= Solution
In the paraxial <thin lens> approximation, let a converging <lens> have <focal length> $f>0$, real-object distance $u>0$, and signed image distance $v$, positive beyond the <lens>. Then
$$
\boxed{\frac1u+\frac1v=\frac1f,\qquad m=-\frac vu.}
$$
A ray parallel to the axis emerges through the far focus; a ray through the optical centre is approximately undeviated. Their intersection gives a real, inverted image for $u>f$: reduced for $u>2f$, equal-sized at $u=2f$, and enlarged for $f<u<2f$. At $u=f$ the emerging rays are parallel. For $u<f$ their backward extensions meet on the object side: $v<0$ and the image is virtual and upright.
The <retina> remains at an approximately fixed distance from the refracting apparatus. <Ocular accommodation> changes the <optical power>. A nearer object increases $1/u$, so a fixed retinal image distance requires larger $1/f$.
For near vision, the <ciliary muscle> contracts and reduces tension in the <zonular fibres>. The elastic <crystalline lens> becomes more rounded, increasing its <optical power>. For distant vision the <ciliary muscle> relaxes, zonular tension flattens the <crystalline lens>, and less power is needed. The <eye>'s several refracting surfaces make the single <thin lens> a qualitative model rather than its full optical prescription.
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