Revise Production and use of ultrasound for Cambridge 9702 (24.1)

Ultrasound refers to sound waves with frequencies greater than 20 kilohertz (kHz), placing them beyond the upper limit of human hearing. For medical imaging, these frequencies are typically much higher, often ranging from 1 to 20 megahertz (MHz). Like all sound waves, ultrasound waves are longitudinal waves, meaning the particles of the medium oscillate parallel to the direction in which the wave energy is travelling.

The heart of an ultrasound system is the transducer, a device capable of both sending out and receiving ultrasound waves. It relies on a special property of certain crystals, known as the piezoelectric effect. When an alternating potential difference is applied across a piezoelectric crystal, it undergoes mechanical deformation, causing it to vibrate at the frequency of the applied voltage. This vibration generates ultrasound pulses.

When ultrasound travels through a medium, its passage can be resisted. This resistance is quantified by acoustic impedance (Z). It's a crucial property that determines how much ultrasound reflects or transmits when it encounters a boundary between two different materials.

$Z = \rho c$

Here, $\rho$ is the density of the medium in kg m$^{-3}$, and $c$ is the speed of sound in the medium in m s$^{-1}$. The unit for acoustic impedance is rayls (kg m$^{-2}$ s$^{-1}$).

Whenever an ultrasound wave encounters a boundary between two media with different acoustic impedances, a portion of the wave is reflected, and the rest is transmitted. The greater the difference in acoustic impedances ($Z_1$ and $Z_2$), the more significant the reflection.

$I_r / I_i = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2$

This formula gives the intensity reflection coefficient, which is the ratio of the reflected intensity ($I_r$) to the incident intensity ($I_i$). If the acoustic impedances of the two media are identical ($Z_1 = Z_2$), there will be no reflection, and the entire wave will be transmitted.

This brings us to a crucial practical aspect of ultrasound imaging: the coupling medium. A layer of gel or oil is applied between the transducer and the patient's skin. Air has an extremely low acoustic impedance compared to soft tissue. Without the gel, the vast majority of ultrasound would be reflected at the skin-air interface, preventing effective imaging. The coupling gel's acoustic impedance is much closer to that of skin, allowing maximum transmission of ultrasound into the body.

As ultrasound waves travel through any medium, their intensity and amplitude gradually decrease. This phenomenon is called attenuation. It's primarily caused by two processes: absorption, where the ultrasound energy is converted into heat within the tissue, and scattering, where the waves are deflected in various directions by small inhomogeneities in the medium.

$I = I_0 e^{-\mu x}$

This formula describes how ultrasound intensity ($I$) decreases exponentially with the distance ($x$) travelled. $I_0$ is the initial intensity, and $\mu$ is the linear attenuation coefficient, which quantifies the rate of intensity loss per unit thickness for a given medium.

A useful concept related to attenuation is the half-value thickness ($x_{1/2}$). This is the specific distance an ultrasound wave must travel through a medium for its intensity to be reduced to exactly half of its initial value. It provides a straightforward way to compare the penetrating power of ultrasound in different tissues.

$x_{1/2} = \ln(2)/\mu$