Past exam of the medicine course of the University of Oxford / 1867 / First examination / Physics / 7 / 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. ThenA 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.