Which physics concept best explains the difference in signal strength between the 5.0 and 6.3 MHz Doppler frequencies?

Which physics concept best explains the difference in signal strength between the 5.0 and 6.3 MHz Doppler frequencies?

• A. The Reynolds number
• B. Poiseuille's law
• C. Bernoulli effect
• D. Rayleigh scattering

Answer: (D)

Rayleigh scattering explains why higher-frequency Doppler signals can be stronger. When the small structures in tissue or particles that scatter the ultrasound are much smaller than the wavelength, the backscattered intensity scales strongly with frequency—roughly with the fourth power of frequency. So increasing from 5.0 to 6.3 MHz shortens the wavelength and makes these tiny inhomogeneities scatter more of the incident energy back toward the transducer, boosting signal strength. Roughly, the strength goes up by about (6.3/5.0)^4, which is around 2.5 times, all else being equal. The other concepts describe flow or pressure–velocity relations and don’t predict how scatter intensity changes with frequency, so they don’t explain this observation.

Rayleigh scattering explains why higher-frequency Doppler signals can be stronger. When the small structures in tissue or particles that scatter the ultrasound are much smaller than the wavelength, the backscattered intensity scales strongly with frequency—roughly with the fourth power of frequency. So increasing from 5.0 to 6.3 MHz shortens the wavelength and makes these tiny inhomogeneities scatter more of the incident energy back toward the transducer, boosting signal strength. Roughly, the strength goes up by about (6.3/5.0)^4, which is around 2.5 times, all else being equal. The other concepts describe flow or pressure–velocity relations and don’t predict how scatter intensity changes with frequency, so they don’t explain this observation.