The purpose of the eye is to form and capture images of the world. The optical components of the eye (cornea and crystalline lens) form an image on the back of the eyeball. The back of the eyeball is covered with an outgrowth of the brain, called the retina, which absorbs the light in the image and turns it into electrical signals, which are then transmitted via a network of neurons along the optic nerve to the brain.
Figure 1 shows a diagram of the eye, with the main optical elements labelled. The eye has four (or depending how you count them, five) refracting surfaces:
The first refracting surface is the convex front of the cornea, which has a positive power. However, the cornea is covered with a thin film of tear fluid (which is another refracting surface, if you count it that way), which provides a smooth surface for the cornea.
The second refracting surface is the concave back of the cornea. The back surface has a negative power.
The third refracting surface is the convex front of the crystalline lens, which has a positive power. The gap between the cornea and the lens is filled with a fluid called the aqueous (which means “watery”) or aqueous humor (“humor” means fluid). The iris, which controls the aperture (i.e. pupil) of the eye, is in front of this surface.
The last refracting surface is the convex back of the lens, which also has a positive power. The final space between the lens and the retina is filled with a jelly-like substance called the vitreous (which means “glassy”) or vitreous humor.
The cornea has a fixed power of about \( +40\text{D} \) , but the power of the lens can vary from about \( +20\text{D} \) up to about \( +30\text{D} \). This variation in lens power is called accommodation and lets us focus on objects at different distances.
Another important optical element of the eye is the iris. The iris is a thin structure lying between the crystalline lens and the cornea. It has a hole in the middle, called the pupil. The iris controls the size of the pupil. The pupil has two functions. One is to alter the amount of light entering the eye (a small pupil admits little light, and a large pupil admits a lot). The main function, however, is to optimize the quality of the image formed on the retina.
The eye is roughly rotationally symmetrical. The (rough) axis of symmetry is called the optic axis ( Figure 2). Rays of light that travel close to the optic axis are paraxial rays. The term “central vision” refers to the area of the image on the retina which is near the optic axis. Central vision can be mostly understood using the paraxial optics that have been covered in Chapters 1 to 6.
The image near where the optic axis hits the retina has the highest image quality. For that reason, the retinal photoreceptors are tightly packed near the central retina. However, the area of greatest photoreceptor density (the fovea) actually sits a little off to one side of the centre of the retina. The visual axis is defined as a line from the fovea through the centre of the crystalline lens. This is about \( 5^o \) away from the optic axis.
The retinal image gets progressively worse as we move away from the optic axis. This area of relatively poor vision is called “Peripheral Vision”.
A model eye is a simplified eye which captures some of the important optical properties of a real eye, and they can be useful for understanding the optical principles of real eyes. One model eye is called LeGrand’s Full Theoretical Eye, shown in Figure 3. This model eye ignores the tear film, and treats the crystalline lens as an object with a single refractive index. In reality, the refractive index of the lens varies from surface to centre (called a graded index) which gives the real crystalline lens a bit more power.
The optical properties of the refracting surfaces of the LeGrand eye are given in Table 1. Powers are calculated using the surface power formula given in Chapter 3. Most of the power of the eye is in the cornea (about \( 40\text{D} \) in total), and the power of the lens surfaces is roughly \( 20\text{D} \).
| Element | Radius | Index to left | Index to right | Surface power |
|---|---|---|---|---|
| Front surface of cornea | 7.8 mm | 1 (air) | 1.3771 | +48.35 D |
| Back surface of cornea | 6.5 mm | 1.3771 | 1.3374 | -6.11 D |
| Front surface of lens |
10.2 mm | 1.3374 | 1.42 | +8.10 D |
| Back surface of lens |
-6 mm | 1.42 | 1.336 | +14.00 D |
We can use the methods described in Chapters 3 and 4 to work out what happens to light as it travels through this model eye, which will be similar to what happens when light travels through a real eye. We work with each surface one at a time.
First, the front surface of the cornea. Let's suppose the light hitting the cornea is coming from a distant object, that is \( V_{in}=0 \):
| Step | Reason | Statement |
|---|---|---|
| 1 | Paraxial Equation | \( n_{in}V_{in}+F_{surface}=n_{out}V_{out} \) |
| 2 | Fill in power, refractive indices from Table 1, row 1 and set \( V_{in}=0 \) |
\( 1\times 0 + 48.35 = 1.3771\times V_{out} \) |
| 3 | Solve for \( V_{out} \) | \( V_{out}=48.35/1.3771=35.11\text{D} \) |
Next, the back surface of the cornea. We need to know what \( V_{in} \) is for the back surface, so we use the stepalong procedure from Chapter 4:
| Step | Reason | Statement |
|---|---|---|
| 1 | \( V_{out} \) from the front surface | \( V_{out}=35.11 \) |
| 2 | Work out distance to point of convergence | \( \text{distance}=1/V_{out}= 0.02848\text{m} \) |
| 3 | Subtract the gap \( 0.00055\text{m}
\) between front and back surfaces |
\( 0.02848-0.00055=0.0279319\text{m} \) |
| 4 | \( V_{in} \) for the back surface is one over this: | \( V_{in}=1/0.0279319 = 35.80\text{D} \) |
From this value of \( V_{in} \) for the back surface of the cornea, we can work out \( V_{out} \) at the back surface. We won't do that step-by-step though; it's quite tedious. Instead, the table(s) below take us through the calculations for all surfaces.
