The human eye forms images on the retina by changing the optical power of its crystalline lens.
This process is called accommodation. When the eye looks at a distant object, the lens is in a more relaxed state and its focal length is slightly longer.
When the eye looks at a nearby object, the ciliary muscles cause the lens to become more curved, increasing its optical power and shortening its focal length so that the image still forms on the retina.
A simple thin-lens model for the eye uses
\[
\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i},
\]
where \(d_o\) is the object distance, \(d_i\) is the image distance from the lens to the retina, and \(f\) is the focal length of the eye’s effective lens system.
In this calculator, the retina distance is entered directly, and the far point and near point determine the relaxed and accommodated states.
The far point is the most distant object point the eye can see clearly.
For a normal relaxed eye, the far point is effectively at infinity.
In that case, the term \(1/d_o\) becomes zero, so the relaxed focal length is simply
\[
f_{rel}=d_i.
\]
If the far point is finite instead of infinite, then the eye needs extra optical power even in the relaxed state.
That is why a finite far point is often used as a simple optical model for myopia-like behavior.
The near point is the closest point that can still be focused sharply.
For a typical young eye, a common textbook value is about \(25\ \text{cm}\).
In the accommodated state, the lens power increases so that the same retina distance is maintained even though the object is much closer.
Optical power is often expressed in diopters, abbreviated D. The definition is
\[
P=\frac{1}{f(\text{m})}.
\]
If the focal length is entered in centimeters, then the power is
\[
P=\frac{100}{f(\text{cm})}.
\]
This is why an eye with focal length about \(1.67\ \text{cm}\) has a relaxed power of about \(60\ \text{D}\), because
\[
P=\frac{100}{1.67}\approx 59.9\ \text{D}.
\]
For the standard sample case with far point at infinity, near point at \(25\ \text{cm}\), and retina distance \(1.67\ \text{cm}\), the relaxed state is
\[
\frac{1}{f_{rel}}=\frac{1}{\infty}+\frac{1}{1.67}=\frac{1}{1.67},
\]
so
\[
f_{rel}=1.67\ \text{cm}, \qquad P_{rel}\approx 60\ \text{D}.
\]
In the accommodated state,
\[
\frac{1}{f_{acc}}=\frac{1}{25}+\frac{1}{1.67}.
\]
This gives
\[
f_{acc}\approx 1.56\ \text{cm},
\qquad
P_{acc}\approx 64\ \text{D}.
\]
The change in power needed to shift from relaxed far vision to the near point is called the
amplitude of accommodation:
\[
\Delta P=P_{acc}-P_{rel}.
\]
In the sample case, that change is about \(4\ \text{D}\).
Even though the focal-length change seems small in centimeters, it corresponds to a noticeable change in diopters because the eye is already a very strong optical system.
The accommodation diagram in this calculator is intentionally schematic. In real eyes, the cornea provides a large fraction of the total optical power, and the crystalline lens fine-tunes the focus.
Also, the eye is not a perfect thin lens and the retina is curved rather than flat.
Still, the thin-lens model captures the core idea: the eye must increase its power for closer objects if the image is to remain on the retina.
At a more advanced level, this topic connects to refractive errors such as myopia and hyperopia, corrective lenses, reduced eye models, and age-related loss of accommodation called presbyopia.
But the basic picture stays the same:
far objects need less optical power, near objects need more, and the eye accommodates by increasing its effective lens power.
