Answer to Question #10024 Submitted to "Ask the Experts"
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The geometry you are using will, indeed, result in a significant amount of backscatter. Johns and Cunningham (Table B-1) shows over 30 percent backscatter (3 mm Al HVL [half-value layer] and 200 cm2 field). You are using a field of over 600 cm2 so the backscatter would be greater than this, i.e., on the order of 43 percent based on extrapolation of Table B-1 data.In addition to backscatter from the lead-aluminum block you will also get scatter from the tabletop, i.e., scatter from the lead-aluminum block to the tabletop and then to the detector. This will increase the scatter significantly and probably cancel out any reduction in scatter at the detector due to the 15 to 18 cm distance. I suspect Siemens is doing their measurement with the detector in air, i.e., suspended off the end of the table so any scatter from the table is eliminated.
It is not unusual with today's fluoroscopy systems to get some penetration through the lead in a test like this. The systems usually drive to 120 kVp or higher and maximum mA. The good news is that the grid is removable. I hope the staff members using the equipment realize that removing the grid during a procedure can reduce the patient and staff doses by about 50 percent. My recommendation would be to start out without the grid and if the images are acceptable then proceed. If the images are of low contrast then insert the grid and continue the procedure. The grid provides minimal benefit in a lot of cases, especially for thinner patients and thinner body parts.
Joel Gray, PhD
Johns HE, Cunningham JR. The physics of radiology. 4th ed. Springfield: Charles C. Thomas; 1983.