(It's worth trying to replicate some of these calculations yourself, just to be sure you know where they come from, but be aware that the values in the table are rounded slightly.)
The full table (Table 6) shows a few interesting things:
In previous chapters, we've thought of the image formed by a thin lens or a paraxial refracting surface as being flat (at least when all the objects are at the same distance) as in Figure 4. In real optical systems, rather than simplified paraxial ones, the image is only flat close to the centre of the image, and is usually a different shape away from the centre.
Quite a lot of careful engineering goes into making sure that the image formed by a camera lens is as flat as possible, because the sensor in a camera (a CCD array or a chemical film) is also flat. If the image is flat, then it will be in focus everywhere on the sensor.
However, the sensor in our eye - the retina - is round, so a flat image isn't desirable for the eye. The shape of the image formed in the eye is called the image shell. The image shell is where all the sharp image points are, when looking at objects that are all the same distance away.
Working out the precise shape of the image shell for a real eye (or even a complicated model eye) can't be done using the paraxial equation, because that is only valid for light rays near the optic axis of the eye. Instead, the image shell must be calculated by simulation, as in Figure 5. Here, a beam of parallel light from a distant point enters the eye and is focused onto the retina. If you drag the light beam up and down with a mouse or your finger, the focused point traces out a curve. Note that the curve only touches the retina exactly at the back of the eyeball, and moves away from it as the light hits the eye at greater angles. However, this is the image shell of the LeGrand model eye, not the real eye.
The model eye calculations show that the vergence of the light leaving the back surface of the lens ( \( 60.25\text{D} \)) is exactly what you need to have the light converge to a point over the distance to the retina ( \( 16.6\text{mm} \)). But that perfect match only happens when you have parallel light ( \( V_{in}=0 \)) entering the eye. What happens if divergent light, say from a near object, enters the eye?
Suppose, for example, an object is placed \( 0.5\text{m} \) in front of the eye. Then the light reaching the cornea is diverging, with a vergence of \( V_{in}=-2\text{D} \). We can work out what happens next, just as we did earlier. I'll spare you the details: the light leaving the back surface of the lens has a vergence of \( V_{out}=57.59\text{D} \); that is, about \( 2.5\text{D} \) lower than the \( 60\text{D} \) in Table 2. The light will then converge to a point \( 17.4\text{mm} \) from the back of the lens, which is nearly a whole millimetre behind the retina. This will produce a blurred image, and it will be very blurred, even though the distance is small. You can simulate this using Figure 6 below: turn the light on and drag the Distance slider to move the object closer to the eye.
But we can clearly see objects that are \( 0.5\text{m} \) away, so how do we do it? If our eye was built like a camera, we would simply move the retina \( 1\text{mm} \) backwards to catch the sharp image. However our eye isn't built that way, and the retina can't move. Instead, what we do is change the power of the eye, specifically the power of the crystalline lens. This is called accommodation.
The natural shape of the crystalline lens in the eye is rather bulged and rounded. The lens is held in place by strands of tissue (suspensory ligaments), which are attached to a ring of muscle on the inside of the eye called the ciliary muscle. When this muscle ring is relaxed, it is close to the inside of the eyeball, and the suspensory ligaments pull on the lens so that it is flatter, and has a low power. This is a bit like a trampoline, where the fabric in the middle is held in tension by the springs connected to the circular frame. When the ciliary muscle ring tightens, the width of the muscle ring gets smaller, so the suspensory ligaments don’t pull on the lens quite so hard, and it settles into its natural more rounded shape. The more rounded the lens, the more curved the lens surfaces are, so this increases the power of the lens. You can see this in ( Figure 6): if you move the Accommodation slider, the lens gets thicker and it bends the light more.
The process of accommodation in the eye adjusts the power of the lens so that the light from a near object is focused on the retina.
It is very difficult to work out the power of the lens when the eye is accommodating. For this reason, we measure accommodation by the divergence of the light hitting the cornea, rather than the change in lens power that accommodation needs to focus the divergent light.
For example, if an object is placed \( 0.333\text{m} \) from a person, the light from the object has a vergence of \( -3\text{D} \) when it hits the person's cornea. If the person can see the object clearly, we say that the person's eye has accommodated by \( 3\text{D} \), since the eye has added enough convergence to the light rays to overcome the intial increased divergence of \( -3\text{D} \). Note that \( 3\text{D} \) of accommodation does not mean that the lens has increased its power by exactly that amount (although it has, roughly).
