Part I*

Image QualityPhotographic QualityGeometric Quality*

Photographic QualityGeometric Quality*

ContrastDensityGeometricDetailDistortionRadiolucentRadio-opaqueTissue densityOptical densityRadiographic contrastSubject contrastmAs reciprocity rulemAs doubling rule15% kVp ruleLong scale contrastShort scale contrastHigh/low contrast

AbsorptionDifferential absorptionStep wedgeGridBeam filtrationAnatomic densityOver-exposedUnder-exposedOver penetratedUnder penetratedsaturation

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*The films or images have different levels of density different shades of gray

X-rays show different features of the body in various shades of gray.

The gray is darkest in those areas that do not absorb X-rays well and allow it to pass through

The images are lighter in dense areas (like bones) that absorb more of the X-rays.

*The radiograph is formed by x-ray photons reaching the image receptor .

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The overall darkening of the image*

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Anatomic DensityBody part/object being x-rayedAtomic #Thickness of part

Optical DensityAmount of x-ray photons reaching the image receptorThe mA appliedThe time appliedAlso referred to as x-ray output*

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mAOne milli-ampere is equal to one thousandth of an ampere.The amount of current supplied to the x-ray tubeRange 10 to 1200 mA

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In seconds

How long x-rays will be produced

0.001 to 6 seconds*

mA X s = mAs*

10 mA1000 mA*

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+ 25%mAS = 25% increase in density100 mAs+50% mAs = 50% increase in density*

Density is like toasttoo much and the toast is burned, too little and it is underdone.

The images differ in density only. Which one looks optimal to you?*

This image was taken at 60 mAs. This Image is overall too dark.

What would you do to fix this image?

This image was taken at 300 mA. What was the time of the exposure?If we wanted to change the mA but keepthe mAs the same, what would we do?*

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Milliamperage second Conversions

Math Review

FractiontoDecimal

1) 1/5

=_______

2) 1/20

=_______

3) 1/8

= _______

4) 1/60

=_______

5) 1/15

=_______

mA x Time (s) = mAs

10) 100x 0.07= _______

11) 100x0.013 = _______

12) 100x0.033 = _______

13) 100x0.25 = _______

14) 100x0.009 = _______

15) 200x0.04 = _______

16) 200x0.07 = _______

17) 200x0.025 = _______

18) 300x0.01 = _______

19) 300x0.08 = _______

20) 300x0.05 = _______

mA x Time (s) = mAs

21) 100x1/8 = _______

22) 100x1/120 = _______

23) 100x1/15 = _______

24) 100x1/40 = _______

25) 100x1/6 = _______

26) 50x1/20 = _______

27) 50x1/120 = _______

28) 50x1/80 = _______

29) 200x1/80 = _______

30) 200x1/12 = _______

31) 300x1/5 = _______

32) 300x1/60 = _______

33) 400x1/80 = _______

34) 400x1/60 = _______

35) 500x1/12 = _______

36) 500x1/20 = _______

37) 600x1/40 = _______

38) 600x1/120 = _______

39) 600x1/25 = _______

40) 600x1/5 = _______

mA or S is unknown

1. 50 mA @ ______ S = 10 mAS

4. _____mA @ 0.1 S = 40 mAS

2. 100 mA @ _______ S = 4 mAS

5. _____ mA @ 0.2 S = 40 mAS

3. 200 mA @ _______ S = 5 mAS

6. _____ mA @ 0.3 = 60 mAS

mAs density changes & kVp contrast changes

1. mAs = 10 Double the optical density(OD) __________ the OD ________

2. mAs = 15 Double the optical density(OD) __________ the OD ________

3. kVp = 50 Narrow the contrast scale ___________ widen contrast ________

4. kVp = 75 Narrow the contrast scale ___________ widen contrast ________

When mA is unknownThe image was shot at 45mAs using a .75second exposure. What is the mA?When s is unknown.The image was shot at 80mAs using the 400mA station. What was the time of exposure?*

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Density(optical density, image density)mAskVpSID Beam FiltrationBeam restrictionBody part thicknessgrids*

15% kVp = doubling of exposure to the film 15% kVp = halving of exposure to the film

15% rule will also change the contrast of the image because kV is the primary method of changing image contrast.Remember : 15% change ( ) KVP has the same effect as doubling or the MAS on density*

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Always collimate smaller than the imagereceptor*

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*A device with lead strips that is placed between the patient and the cassette

Used on larger body parts to reduce the number of scattering photons from reaching the image

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*Too dark Too light

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The difference between the darks and light areas*

Now what changed?....very subtle, often subjective*

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Not very many differencesBetween grays

Also known as high contrast

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CASCADEPatient Interactions*

High kVpPenetrates more easily Causes more graysLow contrast

Low kVp Decreases penetrationCauses more black-white High contrast

Different parts of body attenuate differently

The difference in attenuation is the basis for contrast*

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kVpSubject contrastmAsSIDFiltrationBeam restrictiongrids*

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*The exposure from an x-ray tube operated at 70kVp, 200mAs is 400mR at 36 inches. What will the exposure be at 72 inches?100mR

The x-ray intensity at 40 inches is 450mR. What is the intensity at the edge of the control booth which is 10 feet away?......think carefully50mR

A temporary Chest Unit is set up in an outdoor area. The technique used results in an exposure intensity of 25mR at 72 inches. The area behind the chest stand in which the exposure intensity exceeds 1 mR. How far away from the x-ray tube will this area extend?30 feet

*Use Inverse Square LawThe first exposure value is 400mR. The first distance is 36 inches. The second intensity is what we are looking for. The second distance is 72Square both 72 and 36. Cross multiplyCancel out inches2, multiply, divide?mR= 100mR

*Use the Inverse Square Law. The first intensity is 450mR, the Second intensity is unknown. The first distance is 40 inches. TheSecond distance is 10 feet..Convert feet to inches.So 10 feet is equivalent to 120 inches.Short cut methodCross multiplyCancel units

*Use Inverse Square Law. The first intensity is 25mR, the secondIntensity is 1mR. The first distance is 72 inches, the second distanceUnknown. Cross Multiply

*Due Next week: April 2Download and print from website. Please show your work on the completed assignment

Image production will always be about how many photons reach the image receptor. *Optical density refers to the darkening of an x-ray image

**The difference is not anatomic density because it the same body part. The amount of photons reaching the image receptor has increased from the left film to the right film.**Tube current is equal to the number of electrons flowing from the cathode to the anode per unit time

Optical density is directly proportional to the mAs used to produce image. **The step wedge demonstrated the same number of steps. But the steps on the left are darker than the right. *Only a 20-30% change in density will enable the eye to detect the change. **15% rule: 15% kVp = doubling of exposure to the film 15% kVp = halving of exposure to the film

15% rule will also change the contrast of the image because kV is the primary method of changing image contrast.Remember : 15% change ( ) KVP has the same effect as doubling or the MAS on density

*Could be caused by kVP or mAs. **Sine wave coming into the body.