Calculating nominal power in astigmatism RX

[gosumonkeyninja]

I already posted this question in another thread, put I am dying to get a response and want to make sure as many eyes as possible get a look at this so I can get a speedy answer. I am studying optical theory from notes given to me from a friend of a friend. Due to this some concepts are unclear. I am trying to figure out how to calculate practice problems, but their notes are not complete enough to figure out how to do them and I am unable to find the info in any book or online info. So here is a sample of the questions I am looking at.

Given info:
F1=+1.75
F2=-8.50 @ 135
F2= -12.50 @ 45
Ft=?
Your index is 1.66.


Any ideas? The section is labeled Nominal lens equation with lens maker compensation, if that helps. There are no radii, so I can't use the formula in my book. Also, the equation in the notes I have says: F(nm)=[(n'-1)F(lm)]/0.53 It has no explanation for what these stand for, but I know that n' is the index of refraction and 1 is the constant of air. 0.53 is the conversion used to compensate for different indices. Even with all this information I am at a loss. Can anyone help? I am almost at my wits end!!!

[Judy Canty]

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[Diopterman]

I strongly suggest you invest in "Optical Formulas Tutorial" by Ellen Stoner. You can find it on Amazon.

+1!

This is a book that should be in every Optician's library!

[gosumonkeyninja]

I have that book. I still can't seem to figure out how this formula was derived. Like I said anything similar has to have a radius. I'm not sure how this other equation was derived. I still need help.

[musicvirtuoso]

You are on the right track with the formula. This is a conversion formula to determine refractive power of a lens whose index does not match the 1.53 calibration of a lens clock. The "F(nm)" would of course be the refractive power that you are looking for; the "F(lm)" is the curve read from the lens measure. Since this is not a spherical lens, you would need to apply this formula to both meridians listed. This formula is derived from the lensmaker's formula, which states: F=n' - n/ r where F is the refractive power of the surface, n' is the index of refraction of the secondary medium, n is the index of refraction of the primary medium (which in most cases is going to be air, which has an index of 1), and r is the radius of curvature of the surface. Since two lenses with different curvatures can have the same refractive power (due to differences in their indices of refraction), you can manipulate the lensmaker formula to find the refractive power of a surface when you measure the surface using the lens clock. The equation looks like this intially: n'1 - n1/Flm = n'2 - n2/Fnm (n'1 - index of lens clock or first material and n'2 - index of new material. This is just a reworking of the original lensmaker equations; they are converted to the form r= n' -n/F and then set to be equal to each other since you can make the argument that the refractive powers, F, are the same. You can still set them equal to each other if you convert both equations to the same form, in this case, eliminating the "r" term all together, which is helpful since you don't know this particular piece of information). This can be further simplified to the equation that you wrote earlier. Hope I worded this well...:)

[gosumonkeyninja]

Thank you for helping, I knew it had to be a derived formula. The only thing I am still unclear on is how to combine the two F2s, or am I thinking about this all wrong still? If you could help show me the calculations on my practice problems, I'm sure I could work with it and figure out all the rest.

[optical24/7]

Someone will come along here and correct me if I'm wrong, but i believe the formula to use on the above equation is the refractive power formula. Lens makers formula requires radius, none are given. The FT?= answer they want is the actual power of a lens that clock's the curves you listed. The refractive power formula is;

D marked/D refractive = .53/n refractive - 1

D marked = the lens clock power
D refractive = the actual power of the surface
n refractive = the lenses index of refraction

Given info:
F1=+1.75
F2=-8.50 @ 135
F2= -12.50 @ 45
Ft=?
Your index is 1.66.

1st, find the "clock power" of the base ( -8.50 - +1.75 = -6.75)

Next find your Cyl clock power ( -12.50 - +1.75 = -10.75)

now plug this into your formula;

1.66 - 1 = (.66/.53) x -6.75 = -8.405 This is your sph power

1.66 - 1 = (.66/.53) x -10.75 = -13.386 - 8.405 = 4.98 This is your cyl power

So, Ft = -8.40 -4.98 x 135

You can varify the above with the index of material formula

N = .53 x (true power/clock power) + 1

.53 x (8.405/6.75) + 1 = 1.659 refractive index

.53 x (13.386/10.75) + 1 = 1.659 refractive index

I hope this helped ( and I hope I'm right.:bbg:)

[musicvirtuoso]

Someone will come along here and correct me if I'm wrong, but i believe the formula to use on the above equation is the refractive power formula. Lens makers formula requires radius, none are given. The FT?= answer they want is the actual power of a lens that clock's the curves you listed. The refractive power formula is...

Yes, and no. The so-called "refractive power formula" is also derived from the Lensmaker's formula. Almost the same way that the conversion formula is derived (see above). the "r" term is eliminated when the radii are set equal to each other ie: r(1) = r(2) is the same thing as n'1 - n1/Flm = n'2 - n2/Fnm

Update: I'm not sure if gosumonkeyninja's question was answered about the "F2's". The F2's represent the two curves (as read by the lens clock) on the back surface of the lens in the axis and power meridians; therefore, we can determine the nominal lens power by the process outlined above by Optical 24/7 [of course adding the curve of the the front surface (F1) and the back surface]. So, if the lens had the same index as the calibration of the lens clock, the power of the lens would be: -6.75 -4.00 x 135 (easily calculated using a power cross).

[Darryl Meister]

Optical24/7 has provided you with the correct results.

The main problem that these sample exercises are trying to illustrate is the fact that instruments typically used to measure surface power in a laboratory, such as sag gauges and lens clocks, only measure curvature.

Surface power is equal to the product of the physical curvature (or inverse of the radius) of the lens surface and the difference in refractive index between the lens material and the surrounding medium, typically air (n = 1). In order for an instrument that measures "surface power" to read off a dioptric power value, a refractive index must be assumed, since these tools cannot measure the actual refractive index of the lens material. Most commonly, a "tooling" index of 1.530 is assumed, so these instruments are calibrated for this refractive index.

So, given a convex front lens surface with a radius of curvature of 303 mm (0.303 m), a sag gauge or lens clock would read:




However, unless the refractive index of the lens material just happens to be 1.530, this value will not be equal to the actual refractive power of the surface. In order to determine the actual power of the surface, you must compensate for the difference in refractive index between the tooling index of 1.530 and the actual refractive index (n) of the lens material using a "curve variation factor":



So, given a surface measured by a sag gauge or lens clock of +1.75 D and a lens material with a refractive index of 1.66, the actual refractive power of the surface is given by,




Note that, because the refractive index of the lens material is higher than the tooling index of the instrument, the actual surface power is higher than the measured power. You would simply repeat this conversion for each measurement. Afterward, you can simply add the new values together to determine the refractive power of the lens, assuming that the center thickness is negligible.

Alternatively, if thickness is negligible as it is in your example, you can simply add the initial measurements together to determine the approximate focal power (F_T) of the lens based on the 1.530 tooling index. Then you can simply compensate this focal power using the curve variation factor described above.

So, in your example with a front curve of +1.75 D and a back curve of -8.50 D through one principal meridian in the tooling index of 1.530, the actual focal power of that principal meridian in a lens made from a material with a refractive index of 1.66 would be given by,




This process would then be repeated for the other principal meridian.

Best regards,
Darryl