anatomical areas and identify which echoes are caused by reflection and which are caused by scattering. Note the differences in appearance of the two types of echo.5 Scan a region within a liver from two different directions. Carefully compare the speckle patterns – can you see a difference?
Refraction
You are probably familiar with refraction of light – the bending of the light's path as it passes through different materials (as shown in Figures 2.12 and 2.13).
Figure 2.12 As light passes through a prism its direction of travel changes. This is an example of the refraction of light.
Figure 2.13 Refraction of light coming from an object (darker box) on the bottom of a pool. When a viewer's eye receives the light, the brain processes the information on the assumption that the light has travelled in a straight line (as shown by the broken line), and so it sees the box in the position shown by the lighter box. This is why pools generally look shallower than they really are.
Examples include:
a prism – a piece of glass with a triangular cross-section, often found in a science laboratory (shown in Figure 2.12),
the bending of light as it passes from water to air (see Figure 2.13).
Refraction is also the principle on which optical lenses (such as the one in your camera) are based.
Optical refraction occurs whenever light travels from one medium (e.g. air) into another medium that has a different propagation speed for light (e.g. glass or water). In a similar way, ultrasound is refracted whenever it passes through an interface between tissues with different ultrasound propagation speeds (e.g. from liver tissue to fat).
As with reflection, the geometry is determined by measuring the direction of travel of the ultrasound relative to a line drawn at right angles to the interface.
Figure 2.14 Refraction of ultrasound. The direction of travel of the ultrasound is altered as it passes through the interface between tissues with different ultrasound propagation speeds. In this example the propagation speed is lower in the second tissue (i.e. c2 < c1).
Mathematically the amount of refraction can be determined using Snell's Law:
where θi is the incident angle and θt is the transmitted angle. It is easy to show that if the difference between the two propagation speeds increases then the difference between the two angles will also increase. It can also be shown that if c2 is less than c1 then θt will be less than θi while if c2 is larger than c1 then θt will be larger than θi (see Figures 2.14 and 2.15).
Figure 2.15 Refraction for the situation where c2 is larger than c1. The "bending" of the ultrasound is now in the opposite direction to that shown in Figure 2.14. Note that the beam has been "bent" by just 3.5°.
Figure 2.16 An identical situation to that shown in Figure 2.15 except that the incident angle has increased from 30° to 60°. Note that the effect of refraction is now more pronounced, causing a change of direction of 13°.
Comparing Figures 2.15 and 2.16 shows that the difference between θi and θt also depends on the incident angle θi. As the incident angle increases the bending of the beam increases.
Conversely (see Figure 2.17) when the incident angle is 0° (i.e. when the ultrasound is perpendicular to the interface) the transmitted angle is also 0° and hence no deflection of the beam occurs, regardless of the propagation speeds.
Figure 2.17 With perpendicular incidence (i.e. an incident angle θi = 0°) the beam is not changed in direction at all, regardless of the values of c1 and c2.
When the propagation speed is greater in the second medium than in the first (i.e. c2 > c1, see Figures 2.15 and 2.16) it can be shown that there is a particular value of the incident angle for which the transmitted angle (θt) will be 90° (see Figure 2.18).
Figure 2.18 When the incident angle is equal to the "critical angle" (65° in this case) the transmitted ultrasound is refracted so that the transmitted angle is 90°.
This value of the incident angle is called the critical angle.
What is the significance of the critical angle? Since the ultrasound has a transmitted angle of 90° it is barely entering the second tissue – it is just running along the interface between the two tissues.
What happens when the incident angle is greater than the critical angle? In this case ultrasound is not transmitted into the second tissue at all, and instead all the energy is reflected, as shown in Figure 2.19.
Figure 2.19 When the incident angle exceeds the critical angle, all of the ultrasound energy is reflected from the interface. Notice that the geometry of reflection is then the same as discussed in the previous section, i.e. the reflected angle is equal to the incident angle.
To repeat this point:
If the propagation speed is higher in the second tissue, then a critical angle exists; for incident angles greater than the critical angle, total reflection occurs.
Thus we have seen that there are two separate mechanisms capable of causing total reflection of ultrasound from an interface between two tissues:
when there is a very large difference in the acoustic impedance of the two tissues, or
when the propagation speed in the second tissue is higher and the incident angle exceeds the critical angle.
How do we calculate the critical angle?
Recognising that θt = 90° and so sin θt = 1.0, Snell's Law can be written:
As an example, consider the situation shown in Figures 2.15, 2.16 and 2.18 where c1 = 1450 m/sec (a typical propagation speed for fat) and c2 = 1600 m/sec (a typical value for muscle). Using the equation above you will find that the critical angle
