Transcript for:
Ultrasound Echo Physics

hi learners it's m from sound nerds and this video is the second half of unit six echoes section six b point one what are echoes if you have ever stood in a cave or in a canyon there's a good chance that you've experienced an echo of your voice if you yell into the space the sound wave your voice creates will bounce off a hard surface and return to your ears and ultrasound really isn't just yelling into a space that is your body but the high frequency sound waves created will interact with tissue and some of that sound is reflected like an echo back to the transducer where it is processed and turned into our images without echoes ultrasound imaging would not exist but what are these echoes well as sound enters the body it's going to interact with multiple tissue types and interfaces an interface is an area in which two different media are in contact with one another some interfaces are really large and some are going to be really small however at every interface the sound can do one or more of these things it can be absorbed reflected scattered refracted or transmitted those interactions of the sound in the body are then recorded by the machine and mapped into an image the color or the gray scale that we see is dependent on the strength of the reflection stronger reflectors are going to come back white and when there's no reflection it comes back as black and the different strings in between come back as different levels of gray now we've already discussed absorption as a mechanism of attenuation so we're going to focus more on reflection scattering refracting and transmission as they are more directly related to the sound and creation of echoes when a sound beam is propagating through the body and interacts with a large interface typically those that are larger than a wavelength a reflection is more likely to occur there are two types of reflections specular or at least near specular and diffuse the structures that create specular reflections are more likely going to be large interfaces that are smooth or nearly smooth the echoes that return from these interfaces tend to be stronger and are going to return towards the transducer since specular reflections are stronger and come from smooth edge objects they tend to be displayed as bright white linear echoes a true specular reflection occurs when a wave strikes a boundary that is smooth and flat so think of a mirror and just like in the mirror you cannot see a reflection unless you are directly in front of it so like the mirror the angle at which the wave strikes the surface will matter we'll talk more about this when we talk about the physics of incidence in general though if the sound strikes a specular reflector perpendicularly which means at 90 degrees the echo will return directly back to the transducer this is what creates the really strong bright echo if the sound comes in at an angle though the echo will leave at the same angle and is less likely to return to the transducer most surfaces within the body are not true specular surfaces though they have a little bit of rough edges so we would consider them to be more near specular surfaces with that they tend to be a little bit more forgiving on the direction to which sound will be returned the near specular surfaces are a little more rough and when sound strikes them they will send sound back to the transducer but are a little less angle-dependent however many of these reflectors still return better echoes at 90 degrees so we want to attempt perpendicular insulation when we can some examples of specular reflectors that we see around the body include the diaphragm which if you recall me saying the specular reflectors are going to be linear bright echoes so this is our diaphragm here nice and bright we've got liver on this side lungs would be on this side vessel walls are also specular reflectors this is a carotid artery we can see that the vessel wall is bright white and linear valve leaflets are specular reflectors we can see the valves here and fetal bones are another example of specular reflectors one thing i want to point out about all of these pictures is that look at how the sound is traveling into the body in connection to that specular reflector in our diaphragm example sound is traveling perpendicular to the diaphragm same with the vessel sound is traveling basically perpendicular on these valve leaflets basically perpendicular so the more perpendicular that our specular reflectors sit within the sound beams the stronger their echoes are going to be diffuse reflectors are still large interfaces but they are going to have much rougher surfaces these surfaces can be found in organs or at interfaces diffuse reflectors send most of their echoes back in the direction of the transducer but they are a little bit weaker and a little less predictable on how they are going to travel as they are not always aimed directly at the transducer for the diffuse reflector think about tin foil i know this might sound kind of weird but look at this tin foil ball it is reflecting light actually pretty well and that's kind of what the diffuse reflectors will do as well they're lumpy and bumpy like these but they reflect the sound pretty well and they're going to reflect most of the sound back towards the transducer we'll lose a little bit off to the sides and that's okay because we're going to get tons of information from other sound beams as they enter the body so again as the sound travels into the body it's going to interact with a rougher less smooth surface and as that sound interacts with the reflector some sound is directed back towards the transducer but we're going to lose a little bit off to the sides as well diffuse reflectors are responsible for a lot of the solid tissue we can see within the organ where large interfaces occur for example we've got chamber walls as diffuse reflectors the valve leaflets we consider to be more specular reflectors especially when we're perpendicular to them but we can also see kind of these brighter fuzzier rough surface of the chamber walls those are going to be diffuse reflectors another example is the collecting system within a kidney here we have the kidney parenchyma this brighter part represents the collecting system and we can see that these are relatively bright echoes that are returning to the transducer but they have a very rough surface in there therefore creating a different type of echo than those strong linear echoes we are getting from the specular reflectors we've seen already that diffuse reflectors can cause scattering and we're going to learn more about the small