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

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  • $\begingroup$ your edits are quite helpful, but can I ask why you do edits so frequently? $\endgroup$ – Ubaid Hassan Apr 13 at 21:56
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    $\begingroup$ 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
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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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $20\text{KHz}$ the resolvable distance is $$\frac{{1500{\text{m}}/{\text{s}}}}{{2 \cdot 20{\text{kHz}}}} = 3.75{\text{cm}}.$$


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.

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Higher frequency provides higher resolution.

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  • $\begingroup$ 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

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