American Board of Ophthalmology (ABO) Practice Exam

In the case of a patient with a pupillary distance (PD) of 65 choosing a frame size of 54-18, what is the expected decentration in each eye?

1. 1.5 mm out

2. 3.5 mm out

3. 1.5 mm in

4. 3.5 mm in

The correct answer is: 3.5 mm out

To determine the expected decentration in each eye when a patient has a pupillary distance (PD) of 65 mm and chooses a frame size of 54-18, it is crucial to first understand how decentration relates to PD and the frame specifications. The frame size indicates that the lens width is 54 mm and the bridge width is 18 mm. The total frame width can be calculated as the sum of twice the lens width and the bridge width, which is 54 mm + 18 mm + 54 mm = 126 mm. To find the ideal frame center distance, it should match the PD of the patient. Since the PD is 65 mm, the distance between the centers of the lenses in this frame configuration needs to be at least 65 mm for optimal visual alignment. However, to check for decentration, you take the total frame width and subtract the PD. The centers of the lenses in the standard position of the frame measure approximately 126 mm. Therefore, we perform the following calculation: Total frame width (126 mm) - PD (65 mm) = 61 mm. This is the total excess width that needs to be adjusted for the lenses to align with the pupillary distance