitted from the surface, travels downward, and some of the waves are reflected back when they reach the boundary between two different rock types. Explain how the geologist can use the reflected ultrasound waves to determine the depth of the rock layer boundary, and explain why ultrasound is better suited to this task than visible light.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Ultrasound waves have a frequency above 20 kHz, which is higher than the upper limit of human hearing. When ultrasound reaches a boundary between two media, it is partially reflected back. The speed of sound through rock is approximately 3000 m/s.
Model answer (5 marks)
A geologist emits an ultrasound pulse from the surface. When the pulse reaches the boundary between two rock types, part of the wave is reflected back to the surface.
1. A receiver records the reflected pulse and the time interval between emission and detection.
2. The pulse travels to the boundary and back, so the total distance travelled is twice the depth.
3. Using the known speed of sound in rock (≈3000 m s⁻¹), the depth is calculated: depth = (speed × time)/2.
4. Visible light cannot penetrate opaque rock, so it cannot reach the boundary.
5. Ultrasound is non‑invasive, can travel through solid rock, and allows imaging of subsurface layers without excavation.
1. A receiver records the reflected pulse and the time interval between emission and detection.
2. The pulse travels to the boundary and back, so the total distance travelled is twice the depth.
3. Using the known speed of sound in rock (≈3000 m s⁻¹), the depth is calculated: depth = (speed × time)/2.
4. Visible light cannot penetrate opaque rock, so it cannot reach the boundary.
5. Ultrasound is non‑invasive, can travel through solid rock, and allows imaging of subsurface layers without excavation.
Examiner tips
- Use the term ‘reflected pulse’ and ‘time interval’ to show understanding of the process. Show the calculation: depth = (v × t)/2, noting the factor ½ for the round‑trip. Explain why light fails: opacity of rock. Mention non‑invasive nature to earn the extra point.
Common mistakes
- Forgetting to divide the time by 2 for the round‑trip. Using the speed of light instead of the speed of sound. Saying light can penetrate rock, ignoring its opacity.
Mark scheme (5 marks)
- When ultrasound reaches the boundary between the two rock types, it is partially reflected back to the surface
- A receiver at the surface detects the reflected ultrasound and the time between emission and detection is measured
- Because the speed of the ultrasound through the rock is known (constant), the depth can be calculated using distance = speed × time, with the time halved because the wave travels to the boundary and back
- Visible light cannot penetrate through opaque rock, whereas ultrasound can travel through solid media
- Ultrasound is a non-invasive method that can image beneath surfaces without damaging the rock or requiring excavation
Key terms in this question
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