In the clinical use of ultrasound, transducers are always coupled to the skin by a thin layer of gel or oil, replacing the air that would otherwise exist between the transducer and the skin. (a) Using the values of acoustic impedance given in Table 17.5 calculate the intensity reflection coefficient between transducer material and air. (b) Calculate the intensity reflection coefficient between transducer material and gel (assuming for this problem that its acoustic impedance is identical to that of water). (c) Based on the results of your calculations, explain why the gel is used.

Video Transcript

This is College Physics Answers with Shaun Dychko. We are going to see in this question why there always needs to be some gel in between an ultrasound transducer and a patient's skin. So this is the skin and this is the transducer emitting the ultrasound; there needs to be some goop here that will allow the sound to transmit from the transducer into the patient's skin and then bounce off whatever thing the technician wants to look at such as maybe a fetus growing in there or maybe somebody's heart and so on. Okay! So we can find the ratio of the intensity of the sound that's transmitted compared to what's incident by using this formula here— that's what this intensity reflection coefficient is; it's the intensity that's transmitted divided by the intensity that's incident or actually, it's the intensity reflected I should say because it's the intensity reflection coefficient after all. So at this interface or this boundary between the transducer and this gel is gonna be some incident intensity initially and then there's gonna be some reflection and that's gonna have some intensity as well and the ratio of these two intensities is what this α tells us. Okay! So if this ratio is 1 as it is for when there's air that means all the sound is getting reflected but let's figure out why this is 1. So we plug in the acoustic impedance for the transducer material which we look up in this data table [17.5] and that's 30.8 times 10 to the 6 kilograms per meters squared per second and the acoustic impedance for air is 429 and then for the gel we are told to assume it's the same as that of water which is 1.5 times 10 to the 6 and so we write those pieces of information down. So this is the second material which in part (a) we are assuming is air minus the acoustic impedance of the first material which is the transducer material squared divided by the sum of the acoustic impedance's squared equals this intensity reflection coefficient and that's 1.00 here so all the sound is reflected when you round this to three significant figures anyhow. I mean there's probably some negligible amount that's transmitted but essentially, this device is not going to work because there's no reflected sound from things inside the body because none of this sound gets in there; it's getting all reflected at this interface between the transducer and the air so part (a) there's just air here. Okay! So in part (b), this is when we put in some goop and so now the acoustic impedance of the second material is that of water or the gel and we take that difference, square it and then take the sum of the acoustic impedance's squared and we get 0.823. So that's still quite high— there's a lot reflected— but nevertheless, you know, at least 18 percent or so gets transmitted and then that sound will get reflected by features inside the body and then be able to, you know, tell the technician and the medical professionals what's inside the body. So the gel needs to be used otherwise no sound is transmitted into the patient's body.