Contents

Physics of the Eye

The eye is perhaps the most interesting of all optical instruments. The eye is remarkable in how it forms images and in the richness of detail and color it can detect. However, our eyes commonly need some correction, to reach what is called “normal” vision, but should be called ideal rather than normal. Image formation by our eyes and common vision correction are easy to analyze with the optics discussed in Geometric Optics.

Figure 26.2 shows the basic anatomy of the eye. The cornea and lens form a system that, to a good approximation, acts as a single thin lens. For clear vision, a real image must be projected onto the light-sensitive retina, which lies at a fixed distance from the lens. The lens of the eye adjusts its power to produce an image on the retina for objects at different distances. The center of the image falls on the fovea, which has the greatest density of light receptors and the greatest acuity (sharpness) in the visual field. The variable opening (or pupil) of the eye along with chemical adaptation allows the eye to detect light intensities from the lowest observable to $10^{10} 10^{10} size 12{"10" rSup { size 8{"10"} } } {}$ times greater (without damage). This is an incredible range of detection. Our eyes perform a vast number of functions, such as sense direction, movement, sophisticated colors, and distance. Processing of visual nerve impulses begins with interconnections in the retina and continues in the brain. The optic nerve conveys signals received by the eye to the brain.

The figure depicts the internal structure of an eye with labels. These labels include cornea, iris, aqueous humor, ciliary fibers, lens, vitreous humor, retina, fovea, sclera disc, and optic nerve.

Figure 26.2 The cornea and lens of an eye act together to form a real image on the light-sensing retina, which has its densest concentration of receptors in the fovea and a blind spot over the optic nerve. The power of the lens of an eye is adjustable to provide an image on the retina for varying object distances. Layers of tissues with varying indices of refraction in the lens are shown here. However, they have been omitted from other pictures for clarity.

Refractive indices are crucial to image formation using lenses. Table 26.1 shows refractive indices relevant to the eye. The biggest change in the refractive index, and bending of rays, occurs at the cornea rather than the lens. The ray diagram in Figure 26.3 shows image formation by the cornea and lens of the eye. The rays bend according to the refractive indices provided in Table 26.1. The cornea provides about two-thirds of the power of the eye, owing to the fact that speed of light changes considerably while traveling from air into cornea. The lens provides the remaining power needed to produce an image on the retina. The cornea and lens can be treated as a single thin lens, even though the light rays pass through several layers of material (such as cornea, aqueous humor, several layers in the lens, and vitreous humor), changing direction at each interface. The image formed is much like the one produced by a single convex lens. This is a case 1 image. Images formed in the eye are inverted but the brain inverts them once more to make them seem upright.

Material	Index of Refraction
Water	1.33
Air	1.0
Cornea	1.38
Aqueous humor	1.34
Lens	1.41 average (varies throughout the lens, greatest in center)
Vitreous humor	1.34

Table 26.1 Refractive Indices Relevant to the Eye

Ray diagram in the picture shows the internal structure of an eye and a tree that is taken as an object. An inverted image of the tree is formed on retina with the light rays coming from the top and bottom of the tree; converging most at the cornea and upon entering and exiting the lens. The rays coming from top of the tree are labeled one, two, while the bottom rays are labeled three, four. The inverted image of the tree shows rays labeled three, four at the top and one, two at the bottom.

Figure 26.3 An image is formed on the retina with light rays converging most at the cornea and upon entering and exiting the lens. Rays from the top and bottom of the object are traced and produce an inverted real image on the retina. The distance to the object is drawn smaller than scale.

As noted, the image must fall precisely on the retina to produce clear vision — that is, the image distance $d_{i} d_{i} size 12{d rSub { size 8{i} } } {}$ must equal the lens-to-retina distance. Because the lens-to-retina distance does not change, the image distance $d_{i} d_{i} size 12{d rSub { size 8{i} } } {}$ must be the same for objects at all distances. The eye manages this by varying the power (and focal length) of the lens to accommodate for objects at various distances. The process of adjusting the eye’s focal length is called accommodation. A person with normal (ideal) vision can see objects clearly at distances ranging from 25 cm to essentially infinity. However, although the near point (the shortest distance at which a sharp focus can be obtained) increases with age (becoming meters for some older people), we will consider it to be 25 cm in our treatment here.