non-specular sources that also cause scattering these two reflections are considered backscatter the difference between the two is that diffuse reflectors send unorganized sound mostly back in the direction of the transducer where scattering reflectors send unorganized sound any which random direction that it wants to send it in scattering though is really the backbone of ultrasound it's the echoes that are produced from non-angle-dependent reflectors that make up the majority of the images are going to be expressed with different patterns of interference the pattern of interference is called acoustic speckle and that is why we're able to see the tissue or parenchyma of the organs muscles fat and other connective tissues in the body so basically what happens the sound beam is going to go in and this one is a little upside down so it's actually talking about light but pretend it's the other way so we've got our sound beam coming in and it's going to interact with these small reflectors when the sound beam hits these small reflectors they're both going to create their own wave echoes but as we learned sound can interfere with other sound waves they can deconstruct cancel each other out or they can construct and make bigger waves so we see in this example here lots of destruction occurring where these waves are meeting up it's the patterns that arise from this interference that is displayed to create our organ parenchma a lot of people think that those little dots on the screen the little greens are actually small cells that we can see in the organ or represent something else that's actually in the organ but in reality it's just its sound interference pattern that is being displayed so that speckle look that we get is called the acoustic speckle it's the pattern of the interference that's coming back there are some machines that provide speckle reduction which will kind of smooth out the graininess but when you smooth out the graininess of the acoustic speckle you're also kind of smoothing out some of those specular reflectors that we get as well and therefore our images kind of look like they have like a just a general blur to them so that'll come down to sonographer preference if they like that grainy look versus the smoothed out look if we stop to think about how the body is put together it makes sense that there are so many interfaces and so many small little structures that create our human bodies that create the tissue within our bodies and so there are going to be tons of little small structures in the body and the sound is going to interact with all of them and every interaction that sound has with those little interfaces is going to cause the sound to send back very weak echoes in a bunch of random directions because that's the scattering some are going to be towards the transducer so we're going to be towards the side of the body some are going to be away from the transducer it's just going to send it in random unorganized directions as each of these interfaces interact with the sound they're going to create their own echo which will be a lot of echoes and then those echoes are going to construct and destruct with one another creating that unique pattern to the tissue or to the pathology and then it gets displayed as tiny little grainy dots that make up the parenchyma we mentioned in the attenuation unit that high frequency transducers are more likely to attenuate quicker and that is because in the part of scattering and that is because high frequency waves are going to scatter more than a low frequency wave so here we have our sound interacting with one tiny small little reflector and all these echoes are getting bounced in all sorts of different directions and this is just one tiny little sound wave if we add in thousands more to this picture we're going to get lots of different echoes heading in lots of different directions and that is where that constructive interference and destructive interference is going to come into play they are going to create a pattern that is unique to the tissue that they are coming from so some examples of scattering reflections include really anything that's made up of tissue or a parenchyma so we have our liver tissue here this is all due to scattering so again a lot of people kind of think that all these little graininess all these little dots are the cells itself of the liver or little lobules within the liver it's not true it's just the way that the liver parenchyma interferes with itself and how the machine gets those echoes back to create its own acoustic speckle pattern here's another example this is a testicle again with its own acoustic speckle pattern now there is a special type of scattering that occurs and it's called rayleigh scattering rayleigh scattering is going to occur with very very small reflectors typically going to be with red blood cells when we were talking about normal back scanner we kind of described it as unorganized and random and that it could go in any direction that it felt like rayleigh scattering is a little bit different it is going to be very predictable very organized and it sends sound equally in all directions so we call it omnidirectional and just like scattering rayleigh scattering is going to be very susceptible to the increase of frequency so much to the point that rayleigh scattering is proportional to frequency to the fourth power and that's a lot so we have rayleigh scattering directly proportional to the frequency if we were to increase the frequency so for example just by doubling the frequency rayleigh scattering is going to increase by a factor of 16 because 2 times 2 times 2 times 2 is 16 that's 2 to the fourth if we were to triple frequency the scattering effect would increase by 81 times this is why we want to use low frequency transducers for doppler imaging the echoes from the scatter are already weak they're even weaker from relay scattering we don't want to compromise by using too high of a frequency and not getting good enough signal back from those red blood cells reflection and scattering are responsible for returning echoes from an interface but it's important to remember that not all sound energy is sent back directly to the transducer in fact a really small fraction of the energy is sent back to the transducer where most of the energy is going to continue to move forward and that energy moving forward is considered the transmit wave so the transmit wave is a sound energy that continues beyond the interface without the transmit wave ultrasound really wouldn't be all that useful if the first reflector sent all that sound energy right back to the transducer we