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# 2.6: The Eye

Learning Objectives

By the end of this section, you will be able to:

• Understand the basic physics of how images are formed by the human eye
• Recognize several conditions of impaired vision as well as the optics principles for treating these conditions

# Physics of the Eye

The eye is remarkable in how it forms images and in the richness of detail and color it can detect. However, our eyes often need some correction to reach what is called “normal” vision. Actually, normal vision should be called “ideal” vision because nearly one-half of the human population requires some sort of eyesight correction, so requiring glasses is by no means “abnormal.” Image formation by our eyes and common vision correction can be analyzed with the optics discussed earlier in this chapter.

Figure $$\PageIndex{1}$$ 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 a fixed distance from the lens. The flexible lens of the eye allows it to adjust the radius of curvature of the lens 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 (i.e., the pupil) of the eye, along with chemical adaptation, allows the eye to detect light intensities from the lowest observable to 1010times greater (without damage). This is an incredible range of detection. Processing of visual nerve impulses begins with interconnections in the retina and continues in the brain. The optic nerve conveys the signals received by the eye to the brain.

The indices of refraction in the eye are crucial to its ability to form images. Table $$\PageIndex{1}$$ lists the indices of refraction relevant to the eye. The biggest change in the index of refraction, which is where the light rays are most bent, occurs at the air-cornea interface rather than at the aqueous humor-lens interface. The ray diagram in Figure Figure $$\PageIndex{2}$$ shows image formation by the cornea and lens of the eye. The cornea, which is itself a converging lens with a focal length of approximately 2.3 cm, provides most of the focusing power of the eye. The lens, which is a converging lens with a focal length of about 6.4 cm, provides the finer focus needed to produce a clear 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 (i.e., a real, inverted image). Although images formed in the eye are inverted, the brain inverts them once more to make them seem upright.

Table $$\PageIndex{1}$$: Refractive Indices Relevant to the Eye*This is an average value. The actual index of refraction varies throughout the lens and is greatest in center of the lens.
Material Index of Refraction
Water 1.33
Air 1.0
Cornea 1.38
Aqueous humor 1.34
Lens 1.41*
Vitreous humor 1.34

As noted, the image must fall precisely on the retina to produce clear vision—that is, the image distance di must equal the lens-to-retina distance. Because the lens-to-retina distance does not change, the image distance di must be the same for objects at all distances. The ciliary muscles adjust the shape of the eye lens for focusing on nearby or far objects. By changing the shape of the eye lens, the eye changes the focal length of the lens. This mechanism of the eye is called accommodation.

The nearest point an object can be placed so that the eye can form a clear image on the retina is called the near point of the eye. Similarly, the far point is the farthest distance at which an object is clearly visible. A person with normal vision can see objects clearly at distances ranging from 25 cm to essentially infinity. The near point increases with age, becoming several meters for some older people. In this text, we consider the near point to be 25 cm.

We can use the thin-lens equations to quantitatively examine image formation by the eye. First, we define the optical power of a lens as

$P=\frac{1}{f}$

with the focal length f given in meters. The units of optical power are called “diopters” (D). That is, 1D=1/m,or 1m−1. Optometrists prescribe common eyeglasses and contact lenses in units of diopters. With this definition of optical power, we can rewrite the thin-lens equations as

$P=\frac{1}{d_o}+\frac{1}{d_i}.$

Working with optical power is convenient because, for two or more lenses close together, the effective optical power of the lens system is approximately the sum of the optical power of the individual lenses:

$P_{total}=P_{lens~1}+P_{lens~2}+P_{lens~3}+⋯ \label{sumlens}$

Example $$\PageIndex{1}$$: Effective Focal Length of the Eye

The cornea and eye lens have focal lengths of 2.3 and 6.4 cm, respectively. Find the net focal length and optical power of the eye.

Strategy

The optical powers of the closely spaced lenses add, so $$P_{eye}=P_{cornea}+P_{lens}$$.

Solution

Writing the equation for power in terms of the focal lengths gives

$\frac{1}{f_{eye}}=\frac{1}{f_{cornea}}+\frac{1}{f_{lens}}=\frac{1}{2.3cm}+\frac{1}{6.4cm}$.

Hence, the focal length of the eye (cornea and lens together) is

$f_{eye}=1.69cm.$

The optical power of the eye is

$P_{eye}=\frac{1}{f_{eye}}=\frac{1}{0.0169m}=59D.$

For clear vision, the image distance $$d_i$$ must equal the lens-to-retina distance. Normal vision is possible for objects at distances do=25 cm to infinity. The following example shows how to calculate the image distance for an object placed at the near point of the eye.

Example $$\PageIndex{2}$$: Image of an object placed at the near point

The net focal length of a particular human eye is 1.7 cm. An object is placed at the near point of the eye. How far behind the lens is a focused image formed?

