Question

(a) How far apart are two layers of tissue that produce echoes having round-trip times (used to measure distances) that differ by \[{\rm{0}}{\rm{.750 \mu s}}\]? (b) What minimum frequency must the ultrasound have to see the detail of this small?

Short answer

(a)Two layers of tissue will be far apart by a distance of\[5.77 \times {10^{ - 4}}{\rm{ }}m\]

(b) Minimum frequency will be \[{f_{\min }} = 2.66 \times {10^6}{\rm{ }}Hz\].

Step-by-step solution

Formula for calculation of distance

\[v = \frac{{\Delta d}}{{\Delta t}}\] ….. (1)

Here, \[v\] is the speed of sound, \[\Delta d\] which is the distance between the transmission and reflecting surface which is called depth.

And \[\Delta t\] represents the time when ultrasound waves just fall on the reflecting surface of the tissue.

(a) Distance between two layers of tissue:

Consider the given data as below.

The speed of sound in tissue,\[v = 1540{\rm{ }}{m \mathord{\left/

{\vphantom {m s}} \right.

\kern-\nulldelimiterspace} s}\]

Time delay, \[\Delta t = 0.750{\rm{ }}\mu s = 7.5 \times {10^{ - 7}}{\rm{ }}s\]

This is the time of round trip of sound wave

Here the first trip is completed when the wave starts from the first layer and ends on the second.

And the second trip is completed when this wave is again coming back to the first layer.

Then time to reach the reflective surface will be\[\frac{{\Delta t}}{2}\].

Then actual formula from depth will be

\[v = \frac{{2\Delta d}}{{\Delta t}}\]

The depth\[\Delta d\]will be as follows.

\begin{aligned}2\Delta d &= v\Delta t\\\Delta d &= \frac{{v\Delta t}}{2}\end{aligned}

Now put the values

\begin{aligned}\Delta d &= \frac{{1540 \times 7.5 \times {{10}^{ - 7}}}}{2}\\ &= 5.77 \times {10^{ - 4}}{\rm{ }}m\end{aligned}

Hence, two layers of tissue will be far apart by a distance of \[5.77 \times {10^{ - 4}}{\rm{ }}m\].

(b) Minimum frequency of ultrasound

Ultrasound frequency is defined as the number of ultrasound waves per second. OR The product of frequency and wavelength is the speed of the wave.

The minimum frequency is given as

\[v = {f_{\min }}\lambda \]

Here, \[v\] the speed of sound in tissue,\[{f_{\min }}\] is the minimum frequency, and\[\lambda \] is the wavelength of ultrasound wave that would be equal to the small traveling distance of wave in human tissue and it would be equal to the distance between two layers of tissue.

The speed of tissue,\[v = 1540{\rm{ }}m{\rm{ }}{s^{ - 1}}\]

The wavelength,\[\lambda = 5.77 \times {10^{ - 4}}{\rm{ }}m\]

Then we will get

\begin{aligned}{f_{\min }} = \frac{v}{\lambda }\\ = \frac{{1540}}{{3.77 \times {{10}^{ - 4}}}}\\ = 2.66 \times {10^{6{\rm{ }}}}Hz\end{aligned}

Hence, the minimum frequency of ultrasound will be \[2.66 \times {10^{6{\rm{ }}}}Hz\].