The eye as an optical system
The eye works like a camera with a converging lens. Light passes through the cornea (clear front window), the aqueous humour, the pupil (hole in the iris that controls how much light enters), the lens and the jelly-like vitreous humour, then lands on the retina.
Most of the bending (about +40 D of the total +60 D) happens at the cornea, because the jump from air to cornea is the biggest change in refractive index. The lens adds the rest and can change its shape.
Ray diagrams and the lens equation
We model the eye as one thin converging lens. With the real-is-positive rule: 1/f = 1/u + 1/v and P = 1/f (dioptres, f in metres). The image on the retina is real, inverted and smaller; the brain turns it the right way up.
Accommodation, near point and far point
Accommodation is the eye changing its power to focus at different distances. When the ciliary muscles relax, the suspensory ligaments pull the lens thin: low power, good for far objects. When they contract, the lens gets fatter: more power, good for near objects.
The far point is the furthest point you see clearly (infinity for a normal eye). The near point is the closest (about 25 cm for a young adult). It moves further away with age (presbyopia), because the lens gets stiffer.
Sensitivity, colour and spatial resolution
The retina has two kinds of light detector:
- Rods: about 120 million, mostly away from the centre. Very sensitive, so they work in dim light. Only one type, so no colour.
- Cones: about 6 million, packed in the fovea (centre). Three types with peak response near 420 nm (blue), 530 nm (green) and 560 nm (red). The brain compares their signals to see every colour β this is the trichromatic idea. They need bright light.
Why rods give poor detail
Many rods share one nerve fibre, so the brain cannot tell which rod was hit. Each fovea cone has its own fibre. Two points look separate only if their images fall on two cones with at least one unstimulated cone between them. This gives a best resolution of roughly 1 arc-minute (about 3 Γ 10β»β΄ rad).
Persistence of vision
An image stays for about 1/10 s, so frames shown faster than about 20 per second blend into smooth motion.
Defects of vision and their correction
Myopia (short sight)
Far point is closer than infinity. Rays from far objects focus in front of the retina. Fix: a diverging lens that makes far objects seem to be at the far point. f = β(far point distance).
Hypermetropia (long sight)
Near point is further than 25 cm. Fix: a converging lens that takes an object at 25 cm and forms a virtual image at the person's near point. Use 1/f = 1/u + 1/v with u = +0.25 m and v = β(near point).
Astigmatism
The cornea is curved more in one direction than another, so lines in one direction are blurred. Fix: a cylindrical lens with its axis set at the right angle. A prescription gives power and axis angle.
Combining lenses
Thin lenses in contact: P = Pβ + Pβ.
Try it
Hold a pencil at arm's length and bring it slowly towards one eye. Note where it first goes blurry: that is your near point. Cover one eye and look at a printed word in a dark room as your eyes adjust β you lose colour first. In the 3D, choose short sight, then press glasses and watch the focus move back onto the retina.
Key formulas and definitions
- P = 1/f (P in dioptres D, f in metres)
- 1/f = 1/u + 1/v (real is positive; virtual image distance negative)
- Lenses in contact: P = Pβ + Pβ + β¦
- Myopia correction: f = β(far point distance)
- Hypermetropia: u = 0.25 m, v = β(near point) β 1/f = 1/0.25 β 1/(near point)
- Normal eye: near point β 25 cm, far point = β, total power β +60 D
Worked examples
1. A lens has focal length 25 cm. What is its power?
P = 1/f = 1/0.25 = +4 D.
2. A short-sighted person has a far point of 2.0 m. Find the power of the correcting lens.
The lens must make objects at infinity appear at 2.0 m, so f = β2.0 m. P = 1/(β2.0) = β0.5 D (diverging).
3. A long-sighted person's near point is 1.0 m. What lens lets them read at 25 cm?
u = 0.25 m, v = β1.0 m (virtual image). 1/f = 1/0.25 + 1/(β1.0) = 4 β 1 = 3. P = +3 D, f β 0.33 m (converging).
4. A myopic eye's far point is 50 cm. With the correct lens, what is the new near point if the bare near point was 15 cm?
Lens: f = β0.50 m. Object at u gives virtual image at v = β0.15 m: 1/u = 1/f β 1/v = β2 + 6.67 = 4.67, so u β 0.21 m. New near point β 21 cm.
5. Two thin lenses of +2.5 D and β1.0 D touch. Find the combined focal length.
P = 2.5 β 1.0 = +1.5 D. f = 1/1.5 β 0.67 m.
6. Estimate the smallest gap between two dots you can resolve at 25 cm if the eye resolves 3 Γ 10β»β΄ rad.
Gap = angle Γ distance = 3 Γ 10β»β΄ Γ 0.25 β 7.5 Γ 10β»β΅ m β 0.08 mm.
Common mistakes
- Using f in centimetres in P = 1/f. 20 cm must be 0.20 m, giving 5 D, not 0.05 D.
- Swapping the fixes: myopia needs a diverging (negative) lens, hypermetropia a converging (positive) lens.
- Forgetting the minus sign for the virtual image distance in long-sight questions.
- Saying rods see colour or that cones work best in dim light. It is the other way round.