Meridional magnification formula -- Steve/Daryl

**Jamie Brady** · 12-13-2006, 01:58 PM

Hello all -- This is my first post. Likely one aimed at Darryl or Steve.

I've been asked to surface a lens that is a meridional magnifier without refracting. Sounds to me like a big fat plano lens. I need magnification in varying percentages (ie. 5%, 6%,7%, etc...)

Problem is I am having trouble locating the appropriate formula to achieve my goal. The spec mag formula brings me in the direction, however, not just where I need to be.

Any help would be greatly appreciated.

Jamie
bradyj@ferris.edu

**lensgrinder** · 12-13-2006, 03:43 PM

You need to concentrate on the shape factor of the spectacle magnification formula since you will have a plano surface.

Mshape = 1/(1-(t/n) * F1

Ex. You are looking for a 5% increase

Let's call ((t/n)*F1) = X

Add 1 to the to the percent = 1.05
1.05 * (1-X)=1
X = 0.048

Now pick two common factors that you will use such as thickness and refractive index. What curve do you need?
Let's use 9mm thickness and a 1.5 index
0.048 = (0.009/1.5) * F1
F1 = +8.00D to have a 5% increase in Magnification

Hope this helps.

**Darryl Meister** · 12-13-2006, 06:18 PM

I assume that by "meridional magnifier," you are referring to an iseikonic lens that produces differing amounts of spectacle magnification in the two principal meridians. This will require producing a bitoric lens, with Plus cylinder power on the front surface and Minus cylinder power on the back surface.

The spectacle magnification formula can be used for each principal meridian to determine the magnification difference. If the lens is to be afocal (having no power), you can use the shape component that Lensgrinder referred to earlier in order to determine the difference in front curves necessary to produce the desired difference in magnification between the two meridians.

For hard resin (with an index of 1.500), the percentage change in magnification (dM) due to a change in front curve (dF) is given approximately by:

dM = dF * t / 15

where t is the center thickess in millimeters. (You can effect more of a change in magnification by front curve as the center thickness increases.) You'd want to rearrange to solve for dF:

dF = dM * 15 / t

If you wanted a difference in magnification of, say, 2%, for a lens with an 8 mm center thickness, you would need to have a difference in front curves of roughly:

dF = 2 * 15 / 8
dF = 3.75 D

So, you could use, for instance, a 6.75 D front curve in one meridian and a 3.00 D front curve in the other.

After you've worked on the difference in front curves, you'll need to apply a toric surface on the back to neutralize the power of the front. The back vertex power (Fv) is given approximately by:

Fv = F + B + t/n * F^2

Where F is the front curve, B is the back curve, t is the center thickness (in meters), and n is the refractive index. And, rearranging to solve for the back curve B:

B = Fv - F - t/n * F^2

For an afocal lens, with a back vertex power of 0.00 D, you have:

B = - F - t/n * F^2

You would solve this equation for each principal meridian, using the appropriate front curve. For our example, you would have:

B1 = -6.75 - 0.008/1.500 * (6.75)^2
B1 = -6.99 D

For one meridian, and:

B2 = -3.00 - 0.008/1.500 * (3.00)^2
B2 = -3.05 D

For the other. This lens should produce a meridional difference in magnification of roughly 2% with no back vertex power. You can check your spectacle magnification calculations at OptiCampus Magnification Calculator.