Edexcel A-Level Physics 9PH0
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Waves and Particle Nature of Light
Pulse-echo calculations, resolution and ultrasound
- 1 thing to know
- Papers 2 and 3
Overview
Recall questions on halving the measured time, why an object smaller than the wavelength is invisible, pulse length and resolution, and why the gap between pulses matters.
What you need to know
Answer each one before you reveal it.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Ultrasound travelling at 1540 m s⁻¹ in soft tissue returns from a boundary after 1.3 × 10⁻⁴ s. How deep is the boundary?
Model answer
The depth is 1540 × 1.3 × 10⁻⁴ ÷ 2 = 0.10 m.
- 2
A radar pulse returns from an aircraft 60 μs after it is transmitted. How far away is the aircraft?
Model answer
The distance is 3.0 × 10⁸ × 60 × 10⁻⁶ ÷ 2 = 9.0 × 10³ m, or 9.0 km.
- 3
A ship's sonar pulse travels at 1500 m s⁻¹ in sea water and its echo from the sea bed arrives 0.40 s later. What is the depth of the water?
Model answer
The depth is 1500 × 0.40 ÷ 2 = 300 m.
- 4
Why can a pulse-echo system not detect an object much smaller than the wavelength of the radiation used?
Model answer
The wave simply diffracts around an object that small, so no significant echo is returned for the detector to pick up.
- 5
What is the wavelength of 2.0 MHz ultrasound in tissue where its speed is 1540 m s⁻¹, and what does that imply for the scan?
Model answer
The wavelength is 1540 ÷ (2.0 × 10⁶) = 7.7 × 10⁻⁴ m, so detail finer than about 0.8 mm cannot be resolved.
- 6
Why do medical scans of shallow structures use higher-frequency ultrasound than scans of deep organs?
Model answer
A higher frequency gives a shorter wavelength and therefore finer resolution, but it is absorbed more strongly and so cannot penetrate as far into the body.
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