ocular rx????

[amy88]

HI, Would really appreciate if someone could help on the follwoing question

if the spectacle rx: -4.00/-5.00 x180, what would the ocular rx be?
the answer should be -3.75/-4.25 x 180..

i understand for every 4D change there is -0.25D, but dont understand y the cyl changed to -4.25...

Thanks

[mullo]

It looks to me like the power in the 090° meridian was vertexed also (as I it should have been). The power @ 090° is -9.00 which would vertex to approx -8.00. So you powers in the two principal meridians would be approx -3.75 and -8.00.

Power being approx -3.75 -4.25 x 180 or -8.00 +4.25 x 090.

[Bezza]

Assuming a BVD of approximately 12mm the values you have given for the ocular refraction are correct. The reason for this is that you have a cross-cyl spectacle Rx of -4.00x90/-9.00x180.
Using the effectivity formula to calculate the ocular refraction, K, we have K=Fsp/((1-(d*Fsp))
so for a vertex distance of say 12mm we have for the 90 meridian, K=-4.00/((1-(0.012*-4.00)) = -3.817D
and for the 180 meridian, K=-9/((1-(0.012*-9.00)) = -8.123D
When rounded to the nearest 0.25 dioptres we have -3.75x90/-8.00x180 which transposes to -3.75/-4.25x180