Choosing an ultrasound frequency for a small stone deep in tissue
A sonographer must image a suspected 2.0 mm stone in a duct lying 6.0 cm below the skin. The department has two probes, 2.0 MHz and 5.0 MHz. Soft tissue has speed 1540 m s⁻¹, attenuation 0.50 dB cm⁻¹ MHz⁻¹ and acoustic impedance 1.6 × 10⁶ Rayl; the stone's impedance is 4.0 × 10⁶ Rayl. The scanner's dynamic range is 60 dB, meaning echoes more than 60 dB weaker than the transmitted pulse are lost in noise. A coupling gel of impedance close to tissue is used; without it a thin air layer (impedance about 4 × 10² Rayl) would remain at the skin. The report from a previous clinic states that the 2 MHz probe 'penetrates better and is therefore always the safer choice for this patient'.
1. Calculate the round-trip attenuation and the reflection loss at the stone for each probe, and decide whether each probe can detect an echo within the 60 dB budget at 6.0 cm depth.
Read the worked answer
Attenuation (one way) = 0.50 × depth in cm × frequency in MHz.
- 5.0 MHz: 0.50 × 6.0 × 5.0 = 15 dB one way, so 30 dB round trip.
- 2.0 MHz: 0.50 × 6.0 × 2.0 = 6 dB one way, so 12 dB round trip.
Reflection at the stone. Intensity reflection coefficient: R = [(Z₂ − Z₁)/(Z₂ + Z₁)]² = [(4.0 − 1.6)/(4.0 + 1.6)]² = (2.4/5.6)² = 0.184. In dB: 10 log₁₀ 0.184 = −7.4 dB. The transmitted fraction 0.816 costs a further 10 log₁₀ 0.816 = −0.9 dB each way.
Totals (ignoring the small skin-entry loss, which is the same for both probes):
- 5.0 MHz: 30 + 7.4 + 2 × 0.9 ≈ 39 dB
- 2.0 MHz: 12 + 7.4 + 2 × 0.9 ≈ 21 dB
Both are inside the 60 dB budget, so both probes can detect the echo at 6 cm; the 5 MHz echo has about 21 dB less margin. The 5 MHz probe would become marginal (≈63 dB) at roughly 10 cm depth, which is where the 'penetration' argument in the report would actually bite. At this depth the report's reasoning does not decide the choice.
2. Calculate the wavelength in tissue for each probe and use it, together with the reflection data, to recommend a probe. Explain why the coupling gel is not optional.
Read the worked answer
Wavelengths. λ = c/f:
- 5.0 MHz: 1540/5.0 × 10⁶ = 0.31 mm
- 2.0 MHz: 1540/2.0 × 10⁶ = 0.77 mm
Axial resolution is of the order of half a wavelength (or half the spatial pulse length), and lateral resolution is set by the beam width, which also narrows as frequency rises. A 2.0 mm stone is about 6.5 wavelengths across at 5 MHz but only 2.6 at 2 MHz, so the 5 MHz image resolves its near and far walls as separate echoes; at 2 MHz the walls blur into one and a small stone can be missed or its size misjudged.
Recommendation: use the 5.0 MHz probe. It has adequate (≈21 dB) margin at 6 cm, the reflection coefficient of 0.18 is large enough to give a clear echo, and its resolution is three times better. If the stone proves deeper than 6 cm, or the patient's habitus adds fat, the margin is lost and 2 MHz becomes the fallback - the report's claim is right in principle but misapplied here.
The gel. At a tissue–air interface R = [(1.6 × 10⁶ − 400)/(1.6 × 10⁶ + 400)]² ≈ 0.999, so 99.9% of the intensity is reflected and only about 0.1% (−30 dB) enters the body. That single loss would consume half the dynamic-range budget before any tissue attenuation; the gel replaces air with a matched impedance and lets the beam in.
3. The stone produces a strong echo and also casts a shadow. Explain why the shadow may be the more reliable sign for a 2.0 mm stone, and identify one practical limit on the shadow's usefulness.
Read the worked answer
Why the shadow helps. The stone's impedance mismatch means most of the incident intensity is reflected (R = 0.18) and, more importantly, the remaining beam is strongly absorbed and scattered within the stone rather than transmitted. Beyond the stone the beam is therefore much weaker than in neighbouring tissue, so the region behind it appears dark: a shadow. For a small stone the echo may be only one or two pixels wide and easily confused with speckle, whereas the shadow is a longer, coherent, geometry-linked feature aligned with the beam axis - the eye and the machine's compounding algorithms detect it far more reliably than a 2 mm bright spot. This is why stones are often reported from their shadow.
Practical limit: the shadow is only visible if the stone intercepts a significant part of the beam, i.e. if the stone is comparable to or larger than the beam width at that depth - for a 2 mm stone the beam must be focused to a few millimetres at 6 cm, or the shadow is filled in by surrounding tissue. Also, shadowing depends on the structure behind: bowel gas produces its own 'dirty' shadow, so a shadow behind a small duct stone may be obscured or misinterpreted, and time-gain compensation set too high can brighten the shadow away.