Final answer:
The separation between two layers of tissue producing echoes with round-trip times differing by 0.750 µs is approximately 0.576 mm. The minimum frequency required to see details this small is approximately 1.34 MHz.
Step-by-step explanation:
The separation between two layers of tissue producing echoes with round-trip times differing by 0.750 μs can be determined using the formula:
Separation = Speed of Sound x Round-trip Time / 2
To find the separation in meters, we need to convert the round-trip time from microseconds to seconds:
0.750 μs = 0.750 x 10-6 s
Assuming the speed of sound in tissue is approximately 1540 m/s, we can calculate the separation as follows:
Separation = (1540 m/s) x (0.750 x 10-6 s) / 2
Using the given values, the separation between two layers of tissue is approximately 0.576 mm.
To see details this small, the ultrasound must have a frequency that is inversely proportional to the wavelength of the sound. The minimum frequency required can be calculated using the formula:
Minimum Frequency = Speed of Sound / Wavelength
Assuming the speed of sound in tissue is still 1540 m/s and the wavelength is equal to two times the separation, we can calculate the minimum frequency:
Minimum Frequency = 1540 m/s / (2 x 0.576 x 10-3 m)
Using the given values, the minimum frequency required to see details this small is approximately 1.34 MHz.