Can someone help me please solving this exam question?

**annamisiewicz79** · 01-09-2012, 09:25 AM

A workman using tools calibrated for spectacle crown glass, n=1.523, is instructed to produce the lens
+3.50DS/-1.00DCx180 in toric form with a +8.50 base curve. After working the sphere curve he discoveres that, in error he had used glass of n=1.618. Which toric tool must be used to work on toroidal surface in order that the lens may have finally the correct power?

Thank you for any kind of help.

**rdcoach5** · 01-10-2012, 07:31 AM

Sorry I don't have time to look up this formula but the front surface power comes out to +3.12. Someone will answer this. People who surface lenses do this all day.

**Wes** · 01-10-2012, 08:14 AM

Edit

**annamisiewicz79** · 01-10-2012, 08:24 AM

Thank you.

**Wes** · 01-10-2012, 08:29 AM

If I messed that up, I'm sure someone will be along to correct me. And if I didn't mess that up, well, I'm sure someone will be along to correct me.

See?

**annamisiewicz79** · 01-10-2012, 08:39 AM

I have checked the answers and they are 8.12 and 8.96 and I am stuck again.

**Wes** · 01-10-2012, 08:48 AM

The cylinder difference that's less than 1 diopter makes sense, but a back curve higher than the front curve resulting in a plus lens doesnt. Please recheck everything before I do any more work on this. Shall I assume the +8.5 base curve is meant as a 1.53 tooling curve?

**Wes** · 01-10-2012, 10:25 AM

I get roughly 5.50, 6.37. Too much to type or explain on smartphone.

**annamisiewicz79** · 01-10-2012, 02:22 PM

I have tried everything but I have never got close these answers. Thanks for trying.

**Wes** · 01-10-2012, 02:47 PM

Probably because those answers are wrong.

**Darryl Meister** · 01-13-2012, 12:50 PM

The question is somewhat ambiguous. There is also a reference to working a sphere curve and then using a toroidal tool afterward (I am assuming that this question wasn't written 30 or 40 years ago for a spherical generator or a hand pan).

If we assume that base curve value of +8.50 D refers to the curvature of the lens in the "tooling" index of 1.523, the actual surface power of the base curve becomes (1.618 - 1) / (1.523 - 1) * 8.50 = +10.04 D when using a lens material with an actual refractive index of 1.618. Ignoring lens thickness, since none is mentioned, the back tool curves in the 1.523 tooling index would then need to be -5.54 DS / -6.38 DC (or the nearest available tool curves to these values).

I should add that my results confirm Wes's.

Best regards,
Darryl