Dispensing - Wot I (re)learned 26/11/08

Effectivity! The effective power of a thin lens

If you increase the distance of the lens from the eye the effective power becomes MORE POSITIVE or LESS NEGATIVE

ie the dispensed lens is going to need less plus power if it sits further away from the eye than it did in the trial frame. Or more minus. This ties in with my contact lenses being -4.50/-4.75 when my prescription is closer to -5.00/-5.25.

The optom should quote the BVD from lenses in trial frame to eye if the power is +/-5.00.

What the dispenser needs to do is measure the BVD in the patient's chosen frame (do a bit of fitting first) and then transpose the prescription into cross cyl form, for example

+8.00/+2.00x180 transposed into minus cyl form is
+10.00/-2.00x90 transposed into cross cyl form is
+8.00x90/+10.00x180

Then both the +8.00 and the +10.00 need to be adjusted for the new BVP
Let's say it's 8mm in the trial frame and 15mm in the spec frame.

The focal length of the lens must be reduced by the difference between the trial & spec frame BVP.

d=7mm.

so focal length of the lens = f'-d

effective power = 1/(f'-d) or 1/[(1-F')-d]

so in this example:
@ 90 F=1/[0.125+0.007] = 7.57D
@180 F=1/[0.100+0.007] = 9.34D

So you'd end up giving +7.50x90/+9.25x180 or
+7.50/+1.75x180