Physics Chapter 10 Formulas & Example Problems
A lateral view of the lumbar spine taken at 100 cm SID results in the image of a vertebral body with max. & min. dimensions of 6.4 cm & 4.2 cm, respectively. What is the object size if the vertebral body is 25 cm from the IR?
MF=100/100-25= 100/75= 1.33 Therefore, Object Size is: 6.4/1.33 x 4.2/1.33= 4.81 x 3.16 cm
If a heart measures 12.5 cm at its max. width & its image on a chest radiograph measures 14.7 cm, what is the MF?
MF=14.7 cm/12.5 cm= 1.176
Magnification Factor Formula
MF=Image size/Object size
Magnification Factor w/out object size Formula?
MF=SID/SOD
Focal Spot Blur Formula
SOD/SID=Effective Focal Spot/Focal Spot Blur Focal Spot Blur=(effective focal spot x OID)/SOD
Image Receptor Speed
Speed=1/Exposure in Roentgens to produce an OD of 1.0 plus base & fog.
How much exposure is required to produce an OD of 1.0 above base plus fog density on a 600 speed image receptor?
Speed=1/exposure Exposure=1/speed=1/600=0.00167R=1.7mR
Example of Average Gradient Formula: A radiographic film has a base density of 0.06 & a fog density of 0.11. At what ODs should one evaluate the characteristic curve to determine the film contrast?
The curve should be evaluated at OD 0.25 & 2.0 above base plus fog densities. Therefore, at OD of... OD(1)=0.06+0.11+0.25=0.42 and OD(2)=0.06+0.11+2.0=2.17 If the ODs of 0.42 & 2.17 on the characteristic curve in the preceeding example correspond to LREs of 0.95 & 1.75, What is the average gradient? Avg. Gradient=OD(2) - OD (1)/LRE(2) - LRE(1) 2.17-0.42/1.75 - 0.95=1.75/0.8=2.19
What the OD and LREs mean?
Where OD(2) is the OD of 2.0 plus base & fog densities, OD(1) is the OD of 0.25 plus base & fog densities, & LRE(2) and LRE(1) are the LREs associated with OD(2) & OD(1), respectively.
The lung field of a chest radiograph Transmits only 0.15% of incident light as determined with a densitometer. What is the OD?
0.15%=0.0015 OD=(log10) I(o)/I(t) OD=(log10) 1/0.0015 =log10(666.7) =2.8
Image Receptor Contrast Formula (Average Gradient)
Average Gradient=OD(2)-OD(1)/LRE(1)-LRE(2)
An xray tube target with a 0.6 mm effective focal spot is used to image a calcified nodule estimated to be 8 cm from the anterior chest wall. If the radiograph is taken in a PA projection at 180 cm SID w/a table top to IR separation of 5 cm. What will be the size of the focal spot blur?
Focal Spot Blur=(0.6 mm)(8+5)/180-(8+5) =(0.6mm)(13)/167 =(0.6mm)(0.078) =0.047 mm
Speed vs. mAs Formula
New image receptor speed/Old image receptor speed= Old mAs/New Mas New mAs=Old mAs (Old IR speed)/New IR Speed
A posteroanterior (PA) chest exam requires 120 kVp/8 mAs with a 250 speed image receptor. What radiographic technique should be used with a 400 speed IR?
New mAs=Old mAs (Old IR speed)/New IR speed 8(250)/400=5 mAs New Technique=120 kVp/5 mAs
Optical Density Formula
OD=(log10) I(o)/I(t)
A renal calculus measures 1.2 cm on the radiograph. The SID is 100 cm, and the SOD is estimated at 92 cm. What is the size of the calculus?
Object Size=12(92/100)=1.1 cm
Magnification Factor for object size formula?
Object Size=Image Size (SOD/SID)
Screen film w/an average gradient of 3.1 is used to radiograph a long bone with subject contrast of 4.5. What is the Rad. Contrast?
Rad. Contrast=(3.1)(4.5)=13.95
Radiographic Contrast Formula
Radiographic Contrast=IR contrast X Subject Contrast