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Thread: Help?

  1. #1
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    Help?

    A subject is corrected for near vision at 25cm from the front plane surface of the lens by a plano-convex trial lens +8.00D. The final lens is dispensed in curved form with a +11.25D front curve. What back vertex power must be ordered to so the finall lens duplicates the effect of the trial lens? Assume that both the trial & final lens are a made in glass of n=1.5 and an axial thickness of 6.0mm

  2. #2
    Master OptiBoarder Darryl Meister's Avatar
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    Essentially, you need to determine the difference in near vision effectivity between both sets of lenses for a 25-cm object distance. For objects at near distances, the form and thickness of the lens will influence the effective vergence at the plane of the lens.

    Begin by determining the front curve F1 of the trial lens, given a back vertex power FV of +8.00 D and a back curve F2 of 0.00 D. After some rearranging of the back vertex power formula, we have:



    where t is the center thickness in meters (0.006) and n is the refractive index of the lens material (1.500). The result is approximately +7.75 D. Now calculate the effective vergence of the lens FNEAR, at the plane of the back vertex, for an object distance of 25 cm (or 0.25 m) using:



    which is approximately +3.81 D. Now, given a front curve F1 of +11.25 D, solve for the new back curve F2 that will produce the same vergence FNEAR at 25 cm:



    which is approximately -3.66 D. From this new back curve, the new back vertex power can be calculated normally using the given front curve, refractive index, and center thickness. The new back vertex power FV should be approximately +8.12 D. Since there is not a significant difference in form between the trial lens and final spectacle lens, with only a 3-diopter difference in front curve and an equivalent center thickness, the desired back vertex power is very close to the original value of +8.00 D.
    Last edited by Darryl Meister; 03-23-2009 at 10:53 AM.
    Darryl J. Meister, ABOM

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