Ultrasound Imaging (AQA A Level Physics): Revision Note

Exam code: 7408

Katie M

Written by: Katie M

Reviewed by: Tim

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Specific acoustic impedance

  • The acoustic impedance Z of a medium is defined as:

The product of the speed of the ultrasound in the medium and the density of the medium

  • This quantity describes how much resistance an ultrasound beam encounters as it passes through a tissue

  • Acoustic impedance can be calculated using the equation:

Z = ρc

  • Where:

    • Z = acoustic impedance (kg m-2 s-1)

    • ρ = the density of the material (kg m-3)

    • c = the speed of sound in the material (m s-1)

  • This equation shows:

    • the higher the density of a tissue, the greater the acoustic impedance

    • the faster the ultrasound travels through the material, the greater the acoustic impedance also

  • This is because sound travels faster in denser materials

    • Sound is fastest in solids and slowest in gases

    • The closer the particles in the material, the faster the vibrations can move through the material

  • At the boundary between media of different acoustic impedances, some of the wave energy is reflected and some is transmitted

  • The greater the difference in acoustic impedance between the two media, the greater the reflection and the smaller the transmission

    • Two materials with the same acoustic impedance would give no reflection

    • Two materials with a large difference in values would give much larger reflections

  • Air has an acoustic impedance of Zair = 400 kg m−2 s−1

  • Skin has an acoustic impedance of Zskin = 1.7 × 106 kg m−2 s−1

    • The large difference means ultrasound would be significantly reflected, hence a coupling gel is necessary

    • The coupling gel used has a similar Z value to skin, meaning that very little ultrasound is reflected

Incident and reflected waves in medium Z1, and a refracted (transmitted) wave in Z2, with angles of incidence and refraction measured from the dashed normal.
Refraction and reflection of ultrasound waves at a boundary between two materials with different acoustic impedances (in this case, Z1 < Z2 )

Worked Example

The table shows the speed of sound and acoustic impedance in four different materials.

Medium

Speed of ultrasound / m s-1

Acoustic impedance / kg m-2 s-1

Air

330

4.3 × 102

Gel

1500

1.5 × 106

Soft tissue

1600

1.6 × 106

Bone

4100

7.0 × 106

Use the data in the table to calculate the value for the density of bone.

[2]

Answer:

Step 1: Write down known quantities

  • Acoustic impedance of bone, Z = 7.0 × 106 kg m−2 s−1

  • Speed of ultrasound in bone, c = 4100 m s−1

Step 2: Write out the equation for acoustic impedance

Z = ρc

Step 3: Rearrange for density and calculate

ρ = Zc = 7.0 × 1064100 [1 mark]

ρ = 1700 kg m−3 [1 mark]

Examiner Tips and Tricks

A common mistake is to confuse the c in the acoustic impedance equation for the speed of light - don’t do this!

Reflection & transmission of ultrasound

  • The intensity reflection coefficient α is defined as:

The ratio of the intensity of the reflected wave relative to the incident (transmitted) wave

  • This can be calculated using the fraction:

α = IrIi = (Z2 − Z1)2(Z2 + Z1)2

  • Where:

    • α = intensity reflection coefficient

    • Ir = intensity of the reflected wave (W m-2)

    • Ii = intensity of the incident wave (W m-2)

    • Z1 = acoustic impedance of one material (kg m-2 s-1)

    • Z2 = acoustic impedance of a second material (kg m-2 s-1)

  • This ratio shows:

    • if there is a large difference between the impedance of the two materials, then most of the energy will be reflected

    • if the impedance is the same, then there will be no reflection

Coupling medium

  • When ultrasound is used in medical imaging, a coupler is needed between the transducer and the body

    • This is because the soft tissues of the body are much denser than air

  • If air is present between the transducer and the body, then almost all the ultrasound energy will be reflected

  • To counter this, a coupling gel is placed between the transducer and the body

    • This is because skin and coupling gel have a similar density, so little ultrasound is reflected

  • This is an example of impedance matching, which is defined as when:

Two media have a similar acoustic impedance, resulting in little to no reflection of the ultrasound wave

  • In terms of intensity reflection coefficient, α, between the two media:

    • At lower values of α, the media are impedance matched, so less reflection occurs

    • At higher values of α, the media are not impedance matched, so more reflection occurs

Worked Example

A beam of ultrasound is incident at right-angles to a boundary between two materials, as shown in the diagram.

In medium Z1, incident intensity I travels towards the boundary and reflected intensity IR returns. In medium Z2, transmitted intensity IT travels onwards.

The materials have acoustic impedances of Z1 and Z2. The intensity of the transmitted ultrasound beam is IT, and the reflected intensity is IR.

Medium

Speed of ultrasound / m s-1

Acoustic impedance / kg m-2 s-1

Air

330

4.3 × 102

Gel

1500

1.5 × 106

Soft tissue

1600

1.6 × 106

Bone

4100

7.0 × 106

(a) State the relationship between I, IT and IR. [1]

(b) Use the data from the table to determine the reflection coefficient, α, for a boundary between:

(i) Gel and soft tissue [2]

(ii) Air and soft tissue [2]

(c) Explain why gel is usually put on the skin during medical diagnosis using ultrasound. [2]

Answer:

Part (a)

Step 1: List the known quantities

  • Intensity of incident wave = I

  • Intensity of the transmitted wave = IT

  • Intensity of the reflected wave = IR

Step 2: Relate the quantities:

  • The incident intensity is equal to the sum of the transmitted and reflected intensities:

Incident intensity = Transmitted intensity + Reflected intensity

I = IT + IR [1 mark]

Part (b)(i)

Step 1: List the known quantities

  • Acoustic impedance of gel, Z1 = 1.5 × 106 kg m−2 s−1

  • Acoustic impedance of soft tissue, Z2 = 1.6 × 106 kg m−2 s−1

Step 2: Write down the equation for intensity reflection coefficient α

α = (Z2 − Z1)2(Z2 + Z1)2

Step 3: Calculate the intensity reflection coefficient

α = (1.6 × 106 − 1.5 × 106)2(1.6 × 106 + 1.5 × 106)2 = 0.123.12 = 0.001 [1 mark]

  • This result means that only 0.1% of the incident intensity will be reflected, with the remaining being transmitted [1 mark]

Part (b)(ii)

Step 1: List the known quantities

  • Air, Z1 = 4.3 × 102 kg m−2 s−1

  • Soft tissue, Z2 = 1.6 × 106 kg m−2 s−1

Step 2: Calculate the intensity reflection coefficient

α = (1.6 × 106 − 4.3 × 102)2(1.6 × 106 + 4.3 × 102)2 ≈ (1.6 ×106)2(1.6 ×106)2 ≈ 1 [1 mark]

  • This result means that 100% of the incident intensity will be reflected, with none being transmitted [1 mark]

Part (c)

Why gel is usually put on the skin during medical diagnosis using ultrasound

  • At the air-soft tissue boundary, the intensity reflection coefficient is α ≈ 1

    • Therefore, without gel, there is almost complete reflection - no ultrasound is transmitted through the skin [1 mark]

  • At the gel-soft tissue boundary, the intensity reflection coefficient is α ≈ 0.001

    • Therefore, the gel enables almost complete transmission of the ultrasound through the skin, with very little reflection [1 mark]

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Katie M

Author: Katie M

Expertise: Curriculum Expert

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Tim

Reviewer: Tim

Expertise: Content Creator

Timothy graduated with a first class degree in Mathematics and Physics from the University of Warwick. After working as a postgraduate researcher, Timothy has worked as a content creator for various online revision platforms, creating physics resources for a range of levels and exam boards.