Question

Part Film Distance (PFD) in radiography must be

• A. 0 cm
• B. 80 - 90 cm
• C. 100 - 110 cm
• D. 120 - 130 cm

Hint:

Radiographic Geometry Rules:
Part-Film Distance (PFD OFD) $\rightarrow$ MUST BE ZERO (0 cm) to eliminate penumbra blur.
Focal-Film Distance (FFD SID) $\rightarrow 90 - 100\text{ cm}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Geometric radiographic image sharpness (penumbra reduction): minimizing the Object-Film Distance (OFD PFD) eliminates magnification distortion and geometric blur.
Key Formula or Approach:
\[ \text{Geometric Unsharpness (Penumbra)} = \frac{\text{Focal Spot Size} \times \text{Object-Film Distance (OFD)}}{\text{Focal-Object Distance (FOD)}} \]

Step 2: Detailed Explanation:
Principles of radiographic geometric sharpness:
- Part-Film Distance (PFD Object-Film Distance OFD): The physical distance between the anatomical body part being radiographed and the X-ray cassette/detector.
- Increasing PFD causes divergence of the X-ray beam, producing geometric magnification and wide penumbra borders (geometric unsharpness/blur).
- To maximize radiographic detail and spatial resolution, the anatomical part should be positioned directly in contact with the film cassette, meaning PFD must be 0 cm (or as close to $0\text{ cm}$ as physically possible).
- (Focal-Film Distance FFD is kept at $90 - 100\text{ cm}$).

Step 3: Final Answer:
Therefore, Part Film Distance must be 0 cm, matching option (A).