(a) Find the size of the smallest detail observable in human tissue with 20.0-MHz ultrasound. (b) Is its effective penetration depth great enough to examine the entire eye (about 3.00 cm is needed)? (c) What is the wavelength of such ultrasound in $0^{\circ }\text{C}$ air?

Video Transcript

This is College Physics Answers with Shaun Dychko. Suppose the frequency of some ultrasound being used is 20 megahertz which is 20.0 times 10 to the 6 hertz; question (a) is asking us what is the smallest observable feature given this frequency? Well the smallest observable feature will be about the wavelength of the sound. and the wave equation is the speed of the sound is its frequency multiplied by its wavelength so we can solve for λ by dividing both sides by the frequency. So we know the speed of sound in human tissue is 1540 meters per second so we divide that by 20.0 times 10 to the 6 hertz to get 77 micrometers would be the smallest observable detail. The maximum depth that the ultrasound can penetrate we are told is about 500 times the wavelength. So that's 500 times 7.70 times 10 to the minus 5 meters and that works out to 3.9 centimeters and yes that is deep enough to measure the eye ball since it's greater than 3.0 centimeters and 3.0 centimeters is the amount needed to measure in the eye. And then part (c) asks what would the wavelength of this ultrasound be in air and so the difference is that the speed of sound in air is different and if we look at table [17.1], the speed of sound in air at 0 degrees Celsius is 331 meters per second. So we have 331 divided by the frequency and that is 17 micrometers would be the wavelength in air.