Vertex Distance Compensation

PAGE References to Optical Formulas Tutorial:  (first reference is to edition 1 / second reference is to edition 2).

Vertex Distance Compensation.

Remember when we were talking about having two lenses in an optical system, and I showed you that just by putting space between the two lenses we could change the total power of the optical system without changing the power of the lenses? Adding distance between the lenses increased plus power (or decreased minus power).

A glasses prescription represents the amount of power that needs to be added to the eye's optical system at a particular distance from the eye in order to bring light rays to a point focus at just the right place. Whether this refraction is done in a phoropter or an auto refractor, the distance that the test lenses are from the eye is the refracted vertex distance.

We are not going to go into the way the refractionist has of determining the refracted vertex distances. When the wearer comes to us with prescription in hand, it is assumed that the glasses will fit at about the same distance from the eye that the refractionist used.

For most prescriptions the vertex distance for the glasses can be off by several mm and not create a problem. The glasses would have to have more than 10.00D of power for a 2 mm change to be noticeable to most wearers. For a larger change in vertex distance, such as converting from glasses to contact lenses, you need only have power over 4.00D for a noticeable difference to occur.  I am told that some ABO review books indicate that vertex distance changes should be considered any time the refractive error is  7.00 or more.

The vertex distance for contact lenses is zero. This is the distance from the front of the cornea to the back of the contact lens. For glasses, the measurement is not so simple.

Opticians basically have three ways of measuring the vertex distance.

OK, read through pages (64-70 / 77-84). My recommendation: read it through once for basic understanding, then go back and read it carefully following the examples. Then come back here before you try to do the exercises. While you are reading please ignore the approximation formula. It was important back in the days when we did not have calculators because it was mildly easier to do than the exact formula if you were using paper and pencil. Now that calculators are easy to come by you do not need the approximation. Please learn the exact formula.

Done reading? Did you follow the examples through?

Please go back to the top of page (70 / 83) and reread the Notes, particularly the second one. This is IMPORTANT!!!!!

And what you need for real life is in the "What you really need to know" table on page (69 / 83).

Lets try a few.

  1. The Rx reads: OD -15.00DS, refracted vertex distance 10 mm. The glasses fit at 8 mm vertex distance. What effective power will the wearer experience if you order a -15.00D lens? What power would you order instead to compensate for the change in vertex distance?

    2.   .
    First, decide what SHOULD happen. It is a minus lens moving toward the face, so the lens looses plus power or gains minus power. Therefore the effective power should be stronger than -15.00D, and the compensated power should be weaker than -15.00D.
      . Having trouble getting my answers? Look at the first one. There are a lot of ways to punch this into the calculator, some of which work correctly all of the time. If some of your answers are close to mine, but some are not within 0.01 or 0.02, then here is the way to punch it in:
         -15.00
    De = --------------------------------------
         1 + (+0.002)(-15.00)

    Punch 0.002, "x", 15.00, "", "=", "+", 1, "=", and write down the denominator. Now punch 15.00, "", "", the denominator, "=". This will give you the right answer for effective power.

    When you do compensated power there is just one more plus/minus in there:

         -15.00
    Dc = ------------------------------------
         1 - (+0.002)(-15.00)

    Punch 0.002, "x", 15.00, "", "=", "", "+", 1, "=", and write down the denominator. Now punch 15.00, "", "", the denominator, "=". This will give you the right answer for effective power.
     
     

    • change in vertex distance;  if lens is moved away from the wearer's eye
    		 d, []
    • times original power;  if power is negative
    		 x Dl, []
    • if finding what the new lenses should be
    		 []
    • add 1
    		 + 1
    • write it down, save it in memory, [or press 1/x]
    		 [1/x]
    • original power times what you wrote down or saved in memory,  if original power was negative
    		 x Dl []

     
  2. The Rx reads: OS +12.50DS, refracted vertex distance 9 mm. The glasses fit at 12 mm vertex distance. What effective power will the wearer experience if you order a +12.50D lens? What power would you order instead to compensate for the change in vertex distance?

    3.   .
    First, decide what SHOULD happen. It is a plus lens moving away from the face, so it gains plus power. Therefore the effective power should be stronger than +12.50D, and the compensated power should be weaker than +12.50D.
      .   .
  3. The Rx is OD -8.50 -3.25 x180, refracted at 9 mm. You dispense the glasses. A week later the wearer comes back to you complaining that the street signs look the same as they used to in his old weaker glasses, and the Doctor has assured him that they would be much clearer. What do you do?

    4.   .
    After checking everything else you determine that the glasses were made perfectly, the PD's are correct and the OC's are 3 mm below his pupil center where they were on his old glasses. You used the same base curves, the pantoscopic angle and the face form are all as they should be or as they are in his old glasses. The only real difference is that the new glasses fit at 13 mm, where the old ones were at the same 9 mm as the refraction. [What you really want to do is get him into something that fits closer to his face. Suppose that, for some reason, he does not want that. Maybe his eyelashes have grown?]

    Lets look at the effective power that he is seeing, and then decide what you should order for these glasses that fit too far away from his face.

    First, you must notice that the power in this lens is -8.50D on the 180 meridian and -11.75D on the 90th meridian. You have to determine the effective power in each meridian, then create an Rx again.

    OK, go back to the Optical Formulas Tutorial and do the exercises on pages (67 and 70 / 80-81 and 83-84). Check your answers in the back of the book.  Then read pages 415 - 419 (top left) in System for Ophthalmic Dispensing and do exercises 1-7 on page 424.

    NOTE:  In Systems for Ophthalmic Dispensing Dr. Brooks refers to the adjusted power as effective power, both to determine what power the wearer experiences based on the position of the lens and to determine what power to order to result in what the refractionist intended the wearer to see.  This is because he uses the technique of determining the adjusted focal length of the lens based on where it is positioned.  Either method is correct, and both result in the same answer.  When using the formulas shown in this lesson use the wording in the problem to determine what you should be finding:  if asked what to order (which is the only really important issue here) use the compensated power formula.

    Want some more to do?
     

  4. The Rx is +14.00 -2.00 x 045. Refracted vertex distance is 10 mm. Worn vertex distance is 7 mm.
    5. The page with the answers is here.
  If you are comfortable with this, and like the algebra, you may want to go to this page, which has the derivation of the formula. This is not required reading, so if you don't enjoy algebra, you have my permission to skip it.