Test 1: lecture 4

Question 1

___ The ability to distinguish between tissues with different x-ray attenuation properties

Answer 1

contrast resolution

Question 2

what is subject contrast

Answer 2

the amount of xrays that get through a patient based on what the xrays have to go through

difference in xray intensity that is transmitted through a patient →the amount of attenuation

Question 3

what are three things that contribute to intrinsic subject contract

Answer 3

thickness

density

atomic number (Z)

Question 4

thinner tissue will have ___ attenuation then a thick tissue

Answer 4

less

Question 5

how does density effect xrays

Answer 5

the denser = more attenuation= whiter= more opaque

less dense= low penetration= less attenuation= blacker= less opaque

Question 6

radiolucency

Answer 6

more black- less attenuation- less density

Question 7

photoelectric effect

Answer 7

atomic number ³/energy ³

PE= better contrast when low energy(low kVP) and high atomic number

PE= absorption

Question 8

which one have more PE, which one have more kVP?

Answer 8

PE= absorption= increased contrast

Question 9

if air has a Z of 7.6 why is it more lucent then water with a Z of 7.4

Answer 9

air is more black (more lucent) because its density is so low

Z= atomic number is not the only thing that determines opacity (whiteness)

Question 10

how does contrast work to make better picture

Answer 10

increases atomic number

increases the photoelectric effect= more absorption= more contrast

Question 11

factors that affect scatter

Answer 11

subject thickness

field size (collimation- bigger picture= not as focused= more scatter)

kVP (more energy = more scatter)

Question 12

collimation ___ the field size

Answer 12

decreases

Question 13

___ ability to be able to visualize 2 separate adjacent structures as being separate

Answer 13

spatial resolution

Question 14

4 things that contribute to spatial resolution

Answer 14

blur (sharpness)

magnification

distortion

pixel size

Question 15

how to decrease blur

Answer 15

decrease ODD (object to detector distance)

have object as close to detector as possible

Question 16

blur is due to ___ factors

Answer 16

geometric

closer to the plate = clearer image

Question 17

according to blue, objects look ___ when closer to detector

Answer 17

sharper

Question 18

how to decrease magnification

Answer 18

decrease distance form the plate (ODD)

Question 19

why is L more blury?

Answer 19

L is farther away from the plate, will increase blur and magnification

Question 20

Variation in size/shape of an object (e.g. unequal magnification) on an image when compared to its true size/shape

Answer 20

distortion

Question 21

higher pixel number will have ___

Answer 21

better quality

Question 22

Images are made up of many pixels – entire arrangement is called a ___

Answer 22

matrix

Question 23

increase in matrix size =

Answer 23

more pixels= better picture

Question 24

Subject contrast is the difference in X-ray intensity transmitted through different parts of the subject. Subject contrast does NOT depend on:

Atomic number differences

Density differences

Radiation quality, film thickness and kVp

Photoelectric effect

Thickness differences

Answer 24

Radiation quality, film thickness and kVp

Question 25

Higher kVp techniques will result in... Higher subject contrast No effect in the image quality Lower radiation dose Underexposure Lower subject contrast

Answer 25

Lower subject contrast ## Footnote **increase kVp= more compton scatter= lower contrast** **lower kVp= more PE = more contrast**

Question 26

Which one of the following statements regarding pixels is **_FALSE_**? Pixel stands for picture element A smaller matrix size (e.g. 512 x 512) has smaller pixels compared to a larger matrix (e.g. 1024 x 1024), assuming the digital detector size is equal A higher number of pixels in an image increases spatial resolution An arrangement of pixels is called a matrix

Answer 26

A smaller matrix size (e.g. 512 x 512) has smaller pixels compared to a larger matrix (e.g. 1024 x 1024), assuming the digital detector size is equal (smaller matrix = larger pixels)