Figure 26.4 shows the accommodation of the eye for distant and near vision. Since light rays from a nearby object can diverge and still enter the eye, the lens must be more converging (more powerful) for close vision than for distant vision. To be more converging, the lens is made thicker by the action of the ciliary muscle surrounding it. The eye is most relaxed when viewing distant objects, one reason that microscopes and telescopes are designed to produce distant images. Vision of very distant objects is called totally relaxed, while close vision is termed accommodated, with the closest vision being fully accommodated.

Two cross-sectional views of eye anatomy are shown. In part a of the figure, parallel rays from distant object are entering the eye and are converging on the retina to produce an inverted image of the tree shown above the principle axis. The interior lens of the eye is relaxed and least rounded, given as P small. Distance of image d i is equal to two centimeters, which is the measure of the distance from lens to retina. Distance of object d o is given as very large. In part b of the figure, rays from a button, which is a nearby object, are shown to diverge as they enter the eye. The interior lens of the eye, P large, converges the rays to form an image at retina, below the principle axis. Distance of image d i is equal to two centimeters, which is the measure of distance from lens to retina. Distance of object d o is given as very small.

Figure 26.4 Relaxed and accommodated vision for distant and close objects. (a) Light rays from the same point on a distant object must be nearly parallel while entering the eye and more easily converge to produce an image on the retina. (b) Light rays from a nearby object can diverge more and still enter the eye. A more powerful lens is needed to converge them on the retina than if they were parallel.

We will use the thin lens equations to examine image formation by the eye quantitatively. First, note the power of a lens is given as $p = 1 / f p = 1 / f size 12{p= {1} slash {f} } {}$ , so we rewrite the thin lens equations as

P = \frac{1}{d_{o}} + \frac{1}{d_{i}} P = \frac{1}{d_{o}} + \frac{1}{d_{i}}

26.1

and

\frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} = m. \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} = m.

26.2

We understand that $d_{i} d_{i} size 12{d rSub { size 8{i} } } {}$ must equal the lens-to-retina distance to obtain clear vision, and that normal vision is possible for objects at distances $d_{o} = 25 cm d_{o} = 25 cm$ to infinity.

Take-Home Experiment: The Pupil

Look at the central transparent area of someone’s eye, the pupil, in normal room light. Estimate the diameter of the pupil. Now turn off the lights and darken the room. After a few minutes turn on the lights and promptly estimate the diameter of the pupil. What happens to the pupil as the eye adjusts to the room light? Explain your observations.

The eye can detect an impressive amount of detail, considering how small the image is on the retina. To get some idea of how small the image can be, consider the following example.

Example 26.1

Size of Image on Retina

What is the size of the image on the retina of a $1 . 20 \times 10^{- 2} 1 . 20 \times 10^{- 2} size 12{1 "." "20" times "10" rSup { size 8{ - 2} } } {}$ cm diameter human hair, held at arm’s length (60.0 cm) away? Take the lens-to-retina distance to be 2.00 cm.

Strategy

We want to find the height of the image $h_{i} h_{i}$ , given the height of the object is $h_{o} = 1 . 20 \times 10^{- 2} h_{o} = 1 . 20 \times 10^{- 2} size 12{h rSub { size 8{0} } =1 "." "20" times "10" rSup { size 8{ - 2} } } {}$ cm. We also know that the object is 60.0 cm away, so that $d_{o} = 60.0 cm d_{o} = 60.0 cm size 12{d rSub { size 8{0} } } {}$ . For clear vision, the image distance must equal the lens-to-retina distance, and so $d_{i} = 2.00 cm d_{i} = 2.00 cm size 12{d rSub { size 8{i} } } {}$ . The equation $\frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} = m \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} = m size 12{ { {h rSub { size 8{i} } } over {h rSub { size 8{o} } } } = - { {d rSub { size 8{i} } } over {d rSub { size 8{o} } } } =m} {}$ can be used to find $h_{i} h_{i}$ with the known information.

Solution

The only unknown variable in the equation $\frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} = m \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} = m size 12{ { {h rSub { size 8{i} } } over {h rSub { size 8{o} } } } = - { {d rSub { size 8{i} } } over {d rSub { size 8{o} } } } =m} {}$ is $h_{i} h_{i} size 12{h rSub { size 8{i} } } {}$ :

\frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} . \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} .

