Optics related math
Here's a page with maths related to diopters and glasses.
You don't really need to know any of this stuff to improve your eyesight, but it's good to know for deeper understanding
Contents
Lenses
The image shows a typical converging lens. The three rays drawn each have an interesting characteristic:
 the top ray enters the lens parallel with the optic axis, and so passes through the focal point on the other side
 the middle ray passes through the optical centre of the lens, and is undeviated
 the bottom ray passes through the focal point on the incident side, and so emerges parallel to the optic axis
The thin lens equation
where
 = focal length of lens
 = distance to object ( on the picture)
 = distance to image ( on the picture)
Infinity
The term object at infinity is often used. When is substituted into the thin lens equation, that term vanishes, so that the focal length is then just the image location. For any sufficiently large object distance, the contribution from the reciprocal becomes negligible.
As the object approaches the lens, the outgoing rays converge less and less, until the object reaches the focus of the lens (), and the transmitted light is parallel  we say it "focuses at ", which means it never comes together into a focus. (This can also be described as a virtual image at  see below.)
Virtual image
When the object is brought even closer to the lens () the emerging rays are now diverging. When substituted into the equation, . This is interpreted as a virtual image, behind the lens. This is the mode in which reading glasses (plus lenses) are used  the eye is able to focus on the virtual image which appears to be further away than the real source object.
Diverging lens
A diverging lens behaves in a similar way, but for distant objects  the light rays incident from the source object are refracted outwards so that they are diverging even faster. Again, a virtual focus is said to form behind the lens. The thin lens equation still works as long as you use negative numbers to describe both the (virtual) image location and the focal length.
The corrective lens for myopia is a diverging lens. It works by forming a virtual image of distant objects, and it is this virtual image that the nearsighted eye is able to focus on. (You can also choose to think of it as lens in series with your eye, forming a compound lens with lower power and therefore a longer focal length.)
Diopters are inverse meters
Intuitively, the more powerful a lens is, the more rapidly it can bring incoming light to a focus. So the power is defined as the inverse of the focal length.
See Also Diopters
See Also cm Measurement
Remember that 100cm = 1m.
conversely
Calculating correction
How to calculate the strength of corrective lenses
From 20/20 prescription
The advantage of calculating the strength of differentials as a reduction from full prescription is that you don't have to worry about the cylinder element  just preserve that part.
Assuming you are wearing your full 20/20 correction (or even if you are fortunate to be emmetropic), the eye wants to receive parallel incident light to be relaxed. So we can use a simple geometric argument to calculate the strength of the (converging) lens which should be interposed : an object placed at the focus of a converging lens will result in parallel outgoing light ("focus at infinity"). This parallel light is exactly what the (fully corrected) eye wants to receive.
But rather than actually placing such a lens in front of your full prescription, simply add that value (being careful with signs).
ie if your screen is 50cm away, that corresponds to a power of . So that's the value you'd add to your full prescription (resulting in a lessnegative lens).
From blur horizon of naked eye
Note that this does not take either cylinder (astigmatism) or vertex distance into account.
Your cm measurement gives you the distance the eye can see when it is fully relaxed. You want a (diverging) corrective lens which puts a virtual image of the source object there. So we can solve the thin lens equation to find for an arbitrary source object distance given .
For full correction, that's easy : and so . E.g. if your blur horizon is 20cm you need a 5D correction.
For differentials to use a screen at, say, 50cm, just use . (Which is consistent with the previous version, adding 2D to the calculated full correction of 5D.)
Point of refraction
See also Refraction
Visual acuity
A reminder about some trigonometry / geometry

 so for fixed angle, 'a' is proportional to 'b'
 for very small angles (expressed in radians),
 so at small angles, 'a' is proportional to 'A'
Acuity is a measure of the ability to resolve small details, defined by the "minimum angle of resolution". Normal vision ("20/20", or 1.0) is the ability to resolve features subtending 1 minute of arc. In order to have a biggerisbetter score, the reciprocal of the angle is usually used:
An alternative representation of acuity is "logMAR" ("log_{10} Minimum Angle of Resolution"). This has a biggerisworse direction.
 0.0 corresponds to "20/20" (1.0 or 1 arcminute)
 1.0 is "20/200" (0.1 or 10 arcminutes)
 0.3 is "20/10" (2.0 or 0.5 arcminutes)
On a Snellen chart the letters are defined on a 5x5 grid: the detail to be resolved to distinguish between characters is one pixel of that grid. Here 'b' is the distance to the chart (eg 6m) and 'a' is the height of the letters. It's not critical that the eye is at the bottom or top of a row of letters  as long as the angles remain small, the numerical error is insignificant.
Acuity is reported in the form "distance/letterrow", such as "20/40", etc.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{Acuity} = 5 \times 0.00029 \times \frac{ \text{distance to chart} }{ \text{letter height} } = \frac{ \text{distance to chart} }{690 \times \text{letter height} } } ^{[1]}
Then, for example
 for the "6/12" line, we need Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 690 \times \text{height} = 12m \implies \text{height} = 12m / 690 = 1.73cm}
 for the "20/15" line, we need Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 690 \times \text{height} = 15ft \implies \text{height} = 15 * 12 / 690 = 0.26 \text{inch}}
The denominator can be interpreted as the distance at which the letters subtend 5 minutes of arc.
Average axial length accomodation/rate of change
 The typical emmetropic eye is 25mm
 change in axial length of 1mm = 3D
If someone with typical eyes wanted to adapt say 20/20 to .25 less normalized within 34 months would need to decrease axial length 0.083mm about 0.92microns/day  0.69microns/day average Credit: Mark Podowski
Converting from Glasses to Contact Lens Prescription or viceversa
References
 ↑ The EndMyopia Blog, https://endmyopia.org/usemathtoturnanytextintoyourownimpromptueyechart/