Why use ultrasound for medical imaging? [on hold] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Moderator Election Q&A - Question CollectionIs there an upper frequency limit to ultrasound?How is it possible for an Ultrasound device to correctly interpret a negative density change in tissue?How specifically does an MRI machine build an image from received radio wavesIs it possible to send modulated ultrasound wave from underwater to air?Why the Doppler Ultrasound beam needs to be looking directly down at a pipeIntensity of an ultrasound beam?Dynamic range of ultrasound machine expressed in dBUltrasound wave/beam generationDoppler effect of sound waves in bloodAveraging speed of ultrasound between two differnt boundaires
How come Sam didn't become Lord of Horn Hill?
Circuit to "zoom in" on mV fluctuations of a DC signal?
Is it cost-effective to upgrade an old-ish Giant Escape R3 commuter bike with entry-level branded parts (wheels, drivetrain)?
Is "Reachable Object" really an NP-complete problem?
Is there a kind of relay only consumes power when switching?
Is it a good idea to use CNN to classify 1D signal?
また usage in a dictionary
Is it fair for a professor to grade us on the possession of past papers?
If a VARCHAR(MAX) column is included in an index, is the entire value always stored in the index page(s)?
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
How do I find out the mythology and history of my Fortress?
What does this Jacques Hadamard quote mean?
Can anything be seen from the center of the Boötes void? How dark would it be?
How to tell that you are a giant?
Significance of Cersei's obsession with elephants?
Around usage results
Delete nth line from bottom
Amount of permutations on an NxNxN Rubik's Cube
What does "lightly crushed" mean for cardamon pods?
Why are there no cargo aircraft with "flying wing" design?
Why didn't Eitri join the fight?
Do jazz musicians improvise on the parent scale in addition to the chord-scales?
Should I use a zero-interest credit card for a large one-time purchase?
If a contract sometimes uses the wrong name, is it still valid?
Why use ultrasound for medical imaging? [on hold]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Moderator Election Q&A - Question CollectionIs there an upper frequency limit to ultrasound?How is it possible for an Ultrasound device to correctly interpret a negative density change in tissue?How specifically does an MRI machine build an image from received radio wavesIs it possible to send modulated ultrasound wave from underwater to air?Why the Doppler Ultrasound beam needs to be looking directly down at a pipeIntensity of an ultrasound beam?Dynamic range of ultrasound machine expressed in dBUltrasound wave/beam generationDoppler effect of sound waves in bloodAveraging speed of ultrasound between two differnt boundaires
$begingroup$
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
edited Apr 13 at 21:17
Qmechanic♦
108k122001248
asked Apr 13 at 21:07
Ubaid HassanUbaid Hassan
38014
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.
$endgroup$
put on hold as off-topic by Pieter, GiorgioP, Jon Custer, ZeroTheHero, Kyle Kanos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Pieter, GiorgioP, Jon Custer
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
edited Apr 13 at 21:17
Qmechanic♦
108k122001248
asked Apr 13 at 21:07
Ubaid HassanUbaid Hassan
38014
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.
$endgroup$
put on hold as off-topic by Pieter, GiorgioP, Jon Custer, ZeroTheHero, Kyle Kanos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Pieter, GiorgioP, Jon Custer
$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:56
1
$begingroup$
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♦
Apr 13 at 22:00
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
edited Apr 13 at 21:17
Qmechanic♦
108k122001248
asked Apr 13 at 21:07
Ubaid HassanUbaid Hassan
38014
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.
$endgroup$
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
energy acoustics frequency wavelength medical-physics
edited Apr 13 at 21:17
Qmechanic♦
108k122001248
asked Apr 13 at 21:07
Ubaid HassanUbaid Hassan
38014
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 Apr 13 at 21:17
Qmechanic♦
108k122001248
asked Apr 13 at 21:07
Ubaid HassanUbaid Hassan
38014
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 Apr 13 at 21:17
Qmechanic♦
108k122001248
edited Apr 13 at 21:17
Qmechanic♦
108k122001248
edited Apr 13 at 21:17
Qmechanic♦
108k122001248
108k122001248
asked Apr 13 at 21:07
Ubaid HassanUbaid Hassan
38014
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 Apr 13 at 21:07
Ubaid HassanUbaid Hassan
38014
asked Apr 13 at 21:07
Ubaid HassanUbaid Hassan
38014
38014
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.
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.
