Thick edge substance

[Optometry student]

Hope someone can help with this. Transpose the prescription +2.75DS/+1.50 x 90 to toric form with a -6.00D base curve. This toric lens is of refractive index 1.4985 and edged to a 46x40 oval shape. The thin edge substance is 1.5mm. What is the thick edge substance? (use accurate sag formula)

P.S. I have to idea what to do with the 46x40.

[Darryl Meister]

I happen to know from experience that this particular problem is a bit of a "trick" question, since it assumes that the curves of the lens have not been adjusted to allow for the gain plus power produced by the center thickness. Also, while while the prescription is written in plus cylinder form, the lens is made in minus cylinder form, as indicated by the reference to a "back base curve."

Consequently, this "test" problem is easier to solve than you might at first assume, although solving a "real" problem like this would actually be considerably more difficult. Nevertheless, it is still a time-consuming problem to solve, either way. To keep things straight, it may be easier to transpose the prescription into minus cylinder form: +4.25 -1.50 x 180 before continuing.

The power through the horizontal (180) meridian of this lens is +4.25 D, while the power through the vertical (90) meridian is +2.75 D. The horizontal dimension of the lens is 46 mm, while the vertical dimension is 40 mm. Since the horizontal dimension has the greatest plus power and also the longest dimension, the center thickness of the lens will be determined by the sagittal values of this meridian.

Essentially, you will first need to use toric transposition to determine the front curve and two back curves. With a back base curve of -6.00 D through the horizontal meridian, the front curve would be +10.25 D in order to produce a sphere power of +4.25 DS. The cross curve of the lens would be -7.50 D through the vertical meridian in order to produce the cylinder power of -1.50 DC.

Next, calculate the exact sagitta of the front and back curves through the horizontal and vertical meridians using the exact sagitta formula. You will need to do this for the +10.00 D front surface at 46 mm for the horizontal meridian and again at 40 mm for the vertical meridian. You will need to do this for the -6.00 D back base curve at 46 for the horizontal meridian and for the -7.50 D cross curve at 40 mm for the vertical meridian. The exact sagitta (S) of a lens surface or section of a toric surface is given by:



where r is the radius of curvature of each surface and d is the diameter through each meridian (40 mm and 46 mm). The radius of curvature is given from the surface power of the lens (F) by:



for the front curve and



for the back curve, where n is the refractive index of the lens material (1.4985).

Once you have calculated these four sagittal values, you can then calculate the center thickness and maximum edge thickness. Recall that the minimum edge thickness (M) is 1.5 mm. Since the center thickness is constrained by the horizontal dimension of this lens, the center thickness (CT) will be given by:



And the maximum edge thickness (ET) will be given by:



After solving these equations for each meridian of the lens, you should arrive at approximately 2.81 mm for the maximum edge thickness (ET) of the lens.

Best regards,
Darryl