href="#fb3_img_img_c42f0e9e-5508-542b-9f28-957fb05fb64c.gif" alt="reflection-3"/>
Figure 2.7 (c) Partial reflection of the energy when the two tissues have somewhat different impedances (e.g. a liver tissue - fat interface). A fraction of the energy is reflected and the remainder transmitted. The interface will be seen in the image but it will not be as strong as in the total reflection case.
Mathematically the fraction reflected is given by the following equation, where z1 and z2 are the acoustic impedances in the first and second tissue respectively:
R is called the "reflection coefficient" of the interface (or sometimes the "intensity reflection coefficient"). Since the quantities in brackets are squared, it is easy to show that the reflection coefficient for a given interface will be the same no matter which way the ultrasound is passing through it (i.e. regardless of whether the ultrasound is going from tissue 1 into tissue 2 or vice versa).
Thus, for example, if R = 0.01 this means that 1% of the transmitted energy reaching the interface will be reflected and so 99% will be transmitted into the second tissue. It also means that any echoes returning from within the second tissue will lose 1% of their energy due to reflection as they pass through the interface on their way back to the transducer due to reflection from the interface.
It can also be shown mathematically that R is virtually 1.0 when z1 and z2 are very different (as in Figure 2.7 (a)) and R = 0 when z1 = z2 (as in Figure 2.7 (b)). This will be explored in the exercises at the end of this section.
Since energy cannot be created or destroyed, the sum of the reflected and transmitted energies must always be equal to the incident energy. Taking advantage of this fact, it can be shown mathematically that the "transmission coefficient" T must be given by:
(If you enjoy mathematics, you can check that this is correct.)
So far the discussion of reflection has focussed on the special situation where the ultrasound is incident on the interface at right angles (i.e. at an angle of 90° to the interface); this is termed "perpendicular incidence". What about the more general situation?
When the incidence is not perpendicular, the reflected ultrasound does not travel back to the transducer (see Figure 2.8). As a result the echo from the interface will not be detected and so it will not be seen in the image. (The equations for the reflection and transmission coefficients will also become more complex, but this is beyond the scope of this book.)
Figure 2.8 Geometry for the reflection of ultrasound from a smooth interface. The broken line is a reference line drawn at right angles to the interface. The "incident angle" θi is always equal to the "reflected angle" θr. This is the same geometry as for light reflecting from a mirror and hence it is called "specular" (mirror-like) reflection.
Scattering
The word "scattering" describes the interaction of ultrasound with small structures (such as red cells and capillaries) in the tissues (see Figure 2.9).
Figure 2.9 Scattered energy is sent in all directions. This means that the appearance of a scatterer is independent of the direction of incidence of the ultrasound. This is different to reflection, which is very dependent on the direction of the incident ultrasound.
It differs from reflection in two important ways:
scattered energy is distributed in all directions, whereas reflected ultrasound goes in a single direction;
the scattered energy is generally much weaker than reflected energy and so the echoes due to scattering are generally displayed in the image as low- to mid-level grey tones.
If you look at a typical ultrasound image you will see that the majority of the echo information in the image comes from scattering from within tissue, not reflection from interfaces between different tissues. Thus the nature of the scattered echoes and their appearance in the image are very important.
You will also notice that scattering produces a random granular echo texture in the image. To understand why this happens, consider Figure 2.10.
Figure 2.10 At one instant of time the transmitted pulse will "see" a volume of tissue (shown in light blue); the transducer will receive echoes from any scatterers that are within this volume. In soft tissue there will generally be a large number of scatterers within the volume, and the echo signal seen by the transducer will be the sum of the signals from all these scatterers.
This shows that at each instant the echo signal coming from soft tissue is actually the sum of the echoes from a number of individual scatterers that lie within the ultrasound pulse.
Since these scatterers are randomly positioned relative to each other, their echoes will add together randomly. This causes the echo signal received by the transducer to have a random variation in its amplitude. This phenomenon is termed "speckle" and it gives rise to the "echo texture" that we see in ultrasound images.
Speckle is a random process that is only indirectly related to the distribution of the scatterers. To highlight this, consider Figure 2.11. This is an image of an ultrasound "phantom" – a test object made of a gel material containing scatterers and designed to look like liver tissue when scanned. (The strong white echoes come from "point targets" that are used to check measurement accuracy and other aspects of equipment performance; they will be discussed in chapter 10.)
Figure 2.11 Scan of an ultrasound phantom (test object) showing speckle.
Notice how the echo texture varies with depth. Close to the transducer the texture is quite fine-grained whereas at greater depths it is much coarser. The phantom material, however, is uniform throughout the phantom, highlighting the fact that the speckle does not directly reflect a tissue property.
Suggested activities
1 Calculate the reflection coefficient at a tissue interface for which z1 = 1.5 × 106 and z2 = 1.5 × 104 (i.e. the two impedance values differ by a factor of 100). Note that the reflection coefficient is almost 1 (or 100%) meaning that virtually all the energy is reflected. Repeat the calculation with z1 and z2 interchanged.
2 Calculate the reflection coefficient at a tissue interface for which z1 = 1.5 × 106 and z2 = 1.51 × 106 (i.e. there is a minimal difference between the two impedances). Note that the reflection coefficient is small, which means that very little energy is reflected.
3 Calculate both the reflection coefficient (R) and the transmission coefficient (T) for an interface with z1 = 1.5 × 106 and z2 = 1.8 × 106 (a moderate difference in impedance). Show that (R + T) = 1.
4 Carefully