wouldn't be able to see hardly at all into the body it is a transmit wave that allows us to see deeper and deeper into the body but we do know that as the sound travels it will weaken or attenuate over time as that transmit wave continues from the interface and into a new type of media there is potential for that wave to refract refraction is going to be the change in direction of the transmission as sound travels into a new medium now refraction won't always occur as it really does depend on the angle that the sound comes into the interface and then also the propagation speed of where the sound started and where it's going if the propagation speeds are different and it comes in at an angle refraction will occur a refraction can cause artifacts where anatomy is duplicated side to side because a redirected sound is still going to interact with reflectors just like the transmitted wave that moved forward would have and some of those reflections are going to make it back to the transducer the machine can't tell them apart from real ones and so it'll process it as real echoes section 6b 0.2 rules of the road so now that we know a little bit about echoes and how they're created and that sound will be transmitted at interfaces we need to learn about the physics behind reflection and transmission we're going to do that in sections three and four of this unit but first i want to define a few vocabulary words and i also want to lay out some ground rules behind the physics of reflection and transmission the first vocabulary word that i want to cover with you is called impedance now impedance is going to be the resistance to the propagation of sound in a medium acoustic impedance is one of the determining factors of an interface when two media are adjacent to one another one of the ways that sound can be reflected is if there is a change of impedance from one media to the next the impedance of a medium is calculated and it's calculated using this formula so we have z which represents impedance in the unit of rails equal to density multiplied by the propagation speed to expand on impedance a little bit more and use some relationships that we can pull from this formula we can see that if a medium has increased density then it's going to be harder for sound to travel through it there's going to be more resistance more impedance if a medium has increased propagation speeds it's also going to be harder for sound to travel through again it's causing more resistance to the actual propagation of sound so impedance is determined by the physical characteristics of the medium which in turn affects how the ultrasound will interact at certain interfaces the next two terms that i want to cover are angles we have oblique angles and then we also have a right angle now oblique angles are any angle that isn't 90 degrees 90 degrees refers to the right angle so anything other than 90 is considered an oblique angle there are subsets of oblique angles you can have acute angles which are angles that measure less than 90 and you can have of two singles which are angles that measure more than 90. in our example we have the acute angle on top an obtuse angle on bottom but these are both oblique angles because neither of these are a right angle or an angle at 90 degrees this example on the bottom is indeed a right angle and that is an angle that measures exactly 90 degrees this is going to be created when two planes are perpendicular to one another so that perpendicular means they make a t with one another three more terms that can kind of be grouped together we have incidence normal incidence and oblique incidence incidence in itself is just the angle at which sound strikes a boundary so a sound comes into that interface no matter how it's coming in that's called the incidence but we can further describe the incidence as normal or oblique normal incidence is going to be when the sound beam strikes the boundary at a right angle now normal incidence comes along with some other synonyms we can call it perpendicular incidence orthogonal incidence and 90 degree incidence no matter what it is it is sound coming directly perpendicular to the face of the boundary oblique incidence then is going to be sound coming at any other angle than 90. so most of our sound waves are going to be traveling at oblique instances to our boundaries but as we mentioned back with specular reflectors that normal incidence that 90 degree incidence is really going to produce some really nice crisp reflections from those specular reflectors building on incidents we can talk about a few more pieces of the puzzle when we consider how the beam is coming into the boundary that is the incidence if we compare the line of the sound beam to an imaginary perpendicular line the angle that is created between the two is the incidence angle you might notice this little symbol here this is called theta this is the symbol that typically is used to denote an angle so again we have the sound beam coming in this is the incidence how it's striking that boundary the incidence angle then is that beam compared to this imaginary perpendicular line similar to incidence angle we then have a reflection angle as the beam is reflected back towards the transducer we are going to create a reflection angle and that is going to be between the reflection and that perpendicular line so this is the reflection angle lastly then we have the transmission angle and the transmission angle is going to be sound coming to the boundary at the boundary may be reflected it may continue through or it might do both but the transmission angle refers to the sound that is continuing forward it is the path of the transmission wave in comparison to that perpendicular line that is what creates the transmission angle we're going to talk a lot about medium one impedance one or speed one and compare that to medium two impedance two or speed two medium one is going to be the medium from which the sound is traveling from typically we have the sound coming in this direction medium two then is going to be the medium in which the sound is entering into so medium one has the sound medium two is getting the sound the boundary then is the interface between the two different media represented by this orange line here and here are all the pieces together so we have the incidence creating the incidence angle our reflection creating a reflection angle this is all based on the boundary as sound travels through we get the transmission wave which will create the transmission angle and we need to know that medium one currently has the wave medium two will be getting the transmission all of these