Originally Posted by
TheRobotious
Robert you are correct about the sphere curve. Plus cylinder lenses are ground with two curves on the front, minus cylinder with two curves on the back. The sphere curve is always the spherical curve, as the name suggests. So for plus cylinders, the sphere curve is on the back, for minus cylinders, it is on the front. The base curve is always on the front surface. As lensgrinder says, the base curve may be defined as the weakest curve. For plus cylinders, this means that the weaker curve on the front surface is the base curve (the other being the cross curve). For minus lenses, there is no "weaker" curve on the front, there is only one curvature, which means that the sphere curve is the base curve.
Now on to these questions. The questions are a little confusing depending on how you want to interpret them. The first question asks us to figure out the radii of curvature of the back surface, which suggests the lens is to be ground in minus cylinder form. If you want to keep it in plus cylinder form, then the answer is simple:
Back surface power = +11.00D in all meridians
Therefore we have surface radius in all meridians = -6.727 cm (rearranging surface power formula for radius).
To figure out the front curves, it is easiest to reverse the system. Although at this point you should already expect a negative curvature on the front, which simply isn't done for a +8.50D. Not to mention a negative back curve as well. At this point it is clear that this is not what the question is asking for. Nevertheless, for completeness.
We require vergence leaving the back surface of the lens to be +7.25D@90 and +8.50D@180. Thus, in a reversed system, we begin with -7.25D@90 and -8.50D@180. The vergence after refraction at the back surface if the lens is
90 meridian: -7.25D + 11.00D = +3.75D
180 meridian: -8.50D + 11.00D = +2.50D
The vergence changes by the time we reach the front surface of the lens, such that we have vergences incident on the front surface of:
+3.797D@90 / +2.521D@180 {Using the downstream vergence equation, or changing the desired focal lengths}
Finally, since we measure back vertex power with respect to parallel rays entering the lens, the vergence leaving the front surface of the lens must be zero. Again, we have
Power in 90 meridian = V2 - V1 = 0.00 - (+3.797D) = -3.797D
Power in 180 meridian = V2 - V1 = 0.00 - (+2.521D) = -2.521D
So the power of the front surface of the lens would be approximately -3.80@90 / -2.52@180. The radii are therefore
Radius in 90 meridian (base curve) = -19.49 cm
Radius in 180 meridian (cross curve) = -29.35 cm
As you can see we get a meniscus lens, however it bulges towards the eye, which is absurd. Instead, I interpreted the question as MakeOptics did, and came up with the same answer (using a similar approach).
For question 2, your calculations are correct Robert (just watch the cylinder in the Rx. - should be -1.91D). However, if the lens is ground in minus cylinder form, asking for the sphere curve power is meaningless, since it is given as +10.00D in the question. Again I had the same idea as MakeOptics to convert to a plus cyl, and then you've got the front surface with the weaker base curve as +10.00D and the cross curve as +12.00D (for 10 mm). Regardless if the question wanted the toric base curve, then you would be correct Robert (again just watch out for that second meridian which you've quoted as 5.36 when it should be 5.14 - this will give you and answer of -5.25D and the correct cyl of -1.91D).
Please help. I'm running through some road blocks in my past year exam papers.
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