# How mirror equation can explain farsightedness correction?

I have a friend who has just show me his medical prescription for hyperopia (farsightedness) correction and he needs glasses with 4,25 diopters for that, which seemed to be weird for me because I had learned, from the mirror equation, that the maximum correction possible for hyperopia is 4 diopters:

$$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$$

If we have $0.25m$ for the normal eye distant point and more than $0.25m$ for the farsighted eye distant point (negative sign, because it's a virtual image), then we would have:

$$\frac{1}{f} = \frac{1}{0.25} + \frac{1}{p'} = 4 - \frac{1}{|p'|} \in\quad ]0,4[, \quad\text{since}\quad |p'| \geq 0.25m \quad\text{and}\quad p'<0$$

I did some google search and find out that, indeed, hyperopia can reach values even greater, such as 20 diopters, but I can't find pages where doctors explain that with equations or physics teachers explain how things really work in ophthalmology.

Either I am doing some terrible mistake, or doctors are doing some terrible mistake, or this equation just don't apply to hyperopia at all... Which one is true?

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$$\frac{1}{f_e} + \frac{1}{f_c} = \frac{1}{D}$$
here $1/f_c$ is the optical correction (in diopters when the corresponding focal length $f_c$ is measured in meters), $f_e$ the focal length of the relaxed eye lens, and $D$ the inner diameter of the eyeball. Note that $1/f_c$ is positive when correcting for farsightedness ($D < f_e$), and negative when correcting for nearsightedness ($D > f_e$).