vocabulary words are going to relate back to smooth specular surfaces you can see that our boundary is very flat very smooth and that is how we are going to refer to a lot of the physics behind ultrasound considering a smooth specular surface all these vocabulary words are also areas that are measurable and directions that the waves are taking we learned in our intensity chapter that all waves carry intensity so we consider the intensity that is approaching the boundary to be the incident intensity remember it's going to have a power over an area squared as a unit typically watts per centimeter squared or milliwatts per centimeter squared and for the purposes of this chapter we are also going to consider the intensity of the incident wave being 100 so this is all the energy that's coming into the boundary is found within the incidence beam coming towards the boundary it's going to hold the incidence intensity and be considered 100 of the intensity before it strikes the boundary once that wave strikes the boundary a few things can happen chest reflection can occur just transmission can occur or both can occur focusing on the reflection part of it whatever energy is diverted into the reflection is called the reflection intensity so this is considered the echo the part that is heading back towards the transducer another way to look at the intensity that is reflected back in the reflection wave is something called intensity reflection coefficient and basically that is just saying what percent of the main beam or the incident beam what percent of that has now transferred into the reflection beam just so happens that in clinical imaging most of the time this is very little typically less than one percent if there is transmission then we have the transmission intensity so just like reflection intensity the transmission intensity describes how much from the incidence beam is traveling forward into the next medium after the boundary so after this sound beam interacts with the boundary how much intensity is left in that forward transmission wave and just like reflection we can describe the transmission by a percentage what is the percent of energy that is traveling forward if the reflection holds less than one percent in clinical imaging then 99 or more of the energy is transferred forward so here are all those concepts wrapped up in one diagram again we have our incidence intensity watts per centimeter squared sometimes we consider this holding 100 percent of the power 100 of the intensity this is all before the sound gets to this purple boundary at the purple boundary if there is reflection it will divert some of that incidence intensity it'll take a percentage of it which is the intensity reflection coefficient and some of it will be transferred into the transmission intensity again watts per centimeter squared we can represent it as a percentage as the intensity transmission coefficient this image does show an oblique incidence to this boundary with these three descriptor words know that all three of these are also pertinent to the normal or perpendicular incidents as you'll see in a few slides so next up i have the six rules that i want you to focus on as you're studying the second half of this unit all of these rules will be discussed again as we talk about the physics of normal and oblique incidents but i just want to kind of compile them all together a place that you can easily reference back to them and give you some focus in your studies so rule number one says energy cannot be created or destroyed you probably have heard something very similar to this in previous physics classes because this is the basic conservation of energy law and it's going to apply to intensities and our intensity coefficients this rule is also going to apply to both normal and oblique incidents remember normal incidence is perpendicular 90 degree orthogonal where oblique was anything other than 90 degrees we're going to see these formulas again later but they are directly derived from the fact that ng cannot be created or destroyed so we're going to see that the incident intensity is going to be equal to the reflected intensity and transmitted intensity the energy from the incoming beam has to be split between the reflection and the transmission and if that is true it's also going to be true that 100 of the energy is divided again into the intensity reflection coefficient and the intensity transmission coefficient rule number two tells us that with normal incidence reflection will not occur if the impedances are equal so again normal incidence is that 90 degree perpendicular incidence we will not get any sound heading back to the transducer if the impedances are the same there has to be a difference in impedances for reflection to occur and that is only going to apply to normal incidence we will see that when there are small mismatched impedances there are going to be small reflections where really big mismatched impedances are going to create really big reflections rule number three relates back to actually rule number one and rule number two so with normal incidence 100 transmission will occur if the impedances are the same so rule number one said we can't create or destroy energy so if we have a hundred percent of that energy coming into the boundary and no reflection occurred then all that energy has to keep going forward through the transmission wave that is what this is saying this also is a throwback to rule number two because we just said that if the impedances are the same no reflection can occur which is why it lends itself to the fact that transmission is 100 when impedances are the same rule number four says reflection and transmission with oblique incidents cannot be predicted so we are going to have a very particular set of rules that are going to apply to normal incidents and then we're going to have to throw all those out the window and focus on oblique incidents and the tricky part about reflection and transmission with oblique incidence is that it kind of does what it wants to do there's really no predicting if sound will reflect transmit or do some combination of both rule number five does tell us one thing we do know about oblique incidence that if there is reflection the reflection angle is going to equal the incidence angle so no matter how that beam comes into the boundary with oblique incidence the reflection if it happens is going to leave at the same angle and rule number six there are two criteria that must be met for refraction to occur first one being that there must be oblique incidence the second one is that the two media express two different propagation