MYP · 19 · extended response
An ultrasound image with two hidden boundaries
Physics in society · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Echo timing evidence
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Relative echo amplitude
Horizontal axis (x): Time after transmission / μs
Read the exact data as a table
| Time after transmission / μs | Relative echo amplitude |
|---|---|
| 0 | 1 |
| 5 | 0 |
| 35 | 0 |
| 40 | 0.6 |
| 45 | 0 |
| 65 | 0 |
| 70 | 0.3 |
| 75 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A pulse-echo scanner detects reflected pulses at 40 μs and 70 μs after transmission into tissue. Sound speed is 1500 m s⁻¹. A 5.0 MHz pulse lasts four cycles. At one boundary acoustic impedances are Z1 = 1.5×10⁶ and Z2 = 6.0×10⁶ kg m⁻² s⁻¹. Use reflected intensity fraction R = [(Z2−Z1)/(Z2+Z1)]². Supplied data (Time after transmission / μs; Relative echo amplitude): (0, 1), (5, 0), (35, 0), (40, 0.6), (45, 0), (65, 0), (70, 0.3), (75, 0).
- (a)
Find both boundary depths and their separation, assuming the same wave speed throughout the path.
3 marks - (b)
Estimate axial pulse resolution and the reflection fraction. Explain the role of coupling gel and one access or performance trade-off.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find both boundary depths and their separation, assuming the same wave speed throughout the path.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Each pulse travels to a boundary and back, so depth is d = vt/2.
- 2
At 40 μs, d = 1500(40×10⁻⁶)/2 = 0.030 m. At 70 μs, d = 0.0525 m.
- 3
Their separation is 0.0525−0.030 = 0.0225 m. Different tissue speeds or refraction would make this uniform-speed interpretation approximate.
(b)
Estimate axial pulse resolution and the reflection fraction. Explain the role of coupling gel and one access or performance trade-off.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Four cycles at 5.0 MHz last 4/(5.0×10⁶) = 0.80 μs. The spatial pulse length is 1500(0.80×10⁻⁶) = 1.2 mm; two interfaces need about half this separation, 0.60 mm, to resolve distinct echoes.
- 2
R = [(6.0−1.5)/(6.0+1.5)]² = (0.60)² = 0.36. Amplitude ratios and intensity ratios must not be confused.
- 3
Gel removes an air gap that otherwise strongly reflects sound. Ultrasound avoids ionising radiation and can be portable, but penetration, operator skill and acoustic access limit some examinations; higher frequency can improve resolution while increasing attenuation.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Each pulse travels to a boundary and back, so depth is d = vt/2.
- At 40 μs, d = 1500(40×10⁻⁶)/2 = 0.030 m. At 70 μs, d = 0.0525 m.
- Their separation is 0.0525−0.030 = 0.0225 m. Different tissue speeds or refraction would make this uniform-speed interpretation approximate.
- Four cycles at 5.0 MHz last 4/(5.0×10⁶) = 0.80 μs. The spatial pulse length is 1500(0.80×10⁻⁶) = 1.2 mm; two interfaces need about half this separation, 0.60 mm, to resolve distinct echoes.
- R = [(6.0−1.5)/(6.0+1.5)]² = (0.60)² = 0.36. Amplitude ratios and intensity ratios must not be confused.
- Gel removes an air gap that otherwise strongly reflects sound. Ultrasound avoids ionising radiation and can be portable, but penetration, operator skill and acoustic access limit some examinations; higher frequency can improve resolution while increasing attenuation.
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