MYP Physics · topic MYP-19.05
Ultrasound imaging and access
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
An ultrasound image with two hidden boundaries
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]
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.
Show a hint
Begin with part (a): Find both boundary depths and their separation, assuming the same wave speed throughout the path.
Reveal the worked solution
- 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.
Answer (a) Depths = 30 mm and 52.5 mm; separation = 22.5 mm. (b) Pulse duration = 0.80 μs; axial resolution≈0.60 mm; reflected intensity fraction = 0.36.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.