8. Sharma IP,
A: Between center of two pupil
B: Right temporal pupillary margin to left nasal
pupillary margin
C:Right temporal limbus to left nasal limbus
9. Technique
Steps in Measuring binocular distance PD using a ruler
1.Dispenser position at 40 cm(16 inches)
2.Dispenser closes right eye , patient fixes at dispensers open left eye
3.Dispenser lines up the ruler zero point on the subject’s right pupil, left pupillary
border or left limbus.
4. Dispenser closes left eye , patient fixes at dispensers open right eye
5.Dispenser reads scale directly in line with subject’s left pupil center ,left pupillary
border or left limbus
6.Dispenser closes right eye, opens left; subject fixates on dispenser’s left eye
7.Dispenser checks to make sure that zero point is still correct
Sharma IP,
10. Common difficulties and their
solution
Sharma IP,
Problem Solution
1.Dispenser cannot close one eye Occlude the eye with free hand
2.Subject is strabismic Cover the subjects eye not been
observed
3.Subject is uncooperative child Take a canthus- canthus
measurement
12. Technique
Steps in Measuring monocular distance PD using a ruler
1.Measure the binocular PD using center of pupil as reference point
2.Before moving the ruler, note the reading at the center of the nose. This is
the monocular PD of one eye
3.Stbtract this reading from the binocular reading to obtain the reading of
the other eye.
Sharma IP,
14. Near PD
Required for single vision reading glasses or for
multifocal lenses
Can be either measured or calculated
Sharma IP,
15. Technique
Steps in Measuring near PD using a ruler
1.Dispenser places his/her dominant eye in front of the subject’s nose at the
subjects near working distance
2.Dispenser close the non-dominant eye
3.Subject fixates on dispenser’s open eye.
4.Dispenser places zero point of the PD ruler at the center of subject’s right
pupil.
5.Scale reading at the center of the subject’s left eye is read.
Sharma IP,
16. Calculating near PD
Most commonly used is called three -quarter rule
It states that “for every diopter of dioptric demand, the
optical center of each reading lens, or the geometrical
center of each bifocal addition, should be insert
0.75(three-quarters) mm.”
Dioptric demand =inverse of reading distance in
meters
and is independent of actual bifocal addition power
Sharma IP,
17. Example
A spectacle lens wearer has the following prescription:
OD: -1.00 DS
OS: -1.00 DS
Add: +2.00DS
Distance PD is 64mm.
What is the near PD at 40 cm?
Sharma IP,
18. Solution:
Dioptric demand=1/0.40 =2.50 D
Insert per lens =2.50 x (3/4) =1.9mm
= 2.00mm (approx)
We need to deduct 2 mm from each lens i.e 4 mm from the distance IPD
Near IPD= 64mm- 64 mm
=60mm
Easiest Solution: Deduct 4 mm from distance PD for normal
working distance
Sharma IP,