What is the distance between the optical center and the edge of a Franklin (Executive) bifocal used to calculate image jump?

What is the distance between the optical center and the edge of a Franklin (Executive) bifocal used to calculate image jump?

• A. 0 mm
• B. 1 mm
• C. 2 mm
• D. 5 mm

Answer: (A)

Image jump comes from a mismatch in the optical paths at the boundary between the distance portion and the reading (near) portion. The amount of jump depends on how far the near-edge is from the optical center of the lens—the farther the edge is from the OC, the more vertical shift the edge can introduce when you look through it. In a Franklin (Executive) bifocal, the edge of the reading segment is placed right at the optical center of the lens design. That means the distance between the optical center and the edge is zero millimeters. With zero offset, there’s no additional angular deviation introduced by crossing from distance to near at that edge, so the calculated image jump is zero.

Image jump comes from a mismatch in the optical paths at the boundary between the distance portion and the reading (near) portion. The amount of jump depends on how far the near-edge is from the optical center of the lens—the farther the edge is from the OC, the more vertical shift the edge can introduce when you look through it.

In a Franklin (Executive) bifocal, the edge of the reading segment is placed right at the optical center of the lens design. That means the distance between the optical center and the edge is zero millimeters. With zero offset, there’s no additional angular deviation introduced by crossing from distance to near at that edge, so the calculated image jump is zero.