Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

Answers without the blur.

Just sign up for free and you're in.

Short Answer

In the clinical use of ultrasound, transducers are always coupled to the skin by a thin layer of gel or oil, replacing the air that would otherwise exist between the transducer and the skin. (a) Using the values of acoustic impedance given in Table 17.5 calculate the intensity reflection coefficient between transducer material and air. (b) Calculate the intensity reflection coefficient between transducer material and gel (assuming for this problem that its acoustic impedance is identical to that of water). (c) Based on the results of your calculations, explain why the gel is used.

(a) Intensity reflection coefficient for air to gel boundaries is 0.99.

(b) Intensity reflection coefficient for air to gel boundaries is 0.823.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Formula for the intensity reflection coefficient:

Intensity reflection coefficient (a) is given as the ratio of the intensity of the reflected wave to the intensity of an incident wave.

\[a = \frac{{{{\left( {{z_2} - {z_1}} \right)}^2}}}{{{{\left( {{z_2} + {z_1}} \right)}^2}}}\] ….. (1)

Here, \[{z_2}\] and \[{z_1}\] are acoustic impedances of two boundaries (transmission and reflection boundaries).

Step 2: (a) Calculation for the intensity reflection coefficient:

Consider the given data as below.

The acoustic impedance of transducer (gel or oil) is,

\[{z_2} = 30.8 \times {10^{6{\rm{ }}}}kg{\rm{ }}{m^{ - 2}}{\rm{ }}{s^{ - 1}}\]

The acoustic impedance of air is,

\[{z_1} = 429{\rm{ }}kg{\rm{ }}{m^{ - 2}}{\rm{ }}{s^{ - 1}}\]

Now put these values into equation (1).

\begin{aligned}a &= \frac{{{{\left( {30.8 \times {{10}^6} - 429} \right)}^2}}}{{{{\left( {30.8 \times{{10}^{6{\rm{ }}}} + 429} \right)}^2}}}\\ &= \frac{{9.486136 \times {{10}^{14}}}}{{9.48666 \times {{10}^{14}}}}\\ &= 0.99\end{aligned}

Hence, the value is closed to one tells that boundary of the gel is highly reflective.

Step 3: (b) Calculation for the intensity reflection coefficient

Consider the given data as below.

The acoustic impedance of transducer (gel or oil) is, \[{z_2} = 30.8 \times {10^{6{\rm{ }}}}kg{\rm{ }}{m^{ - 2}}{\rm{ }}{s^{ - 1}}\]

The acoustic impedance of water is, \[{z_1} = 1.5 \times {10^{6{\rm{ }}}}kg{\rm{ }}{m^{ - 2}}{\rm{ }}{s^{ - 1}}\]

Now put these values into equation (1), and you have

\begin{aligned}a &= \frac{{{{\left( {30.8 \times {{10}^6} - 1.5 \times {{10}^{6{\rm{ }}}}} \right)}^2}}}{{{{\left( {30.8 \times {{10}^{6{\rm{ }}}} + 1.5 \times {{10}^{6{\rm{ }}}}} \right)}^2}}}\\ &= \frac{{\left( {858.49 \times {{10}^{12}}} \right)}}{{\left( {1043.29 \times {{10}^{{\rm{12}}}}} \right)}}\\ &= 0.823\end{aligned}

Step 3: (c) Explain, why the gel is used

From the values of the final reflection coefficient, the gel has a high acoustic impedance that provides a high reflection coefficient.

Due to the high reflection coefficient, it will provide a very good reflection that’s why use gel.

Find Study Materials for

Create Study Materials

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Formula for the intensity reflection coefficient:

Step 2: (a) Calculation for the intensity reflection coefficient:

Step 3: (b) Calculation for the intensity reflection coefficient

Step 3: (c) Explain, why the gel is used

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Formula for the intensity reflection coefficient:

Step 2: (a) Calculation for the intensity reflection coefficient:

Step 3: (b) Calculation for the intensity reflection coefficient

Step 3: (c) Explain, why the gel is used

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics