For a FT 28 bifocal, the distance between the optical center and the edge used to calculate image jump is?

For a FT 28 bifocal, the distance between the optical center and the edge used to calculate image jump is?

• A. 5 mm
• B. 4.5 mm
• C. 6 mm
• D. 3.5 mm

Answer: (A)

Image jump happens when your eye shifts from looking through the distance portion to the near portion of a bifocal. The amount of this jump is largely set by how far the edge of the near segment sits from the optical center of the lens—the farther that edge is from the center, the larger the potential vertical displacement of the image. In a standard FT 28 bifocal, the geometry puts the edge of the near segment about 5 mm below the optical center. That 5 mm offset is the lever arm used in calculating the image jump for this frame. So 5 mm is the appropriate distance for this particular frame type; other values would correspond to different frame geometries and aren’t used here.

Image jump happens when your eye shifts from looking through the distance portion to the near portion of a bifocal. The amount of this jump is largely set by how far the edge of the near segment sits from the optical center of the lens—the farther that edge is from the center, the larger the potential vertical displacement of the image.

In a standard FT 28 bifocal, the geometry puts the edge of the near segment about 5 mm below the optical center. That 5 mm offset is the lever arm used in calculating the image jump for this frame. So 5 mm is the appropriate distance for this particular frame type; other values would correspond to different frame geometries and aren’t used here.