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Why use ultrasound for medical imaging?

1

$begingroup$

What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?

energy acoustics frequency wavelength medical-physics

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

19511

New contributor

Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  your edits are quite helpful, but can I ask why you do edits so frequently?
  $endgroup$
  – Ubaid Hassan
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
  $endgroup$
  – dmckee♦
  28 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?

energy acoustics frequency wavelength medical-physics

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

19511

New contributor

Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  your edits are quite helpful, but can I ask why you do edits so frequently?
  $endgroup$
  – Ubaid Hassan
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Tagging is important, it helps people find things, helps the system that auto-identified related content, and enables meaningful analysis of site usage patterns. BUt good tagging takes time, attention to detail, and a minimum level of expertise so that you can identify the relevant tags.
  $endgroup$
  – dmckee♦
  28 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?

energy acoustics frequency wavelength medical-physics

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

19511

New contributor

Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

19511

New contributor

Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

19511

New contributor

Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 1 hour ago

Qmechanic♦

108k122001245

edited 1 hour ago

Qmechanic♦

108k122001245

asked 1 hour ago

Ubaid HassanUbaid Hassan

19511

New contributor

Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Ubaid HassanUbaid Hassan

19511

3 Answers
3

active

oldest

votes

5

$begingroup$

I think the simple answer here is resolution.

Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.

If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by

$$lambda = {c over f} $$

so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....

The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate

$$lambda = 0.001 {rm m} = 1 {rm mm}$$

At 20000 Hz $lambda = 75$ mm

share|cite|improve this answer

edited 54 mins ago

answered 1 hour ago

tomtom

6,39711627

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
  $endgroup$
  – Ubaid Hassan
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
  $endgroup$
  – dmckee♦
  25 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Higher frequency provides higher resolution.

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

$endgroup$

add a comment | 

1

$begingroup$

Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.

The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.

On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
$$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
where $lambda$ corresponds to the wavelength and $d$ is the size of point.

Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.

And as it is known,
$$lambda = c/f,$$
where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.

share|cite|improve this answer

edited 8 mins ago

answered 10 mins ago

DanielTuzesDanielTuzes

1885

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe "enter link description here" is not the link description you want.
  $endgroup$
  – dmckee♦
  9 mins ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

I think the simple answer here is resolution.

Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.

If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by

$$lambda = {c over f} $$

so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....

The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate

$$lambda = 0.001 {rm m} = 1 {rm mm}$$

At 20000 Hz $lambda = 75$ mm

share|cite|improve this answer

edited 54 mins ago

answered 1 hour ago

tomtom

6,39711627

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
  $endgroup$
  – Ubaid Hassan
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
  $endgroup$
  – dmckee♦
  25 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

I think the simple answer here is resolution.

Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.

If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by

$$lambda = {c over f} $$

so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....

The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate

$$lambda = 0.001 {rm m} = 1 {rm mm}$$

At 20000 Hz $lambda = 75$ mm

share|cite|improve this answer

edited 54 mins ago

answered 1 hour ago

tomtom

6,39711627

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
  $endgroup$
  – Ubaid Hassan
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The correct statement is in terms of wavelength. That said, for non-dispersive waves (which is reasonably true for sound) in consistent media (not really true for medical ultra-sounds) the wavelength statements implies the frequency one as a corollary.
  $endgroup$
  – dmckee♦
  25 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

I think the simple answer here is resolution.

Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.

If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by

$$lambda = {c over f} $$

so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....

The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate

$$lambda = 0.001 {rm m} = 1 {rm mm}$$

At 20000 Hz $lambda = 75$ mm

share|cite|improve this answer

edited 54 mins ago

answered 1 hour ago

tomtom

6,39711627

$endgroup$

share|cite|improve this answer

edited 54 mins ago

answered 1 hour ago

tomtom

6,39711627

answered 1 hour ago

tomtom

6,39711627

answered 1 hour ago

tomtom

6,39711627

1

$begingroup$

Higher frequency provides higher resolution.

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

$endgroup$

add a comment | 

1

$begingroup$

Higher frequency provides higher resolution.

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

$endgroup$

add a comment | 

$begingroup$

Higher frequency provides higher resolution.

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

$endgroup$

share|cite|improve this answer

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

answered 1 hour ago

akhmeteliakhmeteli

18.5k21844

1

$begingroup$

Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.

The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.

On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
$$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
where $lambda$ corresponds to the wavelength and $d$ is the size of point.

Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.

And as it is known,
$$lambda = c/f,$$
where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.

share|cite|improve this answer

edited 8 mins ago

answered 10 mins ago

DanielTuzesDanielTuzes

1885

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe "enter link description here" is not the link description you want.
  $endgroup$
  – dmckee♦
  9 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.

The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.

On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
$$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
where $lambda$ corresponds to the wavelength and $d$ is the size of point.

Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.

And as it is known,
$$lambda = c/f,$$
where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.

share|cite|improve this answer

edited 8 mins ago

answered 10 mins ago

DanielTuzesDanielTuzes

1885

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe "enter link description here" is not the link description you want.
  $endgroup$
  – dmckee♦
  9 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.

The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.

On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
$$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
where $lambda$ corresponds to the wavelength and $d$ is the size of point.

Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.

And as it is known,
$$lambda = c/f,$$
where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.

share|cite|improve this answer

edited 8 mins ago

answered 10 mins ago

DanielTuzesDanielTuzes

1885

$endgroup$

share|cite|improve this answer

edited 8 mins ago

answered 10 mins ago

DanielTuzesDanielTuzes

1885

answered 10 mins ago

DanielTuzesDanielTuzes

1885

answered 10 mins ago

DanielTuzesDanielTuzes

1885