Strategy

The near point is 25 cm from the eye, so the object distance is do=25 cm. We determine the image distance from the lens equation:

$\frac{1}{d_i}=\frac{1}{f}−\frac{1}{d_o}.$

Solution

$d_i=(\frac{1}{f}−\frac{1}{d_o})^{−1}$

$=(\frac{1}{1.7cm}−\frac{1}{25cm})^{−1}$

$=1.8cm$

Therefore, the image is formed 1.8 cm behind the lens.

Significance

From the magnification formula, we find $$m=−\frac{1.8cm}{25cm}=−0.073$$. Since m<0, the image is inverted in orientation with respect to the object. From the absolute value of m we see that the image is much smaller than the object; in fact, it is only 7% of the size of the object.

# Vision Correction

The need for some type of vision correction is very common. Typical vision defects are easy to understand with geometric optics, and some are simple to correct. Figure $$\PageIndex{3}$$ illustrates two common vision defects. Nearsightedness, or myopia, is the ability to see near objects, whereas distant objects are blurry. The eye over converges the nearly parallel rays from a distant object, and the rays cross in front of the retina. More divergent rays from a close object are converged on the retina for a clear image. The distance to the farthest object that can be seen clearly is called the far point of the eye (normally the far point is at infinity). Farsightedness, or hyperopia, is the ability to see far objects clearly, whereas near objects are blurry. A farsighted eye does not sufficiently converge the rays from a near object to make the rays meet on the retina.

Since the nearsighted eye over converges light rays, the correction for nearsightedness consists of placing a diverging eyeglass lens in front of the eye, as shown in Figure $$\PageIndex{4}$$. This reduces the optical power of an eye that is too powerful (recall that the focal length of a diverging lens is negative, so its optical power is negative). Another way to understand this correction is that a diverging lens will cause the incoming rays to diverge more to compensate for the excessive convergence caused by the lens system of the eye. The image produced by the diverging eyeglass lens serves as the (optical) object for the eye, and because the eye cannot focus on objects beyond its far point, the diverging lens must form an image of distant (physical) objects at a point that is closer than the far point.

Example $$\PageIndex{3}$$: Correcting Nearsightedness

What optical power of eyeglass lens is needed to correct the vision of a nearsighted person whose far point is 30.0 cm? Assume the corrective lens is fixed 1.50 cm away from the eye.

Strategy

You want this nearsighted person to be able to see distant objects clearly, which means that the eyeglass lens must produce an image 30.0 cm from the eye for an object at infinity. An image 30.0 cm from the eye will be 30.0 cm−1.50 cm=28.5 cm from the eyeglass lens. Therefore, we must have di=−28.5cm when do=$$\infty$$. The image distance is negative because it is on the same side of the eyeglass lens as the object.

Solution

Since di and dodo are known, we can find the optical power of the eyeglass lens by using Equation \ref{sumdiv}:

$P=\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{\infty}+\frac{1}{−0.285m}=−3.51D.$

Significance

The negative optical power indicates a diverging (or concave) lens, as expected. If you examine eyeglasses for nearsighted people, you will find the lenses are thinnest in the center. Additionally, if you examine a prescription for eyeglasses for nearsighted people, you will find that the prescribed optical power is negative and given in units of diopters.

Correcting farsightedness consists simply of using the opposite type of lens as for nearsightedness (i.e., a converging lens), as shown in Figure $$\PageIndex{5}$$.

Such a lens will produce an image of physical objects that are closer than the near point at a distance that is between the near point and the far point, so that the person can see the image clearly. To determine the optical power needed for correction, you must therefore know the person’s near point, as explained in Example $$\PageIndex{4}$$.

Example $$\PageIndex{4}$$: Correcting Farsightedness

What optical power of eyeglass lens is needed to allow a farsighted person, whose near point is 1.00 m, to see an object clearly that is 25.0 cm from the eye? Assume the corrective lens is fixed 1.5 cm from the eye.

Strategy

When an object is 25.0 cm from the person’s eyes, the eyeglass lens must produce an image 1.00 m away (the near point), so that the person can see it clearly. An image 1.00 m from the eye will be 100cm−1.5cm=98.5cm from the eyeglass lens because the eyeglass lens is 1.5 cm from the eye. Therefore, di=−98.5cm, where the minus sign indicates that the image is on the same side of the lens as the object. The object is 25.0cm−1.5cm=23.5cm from the eyeglass lens, so do=23.5cm.

Solution

Since di and dodo are known, we can find the optical power of the eyeglass lens by using Equation \ref{sumlens}:

$P=\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{0.285m}+\frac{1}{-0.985m}=+3.24D. \nonumber$

Significance

The positive optical power indicates a converging (convex) lens, as expected. If you examine eyeglasses of farsighted people, you will find the lenses to be thickest in the center. In addition, prescription eyeglasses for farsighted people have a prescribed optical power that is positive.

## Contributors

• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).