speeds so as that sound comes into the boundary if it comes in obliquely that's meeting one of the criteria the second part though is that there has to be some sort of transmission we don't know what's going to happen with oblique incidents but if transmission occurs and the propagation speeds are different of both of the media then the wave will refract and travel in a different direction so these vocabulary words and these six rules are really going to be the cornerstone of reflection and transmission physics however we do need to take a little bit of a deeper dive into the actual physics of it by going over these rules and these vocabulary words again as we go through the physics hopefully these concepts will stick with you a little bit more and you will be better able to understand how sound is interacting with interfaces in the body now remember all of these interfaces that we are going to discuss during the physics portion are specular services basically they're smooth and they're flat the physics behind diffuse reflectors and all the other lumps and bumps in the body are not in the scope of this we just want to know how sound basically will interact with those strong reflective surfaces so we are going to start out with the physics of normal incidence for section 6v3 we know that when a sound beam comes to an interface at a normal incidence it is a perpendicular incidence or a 90 degree angle to the boundary for a reflection to occur do you remember what must be different it's the impedances the impedance of medium one has to be different from the impedance of medium two for reflection to occur with normal incidence if we have the same impedance of medium 1 and medium 2 no reflection is going to occur if we have a small mismatch between the impedances which really applies to most of the soft tissue in our body we're going to have smaller reflections when we have really large mismatches between impedances we're going to have large reflections a large mismatch example would be the sound coming out of your transducer and into air that is a huge mismatch causes a very large reflection to the point that hardly any sound if any is going to travel forward so recall that as that sound beam is traveling towards the boundary it's going to have an intensity and it really has all the power all the intensity as it heads towards that boundary once it hits that interface some of that energy is going to return as an echo assuming that this media has two different impedances as that perpendicular incidence comes to the boundary some of the intensity will be transferred into the reflection and some is going to be transferred into the transmission when you add the reflection intensity to the transmission intensity they should equal the incident intensity because we cannot create nor destroy any energy when we are talking about intensities then we are focusing on the unit of power over some sort of area squared but we can also look at these as percentages what percentage of the energy went into reflection and what percentage was transferred into transmission so if we consider that the incident beam holds 100 of the energy the percent that is transferred into the reflection is the intensity reflection coefficient abbreviated irc and the percent that is transferred to the transmission is the intensity transmission coefficient or itc when we add the irc percentage and the itc percentage together they should equal 100 if i were to give you the incidence intensity and either reflection intensity or transmission intensity it would actually be very easy to calculate the rest of the variables we can figure out the irc by dividing reflection intensity by incidence intensity just like we can figure out itc by dividing the transmission intensity by the incidence intensity once you have irc or itc you just need to subtract it from 100 to get the other variable so when we are dealing with the intensities of the incidence reflection and transmission all you really need are two of the pieces to be able to figure out the third piece and here's an example of looking at the intensities of the waves and figuring out the irc and the itc so if i told you that the incidence wave has an intensity of 60 milliwatts per centimeter squared and the reflected intensity is 2 milliwatts per centimeter squared we can take 2 divided by 60 and see that it equals 3 percent so the irc in this situation is 3 100 minus 3 leaves us with 97 for the itc now we can confirm that or we could have calculated that first by taking 58 and dividing it by 60. 58 divided by 60 is 97 100 minus 97 would have helped us to calculate the irc with the leftover because remember reflected intensity plus transmission intensity must equal the incident intensity and the irc plus the itc has to equal 100 i don't have any practice examples within this lecture but i do have them included in the activity section so you can try more of this math and double check yourself through the activity link before we leave behind the itc and irc and all the intensity and all of that i do want to show you some math that you might actually have to calculate on your tests or on your boards what you may be given is an intensity of an incident beam so let's say our beam is coming in at 100 milliwatts per centimeters squared with this we are going to have reflection and we are also going to have some sort of transmission let's say for this example that they tell us that our itc is 50 from here you might be asked then to calculate the intensity of the transmission or you may be asked to calculate the irc of the reflection or the intensity of the reflection starting with the itc and calculating intensity all we need to do is take 50 percent of our incident beam so 50 percent of 100 is 50 milliwatts per centimeter squared now the rest is relatively easy knowing that our irc plus our itc has to equal 100 we know then that 50 percent plus 50 is 100 we can then use our irc to calculate our reflection intensity 50 of 100 is also 50 milliwatts per centimeter squared now i use really easy values in this equation because i wanted to just to introduce you to the concept of using your coefficients to calculate your intensities let's try another example with a little bit different numbers remember you can't really use a calculator on the board so the numbers that they are going to give you are going to be relatively straightforward you should be able to do them in your head or with a little bit of scratch paper so this time let's say our incident intensity is 50 milliwatts per centimeter squared and this time it tells us that we have an irc of 20 percent so again