How to measure the strength of a prescription eyeglass lens? 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.
 A: 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:

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.