Ubaid Hassan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Pieter, GiorgioP, Jon Custer, ZeroTheHero, Kyle Kanos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Pieter, GiorgioP, Jon Custer
put on hold as off-topic by Pieter, GiorgioP, Jon Custer, ZeroTheHero, Kyle Kanos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Pieter, GiorgioP, Jon Custer
$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:56
1
$begingroup$
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♦
Apr 13 at 22:00
add a comment |
$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:56
1
$begingroup$
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♦
Apr 13 at 22:00
$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:56
$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:56
1
1
$begingroup$
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♦
Apr 13 at 22:00
$begingroup$
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♦
Apr 13 at 22:00
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
answered Apr 13 at 21:15
tomtom
6,45411628
$endgroup$
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:59
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♦
Apr 13 at 22:03
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 discs (or spheres).
The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, 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 disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.
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.
Estimation
You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20textKHz$ the resolvable distance is
$$frac1500textm/texts2 cdot 20textkHz = 3.75textcm.$$
We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.
answered Apr 13 at 22:17
DanielTuzesDanielTuzes
26116
$endgroup$
add a comment |
$begingroup$
Higher frequency provides higher resolution.
answered Apr 13 at 21:15
akhmeteliakhmeteli
18.5k21844
$endgroup$
$begingroup$
to be honest, it is more of a comment than an answer.
$endgroup$
– Nilay Ghosh
Apr 14 at 16:26
$begingroup$
@NilayGhosh : This is just your opinion. According to the rules (physics.stackexchange.com/help/how-to-answer), "Any answer that gets the asker going in the right direction is helpful, but do try to mention any limitations, assumptions or simplifications in your answer. Brevity is acceptable, but fuller explanations are better."
$endgroup$
– akhmeteli
Apr 14 at 17:20
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
answered Apr 13 at 21:15
tomtom
6,45411628
$endgroup$
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:59
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♦
Apr 13 at 22:03
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
answered Apr 13 at 21:15
tomtom
6,45411628
$endgroup$
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:59
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♦
Apr 13 at 22:03
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
answered Apr 13 at 21:15
tomtom
6,45411628
$endgroup$
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
answered Apr 13 at 21:15
tomtom
6,45411628
edited Apr 13 at 21:34
answered Apr 13 at 21:15
tomtom
6,45411628
answered Apr 13 at 21:15
tomtom
6,45411628
answered Apr 13 at 21:15
tomtom
6,45411628
6,45411628
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:59
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♦
Apr 13 at 22:03
add a comment |
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:59
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♦
Apr 13 at 22:03
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:59
$begingroup$
Is it correct to say wavelength is inversely proportional to resolution? Or that sound wave frequency is directly proportional to resolution?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:59
1
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♦
Apr 13 at 22:03
$begingroup$
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♦
Apr 13 at 22:03
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 discs (or spheres).
The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, 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 disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.
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.
Estimation
You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20textKHz$ the resolvable distance is
$$frac1500textm/texts2 cdot 20textkHz = 3.75textcm.$$
We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.
answered Apr 13 at 22:17
DanielTuzesDanielTuzes
26116
$endgroup$
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 discs (or spheres).
The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, 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 disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.
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.
Estimation
You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20textKHz$ the resolvable distance is
$$frac1500textm/texts2 cdot 20textkHz = 3.75textcm.$$
We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.
answered Apr 13 at 22:17
DanielTuzesDanielTuzes
26116
$endgroup$
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 discs (or spheres).
The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, 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 disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.
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.
Estimation
You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20textKHz$ the resolvable distance is
$$frac1500textm/texts2 cdot 20textkHz = 3.75textcm.$$
We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.
answered Apr 13 at 22:17
DanielTuzesDanielTuzes
26116
$endgroup$
Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 discs (or spheres).
The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, 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 disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.
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.
Estimation
You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20textKHz$ the resolvable distance is
$$frac1500textm/texts2 cdot 20textkHz = 3.75textcm.$$
We considered only 2D imaging here, but similar applies to 3D where you'd like to get information in 3D. In 3D one measures the time difference between the emission and detection of the signal, and using the $v=s/t$ formula, the distance of the object can be calculated.
answered Apr 13 at 22:17
DanielTuzesDanielTuzes
26116
edited Apr 13 at 23:28
answered Apr 13 at 22:17
DanielTuzesDanielTuzes
26116
answered Apr 13 at 22:17
DanielTuzesDanielTuzes
26116
answered Apr 13 at 22:17
DanielTuzesDanielTuzes
26116
26116
add a comment |
add a comment |
$begingroup$
Higher frequency provides higher resolution.
answered Apr 13 at 21:15
akhmeteliakhmeteli
18.5k21844
$endgroup$
$begingroup$
to be honest, it is more of a comment than an answer.