we may be asked to figure out the reflection intensity or we may be asked to figure out the itc or the transmission intensity let's start with the reflection we need to then take 20 of 50 to calculate the intensity of the reflection 20 of 50 is 10 milliwatts per centimeters squared so now we have our reflection intensity and our irc using those values we can calculate the itc and the transmission intensity we know that irc plus itc has to equal 100 so 100 minus 20 gives us 80 percent for our itc now we can either take 80 of 50 to calculate the intensity of transmission or we can use our knowledge that 10 plus the intensity of our transmission must equal the original intensity either way that you decide to solve it you should come up with 40 milliwatts per centimeters squared this goes to show us that as long as we have two pieces of the variables we can figure out that third variable so in this example we had the intensity of the incident and we had the irc of the reflection because we had those two pieces we could calculate everything else so that last example went over how we could calculate irc and itc based on being given discrete values of intensities of the waves but what happens if you're only given the impedances of the media that the sound is traveling through well it gets a little bit more exciting because now we have this formula which if your eyes are melting out of your head right now looking at this formula don't worry quite yet so first remember that this formula is only going to work with perpendicular incidence secondly i nor the boards are going to expect you to take this formula plug in numbers and then answer test question with it what we want to know is do you understand what happens when z2 and z1 are very similar or what happens when z2 and z1 are very different or what happens when z2 and z1 are the same in true solid noise fashion we are going to move forward with plugging in numbers so you can see what happens in all of those scenarios but in the end once you have calculated the irc remember that you can calculate the itc by simply subtracting the irc from 100 percent so let's go ahead and look at some clinically applicable numbers so you can see what's actually happening in the body but again focus on our z2 and z1 similar are z2 and c1 very different or are they the same so our first example has sound that's traveling from liver tissue into the kidney tissue at a perpendicular incidence the liver tissue impedance is 1.64 mega rails and the kidney tissue impedance is 1.62 mega rails so first thing we need to figure out will reflection even occur at this point and the answer is yes we've got perpendicular incidence and the impedances are different very similar but different so knowing that information we're going to go ahead and plug in these numbers to that formula first we're going to plug in medium two's impedance 1.62 which is the kidneys and then we'll plug in medium ones 1.64 for the liver we're going to do this math on a calculator because again you're not going to need these for your boards we're just showing you the example here plug the math in we get something like this work this out anytime you square you're going to end up with a positive fraction that positive fraction then is multiplied by 100 to get a percentage and what we are left with when we calculate all this out is a very tiny little intensity reflection coefficient so this .0037 percent represents the energy that is reflected from the incidence wave a very very very small fraction is going to go into the reflection where most of this is going to continue forward in the transmission wave to get the itc we took 100 and subtracted the irc so when we have very similar or a small mismatch between our incidents we're going to get a very small reflection and most of that energy is going to continue forward so sound is traveling from heart muscle into the lung at normal incidents and there's another keyword normal incidence remember that you need to know your synonyms normal incidence is at a 90 degree angle anyway the heart muscle impedance is 1.74 mega rails and the lung impedance is 0.26 mega rails so let's stop again will reflection even occur and the answer again is yes we've got 90 degree incidence and the impedances are different so let's go ahead and plug in these numbers we're going to plug in media2 which is our lungs 0.26 and our media one impedance which is our heart muscle at 1.74 calculate all that out we're going to get this value we're going to square it and end up with a decimal or a fraction take that decimal or fraction multiply it by 100 to get a percent and we will see that the intensity reflection coefficient percent is 75.69 subtract that from 100 then to calculate the transmission coefficient and we get 24.31 now this is very different from what we saw in the last one our last example had almost 100 of the energy continuing forward in this example we're only sending about 24 percent forward and that is because we have a huge mismatch so that large mismatch is going to end up in a very large reflection of the energy leaving very little to continue forward last example then we've got sound that's traveling from smooth muscle of the spleen into the actual spleen and it is coming in at an orthogonal incidence remember that's another synonym for 90 degrees so the muscle impedance is 1.65 mega rails and the spleen tissue impedance is also 1.65 mega rails so let's stop again will reflection even occur well no it's not going to occur this time the incidence is 90 degrees but our impedances are the same and remember one of our rules was we had to have different impedances for reflection to occur with a 90 degree incidence but we're still going to prove it using our formula so let's go ahead and plug in our numbers we will plug in media 2's impedance in this first spot media 1's impedance in the second spot pull out our calculator do all the math we end up with something like this zero divided by three point three squared is zero zero times one hundred is zero so the irc is zero percent one hundred percent minus zero percent leaves us with an itc of one hundred percent so this proves that no reflection can occur when your impedances are the same and one hundred percent of that energy will continue forward no mismatch is going to mean no reflection so rules number one two and three all applied to our normal or our perpendicular incidence and the physics of the reflection and transmission that is going to occur with that those three rules are going to match up then with our biggest takeaways we are not going to be able to create or lose energy so our