There is a limit to the amount of accommodation that an eye has. You can find this limit by getting closer and closer to an object until you can no longer see it clearly. The closest you can get to an object while still seeing it clearly is called the near point. For example, if a person can clearly see an object placed \( 0.16\text{m} \) in front of their eye, but no closer, then their eye can cope with a divergence of \( 1/(-0.16)=-6.25\text{D} \). The maximum amount of accommodation the patient has is thus \( 6.25\text{D} \).
The near point of the eye in Figure 6 is about \( 0.143\text{m} \), because when you set the Distance slider to that distance, you must use up 100% of the Accommodation to focus the light. So the maximum amount of accommodation that eye has is \( 1/0.143 = 7\text{D} \). The maximum amount of accommodation is called the Amplitude of Accommodation.
Presbyopia is the progressive loss of accommodation as the eye ages. Accommodation relies on the lens being flexible enough to change its shape in response to pulling by the suspensory ligaments. Unfortunately, as we age, the lens loses its flexibility, and the amplitude of accommodation reduces. The name presbyopia means “old sight” since everyone older than about 45 years notices its effects, although it starts younger than that.
If the eye loses accommodation, the near point moves further away from the eye because the near point is related to the total accommodation available. The first sign of presbyopia, then, is when you have to hold objects further away in order to see them clearly; in effect, you have to move the object away until it reaches your near point.
When a person can no longer see near objects clearly, they will probably want reading glasses. These glasses simply add additional power to the eye to compensate for the lack of accommodation that comes with age.
Suppose, for example, a patient has a reading distance of \( 45\text{cm} \). The vergence of the light leaving the pages of a book at that distance is \( 1/(-0.45) = -2.22\text{D} \). If they were young, they could easily accommodate by \( 2.22\text{D} \) to counteract the divergence, and so see the pages of the book clearly. If they are old, however, that accommodation may be completely gone, and they will not be able to focus the pages of the book. However, if we gave them spectacles with power around \( +2\text{D} \), they would then be able to read the book again, because these spectacles would counteract some or all of the divergence of the light from the book.
You can find such spectacles in pound stores or supermarkets, marked with the power (e.g. \( +1 \), \( +2 \), although this is sometimes incorrectly called magnification). However, the centres of these cheap lenses may not line up very well with the centres of the patient’s eyes, and the lens quality is substantially worse than spectacles sold by an optometrist; nonetheless, for a cash-strapped patient, they are far better than nothing.
If the presbyopic patient has no remaining accommodation at all, the power of reading glasses is just one over the distance they will be reading at (or whatever near work they want them for, like painting Warhammer toys). If however the presbyopic patient has some remaining accommodation, a calculation of the power of the reading glasses is not as straightforward, since they will automatically use some (but not all) of their accommodation when doing near work. This automatic accommodation should be taken into account when figuring the power of reading glasses.
Prescribing reading glasses accurately can sometimes be difficult. The problem is that the reading prescription is closely matched to the working distance. For example, a reading prescription of \( +2.5\text{D} \) is for a distance of \( 1/2.5=0.4\text{m} \). It can't be used at distances much different from this, if the patient is presbyopic. This can be a problem if the reading prescription is added onto a distance prescription (for myopia or hyperopia, see the next two chapters), because the distance prescription will often be wrong (occasionally by as much as \( 0.5\text{D} \)). This error is not an issue at distance, but becomes important at near.
The eye, like all optical systems, suffers from the aberrations covered in Chapter 6. The eye reduces the blur caused by aberration by reducing the pupil size when needed. It also reduces aberrations in ways unique to the eye:
This affects the quality of the image over the entire retina. The eye deals with chromatic aberration by simply ignoring many wavelengths. The retina is most sensitive to light of about 580 nanometres (yellow), and is very insensitive to light at the red and blue ends of the visible spectrum. Thus, while the image formed from blue wavelengths (at say 450nm) is quite blurred compared to that formed by yellow (580nm) wavelengths, the retina - particularly the fovea - simply cannot see blue light very well, and doesn’t notice the blue blur much. In addition, the fovea, which has the best resolution of the retina, is covered by a yellowish filter, which screens out much of the blue wavelengths in the visible spectrum.
The eye deals with spherical aberration in two ways. First, the cornea is not actually spherical, but flattens at the edges, so it has a lower power than a spherical surface at the edges. The second way the eye deals with spherical aberration is to ignore the aberrant rays. The rays from the edge of a lens that form the blurred aberrant image will hit the retina at a different angle than the unaberrated (central) rays. The photoreceptors in the fovea are most sensitive to light rays hitting the retina at a particular angle, depending on where the photoreceptor is located. Since the aberrated rays hit at a different angle from the preferred angle of the photoreceptor, the retina is less sensitive to them.
This is only a problem in peripheral vision. The eye deals with this by simply not paying much attention to peripheral vision. If you want to see an object clearly, you move your eyes so the image of that object falls in the central visual area.