$endgroup$
– Nilay Ghosh
Apr 14 at 16:26
$begingroup$
@NilayGhosh : This is just your opinion. According to the rules (physics.stackexchange.com/help/how-to-answer), "Any answer that gets the asker going in the right direction is helpful, but do try to mention any limitations, assumptions or simplifications in your answer. Brevity is acceptable, but fuller explanations are better."
$endgroup$
– akhmeteli
Apr 14 at 17:20
add a comment |
$begingroup$
Higher frequency provides higher resolution.
answered Apr 13 at 21:15
akhmeteliakhmeteli
18.5k21844
$endgroup$
$begingroup$
to be honest, it is more of a comment than an answer.
$endgroup$
– Nilay Ghosh
Apr 14 at 16:26
$begingroup$
@NilayGhosh : This is just your opinion. According to the rules (physics.stackexchange.com/help/how-to-answer), "Any answer that gets the asker going in the right direction is helpful, but do try to mention any limitations, assumptions or simplifications in your answer. Brevity is acceptable, but fuller explanations are better."
$endgroup$
– akhmeteli
Apr 14 at 17:20
add a comment |
$begingroup$
Higher frequency provides higher resolution.
answered Apr 13 at 21:15
akhmeteliakhmeteli
18.5k21844
$endgroup$
Higher frequency provides higher resolution.
answered Apr 13 at 21:15
akhmeteliakhmeteli
18.5k21844
answered Apr 13 at 21:15
akhmeteliakhmeteli
18.5k21844
answered Apr 13 at 21:15
akhmeteliakhmeteli
18.5k21844
answered Apr 13 at 21:15
akhmeteliakhmeteli
18.5k21844
18.5k21844
$begingroup$
to be honest, it is more of a comment than an answer.
$endgroup$
– Nilay Ghosh
Apr 14 at 16:26
$begingroup$
@NilayGhosh : This is just your opinion. According to the rules (physics.stackexchange.com/help/how-to-answer), "Any answer that gets the asker going in the right direction is helpful, but do try to mention any limitations, assumptions or simplifications in your answer. Brevity is acceptable, but fuller explanations are better."
$endgroup$
– akhmeteli
Apr 14 at 17:20
add a comment |
$begingroup$
to be honest, it is more of a comment than an answer.
$endgroup$
– Nilay Ghosh
Apr 14 at 16:26
$begingroup$
@NilayGhosh : This is just your opinion. According to the rules (physics.stackexchange.com/help/how-to-answer), "Any answer that gets the asker going in the right direction is helpful, but do try to mention any limitations, assumptions or simplifications in your answer. Brevity is acceptable, but fuller explanations are better."
$endgroup$
– akhmeteli
Apr 14 at 17:20
$begingroup$
to be honest, it is more of a comment than an answer.
$endgroup$
– Nilay Ghosh
Apr 14 at 16:26
$begingroup$
to be honest, it is more of a comment than an answer.
$endgroup$
– Nilay Ghosh
Apr 14 at 16:26
$begingroup$
@NilayGhosh : This is just your opinion. According to the rules (physics.stackexchange.com/help/how-to-answer), "Any answer that gets the asker going in the right direction is helpful, but do try to mention any limitations, assumptions or simplifications in your answer. Brevity is acceptable, but fuller explanations are better."
$endgroup$
– akhmeteli
Apr 14 at 17:20
$begingroup$
@NilayGhosh : This is just your opinion. According to the rules (physics.stackexchange.com/help/how-to-answer), "Any answer that gets the asker going in the right direction is helpful, but do try to mention any limitations, assumptions or simplifications in your answer. Brevity is acceptable, but fuller explanations are better."
$endgroup$
– akhmeteli
Apr 14 at 17:20
add a comment |
$begingroup$
your edits are quite helpful, but can I ask why you do edits so frequently?
$endgroup$
– Ubaid Hassan
Apr 13 at 21:56
1
$begingroup$
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♦
Apr 13 at 22:00