intensities are going to balance out our reflected intensity plus our transmitted intensity is going to equal our incidence intensity or irc plus itc is going to equal 100 we cannot have reflection unless the impedances of the two media are different when we have small mismatches we're going to have small reflections which means a lot of energy continues through when we have big mismatches we are going to have big reflections meaning much less energy will travel forward recognize that z does represent impedance and that it comes in the unit rails otherwise if you see that z1 z2 squared formula remember that is a way to calculate irc percent in perpendicular incidence let's switch over then to the physics of oblique incidence so when sound arrives at an interface add an oblique angle it's going to behave very differently than at a perpendicular angle perpendicular incidence required a difference of impedances oblique incidence doesn't care about impedances they can be similar they can be completely wildly different they can be the same oblique incidence doesn't care it's going to do its own thing at that interface perpendicular incidence also allowed us to calculate reflection and transmission well that idea is completely out the window as well because oblique incidence reflections might occur or they might not transmission might occur or it might not and some sort of combo of the two might occur we cannot predict what happens with oblique incidence because a weak instance does what it wants so knowing that oblique incidence behaves wildly different than perpendicular instance we really need to think of it as its own entity that follows its own set of rules if reflection occurs remember it might not it will still follow the law of conservation where incident intensity equals reflection intensity plus the transmitted intensity the only difference here though is that we cannot calculate the irc using that impedance formula that's because it doesn't matter with the bleak incidence impedances don't affect don't rule don't dictate how the oblique incidence is going to react at that interface the other thing we know about oblique incidence though is that the reflection is going to leave the boundary at the same angle the incidence beam came in at so this is one kind of steadfast rule that we can say the incidence angle is going to equal the reflection angle we have our incidence beam coming in the angle that the beam creates with that imaginary perpendicular line is the incidence angle the reflection angle then is the beam leaving the boundary and the angle that's created between the imaginary line and the path of the reflection so this angle is equal to this angle if this is 45 degrees this is 45 degrees this is 10 degrees this one's 10 degrees they are going to be the same degree of angulation another rule that we know about oblique incidents is that if transmission occurs again remember it might not but if transmission occurs with oblique incident it will continue in the same direction if the propagation speeds are the same for the media on either side of the boundary so if we have our sound beam coming in through media one it strikes the boundary if a reflection angle occurs it goes off at the same angle if transmission occurs it is going to continue straight forward on its path as long as media one speed is the same propagation speed as media this angle the transmission angle this one here that's created between the transmission path and that perpendicular line will be the exact same as the incidence angle and again that's only going to happen if speed 1 is equal to speed 2 incidence angle equals transmission angle but if the propagation speeds are different the transmission wave is going to bend and head into a different direction and this is called refraction so we've all seen this pencil in water it looks like it's broken and this is because light is moving differently in the air than it does in the water it actually slows it down a little bit and so it looks like we've broken our pencil or bent our pencil and the same idea is true for sound so we have our incidence coming in through a medium through the speed that is dictated by that medium we're going to get to the boundary a reflection might occur and transmission might occur as that transmission occurs into the second medium if it is a different speed than medium one we might see that it bends to the point where it makes a very wide transmission angle or it might bend where it makes a very small transmission angle but as long as it takes a different path that is not the same as the path that it originated in that is called refraction so our last topic of this unit is refraction so we're going to talk about how it occurs and what we need to know to apply to ultrasound physics first things first let's go back to that rule where we learned that for refraction to occur two things must be true we need oblique incidence and we have to have different propagation speeds on either side of the boundary gonna bounce back to this last slide remember we had our incidence angle coming in to the boundary and then at that boundary as the transmission wave went through it is going to bend if speed one does not equal speed two it's going to bend and we're going to learn a little bit more about how it's going to bend depending on the speeds we can see here the same thing happens with light we have a laser coming in through air it strikes a different media that has a different propagation speed and it bends that laser so look it's coming in at an oblique incidence different media and it bends as that laser beam goes through this media it strikes another media add an oblique incidence and it bends again because now it's traveling into a new propagation speed media this is really cool this is what our sound beam is basically doing as well it's going to bend all over the place and interact with different reflectors along its path now to be fair as sound enters into the body and interacts with the soft tissue of the body the propagation speeds of our soft tissue are actually very similar so when you have very similar propagation speeds you will get minimal changes in direction so maybe one or two degrees different but when you have wildly different propagation speeds like between soft tissue and bone what ends up happening is that as soon as that sound hits the bone not only is it going to cause a really strong reflection because it's a specular reflector that strong reflection is going to be sent to the point where it's almost reflecting into itself because it bends so much because there's such a different propagation speed so we're