Is there an easy way ??

[Berno]

Hi everyone!

I've been struggling with this question in a review exam I have and would appreciate greatly your help... what do you think, here it is:

A patient is to be fitted with an RGP lens in one eye.
Preliminary findings are:
Ks 40.86 (8.26) @ 145
43.61 (7.74) @ 55
oc rx +0.25/-1.25x45

a) Based on these findings determine the amount and axis of the residual cylinder

b) A spherical trial lens with BOZR of 8.05 mm and BVP of +1.00 is placed on the eye.
(i) Calculate the resultant tear lens power when this spherical trial lens power is
placed on the eye (3 marks)
(ii) What is the expected sphero-cylinder over refraction when this spherical lens is

worn (4 marks)


thanks a lot~~

[chip anderson]

-2.37 -2.25 x 33(aprox)
Allows for 1/2 cyl against rule in lens of eye. Sphere of -1.37 and -1.00 to compensate for CL BVP and 2.75 residual from tear layer.

[HarryChiling]

Risidual astigmatism is the difference between the rrefractive cylinder power and the corneal cylider power, so:

Corneal Cyl = 43.61 @ 055 / 40.86 @ 145
40.86 - 43.61 = -2.75 x 055

Refractive Cyl = -1.25 x 045

So it would be about -1.50 difference if we were to ignore the slight difference in axis. If we wanted a more accurate measure we would subtract the refractive cyl from the corneal cyl and get -1.63 x 063(use the Darryls Obliquely crossed cyl calculator)

a) -1.63 x 063

If we wanted to calculate the power of the LL (Lacrimal Lens, Tear Lens, etc.) we would take the BOZR of 8.05mm and convert it over to dioptric power, since the BOZR is also the front surface of the LL we will use that as our radius and the index of tears 1.3375 as the index:

(1.3375 - 1) / 0.00805
337.5 / 8.05 (easier form to remember and work this equation)
41.93D

Alrighty then we have the front surface power of our LL, now the back surface of our LL is the same as the K's except since the LL surface will be concave it's (-)

FC_LL = 41.93D
BC_LL = -43.61 @ 055 / -40.86 @ 145

LL_power = FC_LL + BC_LL
41.93 + (-43.61 @ 055 / -40.86 @ 145)
-1.68 @ 055 / 1.07 @ 145

which can also be written as

b)(i) +1.07 -2.75 x 145

To get the expected Spheor Cyl over refraction when this lens is worn you need the difference of the refractive power and the power of the contact lens and lacrimal lens

CL = +1.00
LL = +1.07 -2.75 x 145

Total Lens Power = +2.07 -2.75 x 145

TL = +2.07 -2.75 x 145
Rx = +0.25 -1.25 x 045

TL power - Rx gives you the amount of power the lens is off by
again using Darryls crossed cyl calculator:

+3.04 -3.95 x 142

So if you overrefract you would get

b)(ii)-3.04 +3.95 x 142 -or- +0.91 -3.95 x 052

Unless I goofed somewhere that should about do it, I would double check it as I am rusty with CL's especially RGPs (the pitfalls of being a lab rat)

[Berno]

Thanks a lot mate! that was very detailed and helpful-- cheers:cheers:

[rdcoach5]

Hi everyone!

I've been struggling with this question in a review exam I have and would appreciate greatly your help... what do you think, here it is:

A patient is to be fitted with an RGP lens in one eye.
Preliminary findings are:
Ks 40.86 (8.26) @ 145
43.61 (7.74) @ 55
oc rx +0.25/-1.25x45

a) Based on these findings determine the amount and axis of the residual cylinder

b) A spherical trial lens with BOZR of 8.05 mm and BVP of +1.00 is placed on the eye.
(i) Calculate the resultant tear lens power when this spherical trial lens power is
placed on the eye (3 marks)
(ii) What is the expected sphero-cylinder over refraction when this spherical lens is

worn (4 marks)


thanks a lot~~

Unless I'm mis-reading this, you're fitting 1.00 diop steep which should give an over refraction of -0.75 +1.25@048.The k's spread is 2.75 diop or 1.50 diop more than wanted in the Rx. The axis is not too far off but probably will be around 48-50 degrees
Bob Taylor