Use ultrasound echoes to locate internal boundaries
Explain how reflected ultrasound pulses provide diagnostic information and estimate the depth of a boundary.
Section 24 · Medical physics
A Level · Cambridge International Physics 9702, 2025–2027 syllabus. Topic 24 was checked against Cambridge’s 2028–2030 syllabus; the other remaining syllabus-year changes are still being reconciled. Online study supports the theory but cannot replace supervised practical work, full timed assessment or human subject-teacher review.
Before you begin
- Use exponential functions, wave speed, energy conservation and SI units; convert between joules and electronvolts when required.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why can an ultrasound pulse return from a boundary between tissues?
What you will learn
- Explain how tissue boundaries reflect ultrasound pulses
- Relate echo delay to boundary depth
- State the round-trip assumption in an echo-depth calculation
Echoes mark changes between tissues
A short ultrasound pulse travels through tissue. At a boundary where acoustic properties change, part of the pulse is reflected and part continues. The returning pulse is an echo that the transducer detects.
The echo delay helps locate internal structures. A larger delay means the pulse travelled farther before returning, provided the speed of sound is treated as known. Echo strength also depends on how much of the wave is reflected at the boundary.
The pulse makes a round trip
If an echo returns after time t and sound travels at speed c, the pulse has travelled a total distance ct. The boundary is halfway along this path, so its depth is d = ct/2.
This estimate assumes the wave travels at a uniform speed along the route and that t measures the outward and return journey. If the pulse crosses tissues with different sound speeds, a more detailed model is needed.
Pause and explain
An echo delay measures which travel time?
Worked example
An echo returns 80 μs after a pulse is sent through tissue. Take the sound speed as 1.5 × 10³ m s⁻¹. Estimate the reflecting boundary’s depth.
Show the worked solution
- The 80 μs is a round-trip time, so convert it to 80 × 10⁻⁶ s.
- Total path length = ct = 1.5 × 10³ × 80 × 10⁻⁶ = 0.12 m.
- The boundary is halfway along the path: d = 0.12/2 = 0.060 m.
Answer 0.060 m, or 6.0 cm.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A pulse returns after 60 μs in tissue where c = 1.5 × 10³ m s⁻¹. Find the depth of the reflecting boundary.
Give me a hint
Convert the delay, calculate total distance, then halve it.
Compare my reasoning
- t = 60 × 10⁻⁶ s.
- Round-trip distance = ct = 1.5 × 10³ × 60 × 10⁻⁶ = 0.090 m.
- Depth = 0.090/2 = 0.045 m.
0.045 m, or 4.5 cm.
Look for these in your work
- Converts μs to s
- Uses the round-trip distance
- Divides by two for depth
Two tissue boundaries produce echoes at 40 μs and 100 μs. If sound speed is uniform, which boundary is deeper, and by what factor?
Give me a hint
Depth is proportional to round-trip echo time for a fixed sound speed.
Compare my reasoning
- The later echo has the longer travel time.
- The time ratio is 100/40 = 2.5.
- With the same sound speed, the depth ratio is also 2.5.
The 100 μs boundary is deeper, by a factor of 2.5.
Look for these in your work
- Identifies the later echo as deeper
- Compares the delays
- Uses proportionality under the stated uniform-speed assumption
A scan assumes c = 1.5 × 10³ m s⁻¹ and reports a depth of 3.0 cm from an echo delay. Calculate the echo delay and state the path length travelled by the pulse.
Give me a hint
The depth is one-way; the pulse path is twice this distance.
Compare my reasoning
- Round-trip distance = 2d = 2 × 0.030 = 0.060 m.
- t = distance/speed = 0.060/(1.5 × 10³) s.
- t = 4.0 × 10⁻⁵ s = 40 μs.
The pulse travelled 0.060 m in total and the echo delay is 40 μs.
Look for these in your work
- Doubles the boundary depth
- Uses t = s/c
- Converts seconds to microseconds correctly
Common mistakes
- Using d = ct without dividing by two
- Forgetting to convert microseconds to seconds
- Treating echo strength as a direct measurement of depth
What can you explain now?
Why can an ultrasound pulse return from a boundary between tissues?
Compare with the explanation
Part of the wave is reflected when the acoustic properties change
A change at the boundary produces a reflected pulse.
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Make it stick
Why is the measured ultrasound echo path twice the depth of the boundary?
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