going to focus more on some minor changes so we can see the differences and what's more clinically applicable to what we're seeing in ultrasound if we know the propagation speed of the two media that we're dealing with and the angle to which the incidence beam has interacted with the boundary we can actually calculate what's going to happen in the second medium and we can do that with snell's law so snellslaw is going to take a look at the angles of incident and the angle of transmission and kind of compare that ratio to the ratio of the propagation speeds in medium 2 and in medium one another way we can write this formula is to write it out like this we have the transmission angle equal to the incident angle multiplied by the medium two speed divided by medium one speed now again if you're looking at this and thinking uh what am i supposed to do with this formula now do i have to use this do i have to memorize this no you don't but again we're going to use this formula to help us see the relationships of how the mediums can change the angle of transmission we will see that when we increase the speed of medium 2 that it is going to increase the transmission angle if we decrease the speed of medium two compared to medium one then we decrease the transmission angle so it actually works out really nicely but this is the math behind it just going to show you that even though it seems like magic it's not there's a lot of physics and math behind it all if you're not familiar with the s-i-n abbreviation that is sine and that is a trigonometry derived value that just really kind of helps us to put another functional value on the angle to help us figure out a degree angle that we're more comfortable with using like 45 degrees 37 degrees that kind of stuff so again these are not formulas that you're going to be expected to calculate for a test or for your boards but these are formulas that are going to help us to understand the core of what we need to know for ultrasound physics snell's law would help us calculate exactly what would happen with the transmission angle given any variance in speeds of the medium but that's not what we need to focus on there are only three scenarios that i want you to pay attention to and to know the first scenario is when medium one speed is equal to medium two speed we will see with that that incidence angle is going to be equal to transmission angle there is no refraction so i have this in the picture here we have liver with a propagation speed of 1570 meters per second and we have the spleen with 1570 meters per second so sound is coming from medium one into medium two and they are the same speed the incidence angle is depicted by this blue triangle and the transmission angle is seen here with the yellow triangle incidence angle is 45 degrees transmission angle is 45 degrees there is no refraction in this scenario a beam comes in at an oblique incidence to a boundary if reflection occurs that is going to leave at the same angle if transmission occurs it is going to continue forward on the same path and that again is because medium one speed is equal to medium two speed when we take a closer look at all the angles that these waves have created we'll see that the incidence angle is 45 degrees and following our rules of oblique incidence we know then that the reflection angle is also 45 degrees but more importantly for our discussion of refraction the transmission angle is also 45 degrees so when medium 1 speed equals medium 2 speed incidence angle equals transmission angle now the second scenario is when the speed of medium one is greater than the speed of medium two when we see that scenario we'll know then that the incidence angle is going to be greater than the transmission angle again we can see it here we have the incidence angle at 45 degrees and our transmission angle is only 17 degrees when the sound from medium one with a fast speed came into the medium 2 with a slow propagation speed it caused refraction to occur creating a smaller transmission angle so again the beam is going to come in at an oblique incidence if reflection occurs it'll go off into the other direction and if transmission occurs it is going to bend remember we're going from a high speed to a low speed looking at those angles again we have incidence at 45 degrees that means our reflection is also 45 degrees but more importantly again we see that the transmission angle is smaller than the incidence angle at 17 degrees so medium 1 is greater than medium 2 speed incidence angle is greater than transmission angle the last scenario that we need to be aware of then is what happens when the speed of medium one is less than the speed of medium two so here we have fat 1440 meters per second and tendon which has a propagation speed of 1750 meters per second because medium 1 speed is less than medium 2 speed we should see the incidence angle being less than the transmission angle so again the incidence angle comes in at an oblique incidence if reflection occurs it'll go off the other direction if transmission occurs in this scenario it is going to refract causing a larger transmission angle let's go ahead and look at those angles in a little bit more detail again we started with an incidence angle of 45 degrees meaning our reflection angle is 45 degrees but then we see that our transmission angle has increased to 59 degrees so when medium one speed is less than medium two speed we expect to see that the incidence angle is less than the transmission angle and that is it those are the three scenarios that you need to know about refraction knowing that they are derived from snell's law and refraction will only occur when there is oblique incidence and a change in propagation speed on either side of the boundary and that brings us to the end of unit 6b also completing all of unit 6. there were a lot of formulas in this unit and i did not do any practice moments with them on purpose because i want you to focus more on what the equations imply versus how to actually calculate irc based on impedance or the refraction based on snell's law the only real values that you'll be responsible for calculating in regards to this unit are the irc itc and the intensity values it's going to be relatively straightforward math but i do have some practice questions in your activities regarding those formulas as well as some of the other ones if you just want some practice time with them but for the most part focus on the vocabulary focus on the rules that i gave you and be able to identify the principles that work in regards to reflection and transmission in both normal and oblique incidents