1. ## Vertex distance problem

And further, if I recall correctly, for a plus lens, the closer a lens is to a specified focal point, the more plus power is needed for that lens to achieve the same specified focal point. In that case if we know the original lens have a power of +8.50D, then its focal point is = 1/8.5 = .1176m or 11.76 cm. Wouldn't that mean for the second lens that is sitting 5mm away from the eye is also 5mm closer to the focal point of the first lens, so we would find that the second lens needs a focal length of 11.76cm - .5cm = 11.16cm, then the power of the second lens would = 1/.1116 = +8.96D?

Am I not seeing something in the problem ?
Thank you!

2. The reference points are the corneal plane, the doctor's (refracted) vertex distance, and the wearer's vertex distance.

Remember CAP = Closer Add Plus.

In the example provided, the fitted lens rests 5mm farther away from the corneal plane than the refraction distance. Decrease the plus power to compensate.

If there is astigmatism, solve for both principal meridians.

Hope this helps,

Robert Martellaro

3. Hey Robert,

It became much more apparent once I drew out what you said. Thanks!

4. Originally Posted by azhanli22
Hey Robert,

It became much more apparent once I drew out what you said. Thanks!

This may be of interest:

https://www.odob.health.nz/wp-conten...f-D-Wilson.pdf

Best regards,

Robert Martellaro

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