26.3

Rearranging to isolate $h_{i} h_{i} size 12{h rSub { size 8{i} } } {}$ yields

h_{i} = - h_{o} \cdot \frac{d_{i}}{d_{o}} . h_{i} = - h_{o} \cdot \frac{d_{i}}{d_{o}} .

26.4

Substituting the known values gives

\begin{array}{l} h_{i} & = & - (1.20 \times 10^{- 2} cm) \frac{2.00 cm}{60.0 cm} \\ = & - 4.00 \times 10^{- 4} cm . \end{array} \begin{array}{l} h_{i} & = & - (1.20 \times 10^{- 2} \\ = - 4.00 \times 10^{- 4} \end{array}

26.5

Discussion

This truly small image is not the smallest discernible—that is, the limit to visual acuity is even smaller than this. Limitations on visual acuity have to do with the wave properties of light and will be discussed in the next chapter. Some limitation is also due to the inherent anatomy of the eye and processing that occurs in our brain.

Example 26.2

Power Range of the Eye

Calculate the power of the eye when viewing objects at the greatest and smallest distances possible with normal vision, assuming a lens-to-retina distance of 2.00 cm (a typical value).

Strategy

For clear vision, the image must be on the retina, and so $d_{i} = 2.00 cm d_{i} = 2.00 cm$ here. For distant vision, $d_{o} \approx ∞ d_{o} \approx ∞$ , and for close vision, $d_{o} = 25.0 cm d_{o} = 25.0 cm$ , as discussed earlier. The equation $P = \frac{1}{d_{o}} + \frac{1}{d_{i}} P = \frac{1}{d_{o}} + \frac{1}{d_{i}}$ as written just above, can be used directly to solve for $P P$ in both cases, since we know $d_{i} d_{i}$ and $d_{o} d_{o} size 12{d rSub { size 8{0} } } {}$ . Power has units of diopters, where $1 D = 1/m 1 D = 1/m size 12{"1D"="1/m"} {}$ , and so we should express all distances in meters.

Solution

For distant vision,

P = \frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{\infty} + \frac{1}{0.0200 m} . P = \frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{\infty} + \frac{1}{0.0200 m} .

26.6

Since $1 / \infty = 0 1 / \infty = 0 size 12{1/ infinity =0} {}$ , this gives

P = 0 + 50 . 0 / m = 50.0 D (distant vision). P = 0 + 50 . 0 / m = 50.0 D (distant vision).

26.7

Now, for close vision,

\begin{array}{l} P & = & \frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{0.250 m} + \frac{1}{0.0200 m} \\ = & \frac{4.00}{m} + \frac{50.0}{m} = 4.00 D + 50.0 D \\ = & 54.0 D (close vision). \end{array} \begin{array}{l} P & = & \frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{0.250 m} + \frac{1}{0.0200 m} \\ = \frac{4.00}{m} + \frac{50.0}{m} = 4.00 D + 50.0 D \\ = 54.0 D (close vision). \end{array} alignl { stack {
 size 12{P= {  {1}  over  {d rSub { size 8{o} } } } + {  {1}  over  {d rSub { size 8{i} } } } = {  {1}  over  {0 "." "250 m"} } + {  {1}  over  {0 "." "0200 m"} } }  {} #
= {  {4 "." "00"}  over  {m} } + {  {"50" "." 0}  over  {m} } =4 "." "00"D+"50" "." 0D {} #
="54" "." 0D {}
} } {}

26.8

Discussion

For an eye with this typical 2.00 cm lens-to-retina distance, the power of the eye ranges from 50.0 D (for distant totally relaxed vision) to 54.0 D (for close fully accommodated vision), which is an 8% increase. This increase in power for close vision is consistent with the preceding discussion and the ray tracing in Figure 26.4. An 8% ability to accommodate is considered normal but is typical for people who are about 40 years old. Younger people have greater accommodation ability, whereas older people gradually lose the ability to accommodate. When an optometrist identifies accommodation as a problem in elder people, it is most likely due to stiffening of the lens. The lens of the eye changes with age in ways that tend to preserve the ability to see distant objects clearly but do not allow the eye to accommodate for close vision, a condition called presbyopia (literally, elder eye). To correct this vision defect, we place a converging, positive power lens in front of the eye, such as found in reading glasses. Commonly available reading glasses are rated by their power in diopters, typically ranging from 1.0 to 3.5 D.