# 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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## 1 Answer

Extreme hyperopia would correspond to your eye lens in relaxed conditions being close to an optical flat. In such a case you would need a contact lens with a focal length of about 25 mm (typical human eyeball diameter). This corresponds to a lens of 1000/25 = 40 diopters.

In other words: a farsighted eyeless requiring 20 diopter correction has a focal length at relaxed conditions of about twice the required (25 mm) focal length.

A simple (approximate) means to estimate the diopters of correction needed, is to use:

$$\frac{1}{f_e} + \frac{1}{f_c} = \frac{1}{D}$$

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 ($D > f_e$).

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Thank you! But then why there are plenty of questions in physics books who force us to calculate corrections considering the distant point and not the average diameter of the human eyeball? (and I'll try not to cite any book here) –  rafa Aug 13 '13 at 15:33