Physics › Medical physics › The physics of the eye
The physics of the eye
A camera focuses by sliding its lens. The eye cannot: its screen is glued 23 mm behind the lens and will not move. So the lens changes shape instead, and when it can no longer change enough, an optician sells you the difference in dioptres.
Builds on Lenses and images and Refraction and total internal reflection.
IN THIS TOPIC
- Trace rays through the eye and say what kind of image lands on the retina.
- Explain accommodation as a change of power at fixed image distance, and use 1/u + 1/v = 1/f to find the eye's focal length.
- Find the power in dioptres of the lens that corrects myopia or hypermetropia.
- Say what astigmatism is and read the three numbers of its prescription.
- Explain the eye's spectral response and its spatial resolution in terms of rods and cones.
WHAT YOU PROBABLY THINK
The eye focuses on near and far objects by moving its lens, the way a camera does.
One converging system, one fixed screen
Light entering the eye is bent twice. Most of the bending happens at the cornea, the curved front window, because that is where light crosses the largest change of refractive index. The lens behind it makes the smaller adjustment, and the iris opens and closes the pupil to control how much light gets in. For calculations the whole arrangement is treated as one thin converging lens, with the retina acting as the screen about 23 mm behind it.
So the image on the retina is real, inverted and diminished. Nothing later turns it the right way up. The brain simply learns to read it, which is why the standard question asks about the image and not about what you experience.
Three equations from the lenses lesson do all the arithmetic here. Power and focal length are two names for one property,
with f in metres and P in dioptres. Object distance, image distance and focal length are tied together by
in the real is positive convention, and the size of the image follows from
Accommodation: the power that changes
Here is what the opening claim gets wrong. The retina cannot move, so v is stuck at about 23 mm whatever you look at. Read 1/u + 1/v = 1/f with v fixed and one thing has to give. When u changes, f must change with it, and the eye must supply a different power for every distance. Ciliary muscles do it by squeezing the lens into a fatter, more strongly converging shape. That is accommodation.
Two distances mark the limits of what accommodation can reach. The far point is the furthest object the eye can focus, at infinity for a healthy eye, and the eye is fully relaxed there. The near point is the closest, taken as 0.25 m for a normal adult eye and reached with the lens working hardest. That 0.25 m is also called the least distance of distinct vision, and it is the distance every reading calculation assumes.
WORKED EXAMPLE
How much power does accommodation supply?
An eye has its retina 23 mm behind the lens and its far point at infinity. Find the power of the relaxed eye, and the power it needs to focus on a page 0.25 m away.
Relaxed: u is infinite, so 1/u is zero and 1/f = 1/0.023. The focal length is 23 mm and P = 43 D.
On the page: 1/f = 1/0.25 + 1/0.023 = 4.0 + 43.5 = 47.5, so P = 48 D and f = 21 mm.
Accommodation therefore supplies 47.5 − 43.5 = 4.0 D, which is exactly 1/0.25. The extra power an eye needs is always one over the object distance, because the 1/v term never moves.
The trap is writing v = 0.25 m. The object is at 0.25 m; the image is on the retina, where it always is.
Two defects, two lenses
Myopia, short sight, is the eye that is too powerful for its own length, so parallel rays from a distant object converge in front of the retina and arrive at it already spreading again. Distant objects blur. The eye's far point is no longer at infinity but at some finite distance, and near vision is usually fine.
Hypermetropia, long sight, is the opposite. The eye has too little power for its length, so rays from a close object would meet behind the retina. Close objects blur, and the near point has moved further out than 0.25 m.
A correcting lens is not asked to form an image on the retina. It is asked to move the object to somewhere the eye can already cope with, and the image it makes is the object the eye then works on. That image is virtual, on the same side as the object, so it carries a negative v.
WORKED EXAMPLE
The lens for a short-sighted eye
A myopic eye has its far point 0.40 m away. Find the power of the spectacle lens that lets it see distant objects.
The lens must take an object at infinity and place its image at the far point, so u is infinite and v = −0.40 m.
1/f = 0 + 1/(−0.40) = −2.5, so P = −2.5 D, a diverging lens.
Drop the minus sign on v and the answer comes out +2.5 D. That lens would converge the light still earlier and make the sight worse, so the sign is the whole question.
WORKED EXAMPLE
The lens for a long-sighted eye
A hypermetropic eye has its near point 1.0 m away. Find the power of the lens that lets it read at the normal 0.25 m.
The lens must take the page at u = 0.25 m and form a virtual image at the eye's own near point, so v = −1.0 m.
1/f = 1/0.25 + 1/(−1.0) = 4.0 − 1.0 = 3.0, so P = +3.0 D, a converging lens.
Note which distance is which. The near point belongs to the eye and becomes v; where the patient wants to read becomes u.
