SonographyAcademy

Curriculum · Ultrasound Physics & Instrumentation

Wave–Tissue Interaction: Reflection, Refraction, Scattering & Attenuation

How the pulse interacts with tissue: specular vs diffuse reflection and the reflection coefficient, Snell's law, Rayleigh scattering (f⁴), and attenuation (0.5 dB/cm/MHz) with the half-value layer.

~35 min · level: foundation · SPI draft — pending clinical review

Learning objectives

  • Distinguish specular from diffuse reflection and compute the intensity reflection coefficient.
  • State the two conditions required for refraction and apply Snell's law qualitatively.
  • Explain Rayleigh scattering and its f⁴ frequency dependence.
  • Quantify attenuation (0.5 dB/cm/MHz) and define the half-value layer.

As the pulse meets boundaries and inhomogeneities, energy is reflected, refracted, scattered, and absorbed. Reflection and scatter create the image; absorption and redirection are losses (attenuation) and artifact sources.

Reflection occurs at boundaries of differing acoustic impedance. Specular reflection comes off large, smooth interfaces (diaphragm, bladder wall) and is angle-dependent — strongest near perpendicular incidence, which is why these structures dim when insonated obliquely. Diffuse reflection comes off rough interfaces and scatters in many directions, so it is less angle-dependent. The fraction of intensity reflected at normal incidence is:

R=(Z2−Z1Z2+Z1)2T=1−RR = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^{2} \qquad T = 1 - R
Intensity reflection and transmission coefficients (normal incidence).

If Z1=Z2Z_1 = Z_2, then R=0R = 0 — the boundary is invisible. Soft-tissue/soft-tissue boundaries have tiny RR (often < 1%), so most energy transmits and deep structures stay visible. Soft-tissue/air (R≈99.9%R \approx 99.9\%) and soft-tissue/bone (R≈30R \approx 30–40%40\%) reflect almost everything → shadowing beyond gas and bone.

Worked example — reflection coefficient

Estimate the fraction of intensity reflected at a fat (Z≈1.34Z \approx 1.34 Mrayl) / muscle (Z≈1.71Z \approx 1.71 Mrayl) interface.

Solution.

R=(1.71−1.341.71+1.34)2=(0.373.05)2=(0.121)2≈0.0147R = \left(\dfrac{1.71 - 1.34}{1.71 + 1.34}\right)^2 = \left(\dfrac{0.37}{3.05}\right)^2 = (0.121)^2 \approx 0.0147, i.e., about 1.5% reflects and ~98.5% transmits — typical of soft-tissue boundaries, which is why we can image deep to them.

Refraction is bending of the transmitted beam. It requires two conditions simultaneously: (1) oblique incidence and (2) a difference in propagation speed across the boundary. Perpendicular incidence produces no refraction even with a speed mismatch.

sin⁡θtsin⁡θi=c2c1(Snell’s law)\frac{\sin\theta_t}{\sin\theta_i} = \frac{c_2}{c_1} \quad\text{(Snell's law)}
Refraction at an oblique interface with differing speeds.

Scattering redirects sound in many directions off structures small relative to λ\lambda (organ parenchyma, blood cells) and creates the granular echotexture (speckle). For scatterers far smaller than λ\lambda (e.g., red blood cells, ~7 µm), Rayleigh scattering applies, with a steep frequency dependence:

Iscattered∝f4I_{\text{scattered}} \propto f^{4}
Rayleigh scattering rises with the fourth power of frequency.

Attenuation is the progressive loss of intensity with distance, from absorption (conversion to heat — the dominant mechanism and the basis of thermal bioeffects), reflection, and scattering. In soft tissue:

Attenuation (dB)=0.5 dBcm\cdotpMHz×f(MHz)×depth(cm)\text{Attenuation (dB)} = 0.5\,\frac{\text{dB}}{\text{cm·MHz}} \times f(\text{MHz}) \times \text{depth(cm)}
Soft-tissue attenuation rule of thumb (one-way).

Because attenuation scales with frequency, higher frequency means shallower penetration — the other half of the resolution-vs-penetration tradeoff. The half-value layer (HVL) is the depth over which intensity falls to one-half (−3 dB); HVL decreases as frequency increases.

Worked example — attenuation

How much one-way attenuation does a 5 MHz beam accumulate over 4 cm of soft tissue?

Solution.

0.5×5×4=10 dB0.5 \times 5 \times 4 = 10\text{ dB} one-way (round-trip ≈ 20 dB). This is why a 5 MHz beam struggles to image very deep structures, and why deep abdominal work uses 2–3.5 MHz.

Key takeaways

  • The intensity reflection coefficient at normal incidence is R = ((Z2 - Z1)/(Z2 + Z1))^2, so identical impedances give R = 0 (invisible boundary), soft-tissue/air reflects about 99.9%, and soft-tissue/bone about 30-40%, causing shadowing beyond gas and bone.
  • Specular reflection off large smooth interfaces is strongly angle-dependent (brightest near perpendicular incidence and dimming when insonated obliquely), whereas diffuse reflection off rough interfaces scatters in many directions and is less angle-dependent.
  • Refraction (Snell's law) requires BOTH oblique incidence AND a propagation-speed difference at the boundary; either condition alone produces no beam bending, and refraction causes lateral misregistration and edge shadowing.
  • Rayleigh scattering from structures much smaller than the wavelength (e.g., red blood cells ~7 um) scales as I proportional to f^4, which is why blood returns far more signal at higher frequency.
  • Soft-tissue attenuation follows 0.5 dB/cm/MHz (one-way), driven mainly by absorption, so higher frequency means shallower penetration; the half-value layer is the depth at which intensity falls by half (-3 dB) and decreases as frequency rises.

Check your understanding

Registry-style items with worked rationales.

1Two tissues have identical acoustic impedance. The intensity reflected at their boundary is:application

2Refraction of the ultrasound beam requires:recall

3A 3 MHz beam travels 6 cm one-way through soft tissue. Approximate the attenuation.application

4Scattering from red blood cells increases with frequency according to which relationship?recall

Go deeper — trusted free resources

Hand-picked, verified links to authoritative open resources. Opens in a new tab.

Radiopaedia
Acoustic Impedance (Ultrasound)

Defines acoustic impedance (Z = density x speed of sound) and explains why impedance mismatch at tissue interfaces governs how much sound is reflected versus transmitted.

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Radiopaedia
Reflection (Ultrasound)

Covers specular versus diffuse reflection at impedance boundaries and how the angle of incidence and refraction relate to the impedance difference between tissues.

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Radiopaedia
Scattering (Ultrasound)

Explains scattering from structures smaller than the wavelength (e.g. red blood cells, organ parenchyma) and distinguishes diffuse reflectors from specular reflectors.

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Radiopaedia
Attenuation (Ultrasound)

Details how beam amplitude decays with depth through absorption, reflection, and scattering, and why higher-frequency transducers attenuate faster and limit penetration.

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References

  1. Edelman SK. Understanding Ultrasound Physics. 4th ed. ESP Inc.; 2012.
  2. Kremkau FW. Sonography Principles and Instruments. 9th ed. Elsevier; 2016.
  3. Ultrasound Physics and Instrumentation. StatPearls, NCBI Bookshelf.
  4. AIUM Practice Parameters and bioeffects statements.