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If I am to confirm that a lens has diopter +2.50, I might seek a reference lens (one that I know is +2.50) and position both away from a sheet of paper in plain sunlight, then confirm that the focal distance is the same.

How do I do the same with a lens that is -2.50, perhaps even without having another reference -2.50 lens?

It's not a theoretical question. Rather than select a frame for eyeglasses by driving and putting on face masks during the quarantine, I selected one using an online store's software. But I'm unable to estimate the competence of the optician by meeting them, hence the question.

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z Jul 17 '20 at 0:45
  • $\begingroup$ @Sam Maybe you can change the title to something like: How to measure the strength of a prescription eyeglass lens? Then it will be more easily found in the future by people with a similar question. $\endgroup$ – Han-Kwang Nienhuys Jul 17 '20 at 9:45
  • $\begingroup$ @Han-KwangNienhuys Done. Are you willing to accept critique of your edit of the answer? Have you read the 5th edition of a book, when you were entirely familiar with the contents of the 1st edition, and wondered why the author didn't put the new material into a separate book? I feel the same about your answer. If someone is here for the "tome 1 answer", they should see it easily. $\endgroup$ – Sam Jul 17 '20 at 9:48
  • $\begingroup$ @Han-KwangNienhuys Also: the number 2 arises in your answer; I changed the diopters to +/-2.5 to enhance clarity of the question (since we're proceeding by a "worked example" rather than by variable, this is rather necessary). Could you adjust the answer accordingly? $\endgroup$ – Sam Jul 17 '20 at 9:52
  • $\begingroup$ I'm not quite following you. The first paragraph of the answer didn't change (the "1st ed."); what does it help to change the case from +/-2.00 to +/-2.50? The part about the cylinders is to address a comment by someone else. I can add sketches for the cylinder-free +2.00 and -2.00 lenses if you think that helps. $\endgroup$ – Han-Kwang Nienhuys Jul 17 '20 at 10:05
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For a positive lens: focus sunlight (for safety, preferably on dark, inflammable object such as a brick) and measure the distance in meters from the lens to the brick. The optical power of the lens in diopters is one divided by the focal length. In the case of a +2.50 dpt lens, the focal length will be 0.40 m.

For a negative lens, you can use white paper. Rather than a point focus, you'll see a bright spot that is larger than the lens, but has the same shape. The distance where the spot is exactly two times the lens size is the focal length, apart from the minus sign. For a -2.50 dpt lens, this will happen at 0.40 m. It helps if you draw the outline of the lens at twice the size on the paper; if the lens is a rectangle of size 5 cm x 4 cm, then draw a 10 cm x 8 cm rectangle on the paper and find the distance where the spot fills the drawn rectangle.

Many eyeglass prescriptions include a cylinder. In that case, it's most practical with a round lens; if the lens isn't, you can use paper with a circular cutout and hold it against the lens.) You will find different distances for the long and short axis of the ellipse, and correspondingly two powers (in diopters). One of the two is the nominal spherical power; the difference is the cylindrical power. See the sketch below:

Measurement of the power of an astigmatic prescription lens

Suppose that you find a horizontal focal length $f_H=-0.50~\mathrm{m}$ and a vertical focal length $f_V=-0.44~\mathrm{m}$. The corresponding powers are $P_H=-2.00~\mathrm{dpt}$, $P_V=-2.25~\mathrm{dpt}$. This could be written as a prescription SPH=-2.00, CYL=-0.25 or as SPH=-2.25, CYL=+0.25.

The cylinder needs an axis angle as well; the convention is that 90 deg corresponds to a vertical cylinder axis. For this example, it could be a horizontal (180 deg) negative cylinder or a vertical (90 deg) positive cylinder, so the prescription is either SPH/CYL/AX -2.00/-0.25/180 or -2.25/+0.25/90. Which of the two is written on your prescription depends on how and where the prescriber was trained.

Final note: if you don't have sunny weather, you can use a household lamp (preferably a small LED spotlight) at a large distance, for example $L=5$ m. You'll need to add $1/L$ to the power (in dpt) measurements. For the 2.50 dpt example and a lamp at 5 m, you'll find the focus at 0.43 m; the lens power is then $P=1/0.43 + 1/5=2.5$ dpt.

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  • $\begingroup$ Using negative unity was silly. I forgot that diopters involve an inverse. Let me change the question (and the answer along) to make both clearer. $\endgroup$ – Sam Jul 17 '20 at 0:19
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    $\begingroup$ Unless the cylindrical part of the prescription is zero, this won't work very well because the focal length of the lens will be different in different planes and it may even be convex in one plane and concave in another.. The negative prescription means you are myopic, so the practical test is to check if distant objects are in focus when your eyes are "relaxed". If you are questioning the competence of the professional who made your glasses, you should also question the competence of the professional who did the eye test - you are assuming that -2.0 is the correct prescription! $\endgroup$ – alephzero Jul 17 '20 at 0:32
  • $\begingroup$ @alephzero Agreed. The real test of competency is whether you can see in focus with your glasses. $\endgroup$ – Gilbert Jul 17 '20 at 2:21
  • $\begingroup$ @alephzero you can use the same procedure for a cylindrical lens; I updated the question. Judging the prescription by "when your eyes are relaxed" is very difficult unless you're well into the reading-glasses age (50+), because the accommodation reflex is mostly involuntary. That's why an optometrist will gradually try more negative lenses until you start responding by "I don't see a difference". $\endgroup$ – Han-Kwang Nienhuys Jul 17 '20 at 7:50

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