Astigmatism is a different fault. The cornea, or sometimes the lens, is not the same shape in every direction across its face, so it has more power along one axis than along the one at right angles to it. A point object cannot be brought to a point image in both directions at once, and the patient sees lines sharp in one orientation and blurred in the other.
The cure is a lens carrying matching extra power along one axis only, so a prescription needs three numbers rather than one. SPH is the ordinary spherical power in dioptres. CYL is the extra cylindrical power, again in dioptres. AXIS is an angle in degrees from 0 to 180 saying which direction the cylinder acts along. A prescription reading −2.50 / −0.75 × 90 is a myopic eye whose astigmatism is corrected on the vertical axis.
The retina as a photodetector
As a detector the eye is answering two different questions at once, and it has two kinds of cell to answer them. Cones work in bright light, come in three types responding to different bands of wavelength, and between them give colour vision. Rods work in dim light, come in one type only, and so report brightness and nothing about colour.
That figure is the eye's spectral response, and it is the reason the visible spectrum ends where it does. Sensitivity peaks in the yellow-green, near 555 nm in good light, and falls to nothing outside roughly 400 nm to 700 nm. The rods peak at a shorter wavelength, near 500 nm, so as the light fades the eye's best colour shifts towards blue and reds go dark first.
Spatial resolution is decided by how the cells are wired, not by how many there are. Cones crowd into the fovea, a small patch on the axis of the eye, and each one there has effectively its own nerve fibre. Two lit cones with an unlit one between them therefore arrive at the brain as two separate signals, and detail survives. Rods lie outside the fovea and many of them share a single fibre, so their signals are added together before they leave the eye.
That sharing is a bargain, not a fault. Adding the light caught by many rods makes a signal large enough to notice when almost no light is arriving, which is why the dark-adapted eye can work by starlight. What it costs is detail, because the brain cannot tell which rod in the group was lit. Cones make the opposite trade, and the eye's practical resolution of about one minute of arc, roughly 3 × 10−4 rad, is a fovea figure. Look slightly to one side of a faint star and it brightens, because you have moved its image off the fovea and onto the rods.
THE EXAM BIT
- Say real, inverted and diminished for the retinal image. All three words are separately creditable and the whole phrase takes one line.
- Accommodation answers must name what stays fixed. The image distance is the length of the eyeball and cannot change, so the focal length and hence the power must change instead, and the ciliary muscles change the shape of the lens to do it.
- In every eye calculation the retina is the image, so v is the eyeball length and u is the object. Swapping them is the commonest way to lose all four marks.
- Correcting-lens questions turn on one virtual image. Myopia: object at infinity, image at the far point, v negative, so the power comes out negative. Hypermetropia: object at 0.25 m, image at the patient's near point, v negative again, and the power comes out positive.
- Convert to metres before taking any reciprocal. A dioptre is one per metre, so 23 mm has to be 0.023 m, and a power quoted per centimetre is worth no marks.
- For astigmatism, AQA asks for the format of the prescription rather than the optics. Three numbers: spherical power, cylindrical power, and the axis in degrees.
- Resolution answers must reach the nerve fibres. Cones in the fovea have one fibre each, so nearby cones stay separate; many rods share a fibre, so their signals are combined and cannot be told apart.
CHECK YOURSELF
An eye has its retina 23 mm behind its lens. Its far point is 0.50 m away. Find the power of the relaxed eye, state which defect this is, and find the power of the correcting lens.
Show a hint
Relaxed means focused on the far point. The correcting lens takes an object at infinity to that same far point.
Show the answer
Relaxed on the far point: 1/f = 1/0.50 + 1/0.023 = 2.0 + 43.5 = 45.5, so P = 46 D.
A far point at a finite 0.50 m instead of infinity is myopia. The relaxed eye is more powerful than its length needs, so distant objects focus in front of the retina.
The correcting lens must place an object at infinity at the far point: u infinite, v = −0.50 m, so 1/f = −2.0 and P = −2.0 D, a diverging lens.
Check the sense of it. The eye is 2.0 D too strong for infinity, and the lens takes exactly 2.0 D away.
The image distance in the eye is fixed, so focusing changes the power, not the position of the lens.
Power in dioptres needs the focal length in metres, and a diverging lens has a negative power.
Myopia is a finite far point corrected by a diverging lens; hypermetropia is a distant near point corrected by a converging lens.
Cones give colour and detail in bright light; rods give sensitivity in the dark and lose the detail by sharing nerve fibres.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Trace rays through the eye and say what kind of image lands on the retina.
- Explain accommodation as a change of power at fixed image distance, and use 1/u + 1/v = 1/f to find the eye's focal length.
- Find the power in dioptres of the lens that corrects myopia or hypermetropia.
- Say what astigmatism is and read the three numbers of its prescription.
- Explain the eye's spectral response and its spatial resolution in terms of rods and cones.
Open the full revision checklist to track your progress across the whole unit.