27.6 Ultrasound

Question 1

What is ultrasound and why can’t we hear it

Answer 1

U,traosund js any sound frewuency grewter than 20kHX

Human hearing is between 20 and 20KHz, and so we can’t hear anything above that , so we can’t hear ultrasound

Question 2

What are ultrasound used for

Answer 2

In medical = imaging , such as imaging a baby, non invasive lay and using no sort of high frequency photons

But ultrasound can be used in general to find cracks in things etc

Question 3

Why ultrasound advantage to image a baby

Answer 3

Because non ionising, non invasive AND CHEAP AND QUICK TO DO

Question 4

Simple how does an ultrasound transducer work

Answer 4

Changes electrical energy to generate sound waves, and converts sound waves into electrical energy as signals

Here it’s done using the piezoelectric effect

Question 5

What is the pizoelevgric effect

Answer 5

When crystal like quarts is compressed / stretched, it can produce an emf which can be processed to make an image

Similarly when an emf is run across the ends of a crystal, it will cause it to stretch and Co press accordingly , which produces sound waves

Question 6

How does ultrasound work

Answer 6

Transducer produces a PULSE of ultrasound waves of specific frequency and at specific frequency
- based on the different types of media, ultrasound waves will reflect with different intensities
- the time taken for an ultrasound wave to return to the transducer ( which is found as it causes an emf to be produced) can then be used to find the distance to this boundary
- and a computer can use this information to produce an image of whatever is being modelled

Question 7

When does a reflection of ultraodund, or any wave really occur

Answer 7

At a boundary, some reflection, some tramsitwnce ( which will be refracted)

Question 8

How would ultrasound scan for a retina look like

Answer 8

Transducer emits ultrasound waves through the eye
- at any boundary, there will be a reflection , and so ultrasound will arrive at the transducer , which produces and emf based in the intensity of the ultrasound
- thr intnesity will fall because part of the wave transmits etc , but we only itneredted in the time
- differnt boundaries will reflect accordingly, and take longer as further away
- lomg as we know our boundaries, we can find the distance if we know the average speed, and hence the distance from the lens to the retina

Question 9

M

Question 10

Ultrasound A scan

Answer 10

• a single transducer is used to image alomg a STRAIGHT LINE
• ultrasound is sent in a single direction and time is taken for reflections recorded

Used to measure retina distance, bone thickness etv

Question 11

B scan

Answer 11

Produces a 2D image in a screen
- transducer is moved over a patients skin, sending ultrasound in different directions
- for each position of the ultrasound, the computer records a row of dots based on the reflections
- after moving across the skin, this can produce a 2D image of dots
- each dot represents a boundary as that’s the point of reflection

The brightness of the dots are proportional to the intensity of reflected ultrasound wave, which depends on the boundaries, and hence an image is produced

Question 12

Difference between A and B scan

Answer 12

A scan happens in one direction , along one line of axis and is used to measure distance,e and does not produce an image

B SCAN is where transducer emits ultrasound in series of directions. Where reflections have occurred the computer produces a row of dots based on each positon of the transducer, where the brightness is proprtinsl to the intensity of reflected wave. After moving across the skin, the series of dots form a 2 d image

Question 13

How to determine frequency of emission

Answer 13

You want all the waves to reach the max depth before next wave emitted
So find frewuency, time period, muktilky. By speed and then see distance

If ti can reach max depth before next it’s good

Question 14

Acoustic impedance

Answer 14

Is defined as density if the substance x speed of UKTRASOUND in it

Which allows us to see how much something will be reflected intensity wise but only for ultrasound

Question 15

Why does intensity of reflected percentage EQUATUON show that the more different the acoustic impedances are the more is reflected

Answer 15

The more different they are, the higher the numerator

The more similar they are, the more the numerator tends to 0

Now if they are the same, it is 0, hence 0% of the wave is reflected, and so when entering similar acoustic impedances, barely anythign is reflected all transmitted

Question 16

How to make somehting soundproof then

Answer 16

Have something with gradually changing scosutic impedances.

This ensures minimal reflection, so it’s not an echo, because one way is to make impedance so high, but we don’t want an echo

The way to grwdiaully increade acoustic impedance means the sound is transmitted, but small reflections keep happenign, whcih reduces the final intensity of sound waves being emitted, so is now soundproof , as the person OPPSITE barely hears, and so do you

Each time it gradually changes impedance it looses energy such thst by the end it doesn’t have much energy energy at all

Question 17

Transmitted intensity

Answer 17

= I0 - air

Question 18

Hat happens when transducer placed against skin

Answer 18

The air skin boundary infect have high differences in acoustic impedances, hence a LARGE proportion of the ultrasound wave will be reflected, and so w can’t even do thr b scan

Hence we use acoustic matching, the technique where we apply COUPLING GEL between the skin and the transducer which makes the boundary gel skin.

But as the acoustic impedance are very similar, this means barely any u,tradlund waves will be reflected upon entering the skin, such thst we can actually scan

Otherwise 99% of intensity is reflected

Question 19

Recall what is Doppler shift

Answer 19

The apparent shift in wavelength / frewuency due to the relative motion between the object and the observer

Such thst if it goes away the wavelengths appear to stretch and thus redshift, coming towards you appear to reduce and thus blue shift

Question 20

How can Doppler shift be used with blood

Answer 20

The transducer is aimed at an angle to the skin

Based on the blood flow, the reflected waves as they reflect off blood cells will be Doppler shifted
- where blood travelling away, red shift, towards, blue shift
- regardless the change in frequency is proprtinsl to the speed of the blood
- hence usign EQUATUIN, where Doppler shift normally v/c, now it’s v/ c of the ultrasound

We can find the speed of the blood

They are being reflected by iron rich blood cells

Question 21

Why can’t you induce it normal to blood flow

Answer 21

Bevause cos 90 = 0, hence we would not be ab,e to find the speed of thr blood

Question 22

Advantages of ultraodund

Answer 22

Not ionising

And higher resolution between tissue and blood

Question 23

What do we say about beta and gamma radiation

Answer 23

We say we don’t use beta at all because ionising, and not penetrating enough to be detected

Question 24

Choosing between fatter spaced lead tubes or thin for collimator

Answer 24

Thin, as it reduces the SPREAD OF ANGLES of gamma photons that can make it to the scintillator, hence increases CERTAINTY about the exact postiom of gamma photons came from , direcltu below, and hence produces a SHARPER image

Blur occurs when the gamma photons postion is not well defined, as coming from different places messes up the image

Question 25

Hey is cat scan high quality

Answer 25

Prodigies 3 d images, where similar art euwtion coeffeint material can be differentiated by