chapter 20 human vision - doane...

Physics Including Human Applications

Chapter 20- Human Vision

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Chapter 20 HUMAN VISION

GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Human Vision Characterize the physical parameters that are significant in human vision. Visual Defects Explain the causes and corrections for the visual defects of myopia, hypermetropia, presbyopia and astigmatism. Characteristics of Vision Define the following terms:

visual acuity accommodation scotopic vision dichromat photopic vision

Corrective Lenses Solve problems involving visual defects and their corrections using lenses.

PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 18, Optical Elements, and Chapter 19, Wave Properties of Light.



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Chapter 20 HUMAN VISION

20.1 Introduction From your own experience, what descriptive statements can you make concerning the following aspects of your vision: sensitivity for low light levels, response to different colors, ability to discriminate between two objects at a distance, and the reliability of visual depth perception? The human eye is our main source of information about our external environment. It responds to light and supplies messages to the brain. In this chapter we will explore the human visual system and the physical principles behind its operation. 20.2 The Powerful Eye In this chapter we will examine one of the most amazing optical systems you will ever encounter. The human eye is an optical system of unsurpassed versatility. Its sensitivity is so great that it apparently detects single packets of light, yet visual information is processed only when sufficient numbers of these packets are detected coincidently to produce a visual pattern. This same system is capable of handling intensities more than 109 times the minimum energy threshold. Let us look at the physics behind the performance of the human eye. The eye is analogous to a camera in its basic operation Figure 20.1. There are three basic parts of the eye: a light focusing system, an automatic aperture system, and a light sensitive detection system. In the light focusing system, the cornea and the lens account for the refractive power of the eye. Most of this power is due to the curvature of the cornea, but the muscles controlling the curvature of the lens provide the flexible focusing power of the eye. These eye, or ciliary, muscles permit the normal eye to focus on the retina light from objects very distant, essentially parallel rays of light, as well as light from objects as close as the normal near point of 25 cm from the eye. Measure your near point.

Figure 20.1

Diagram of the human eye.



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Move this page toward your eyes until the images of the letters begin to blur. The distance from your eye to the page at the nearest point of clear vision is your near-point vision. You have greater eye accommodation than normal if your near-point distance is less than 25 cm. The power of a lens is a measure of its ability to bend rays of light. A lens of high power has a small focal length. The numerical value for the power of a lens in diopters is computed by taking the reciprocal of the focal length of a lens when the focal length is measured in meters. For example, a lens of 5-cm focal length has a power of 20 diopters. The human eye has a variable focal length so the eye can change its power to focus on objects at various distances from the eye. Let us calculate the accommodation ability of a typical human eye. If the distance from the eye lens to the retina is given by l (about 3 cm for a human eye), the power of the eye lens for distant objects, symbolized by P infinity P∞, can be calculated by using the thin lens formula from Chapter 18, 1/p + 1/q = 1/ f (18.13) where p is the distance to the object, q is the distance to the image, and f is the focal length of the lens. For distant objects the reciprocal of the object distance 1/p is approximately zero, so the distance from the eye lens to the image on the retina is equal to the focal length of the eye. Therefore, the power for the eye lens for distant objects P∞, P∞ = 1/f = 1/l ≅ 30 diopters (20.1) where l is the distance from the eye lens to the retina. The power of the eye for near objects, Pn, can be similarly calculated, Pn = 1/f = 1/p + 1/q (20.2) where p is the distance to an object at your near point (say, 25 cm) and q is the distance from the eye lens to the retinal, Pn = 1/near point + 1/l = 1/0.25 + 1/l = 4 + P∞ ≅ 34 diopters (20.3) Hence, the change in the power of a human eye lens to accommodate objects at various distances from the eye is about 4 diopters. Design a simple experiment using a meter stick to measure the ability of your eyes to accommodate objects at various distances from your eye. If you wear eyeglasses or contact lenses, perform this experiment both with and without your glasses or lenses. The automatic aperture system involves a negative feedback system through the brain to the iris Figure 20.1. The system automatically adjusts the opening size of the iris in response to the ambient level of light intensity. In response to high intensity light, the iris stops down the opening to minimize damage to the detector system and to provide the proper intensity levels to the detecting system. The response to low light levels, referred to as dark adaption, takes up to 45 minutes. It provides the eye with its maximum entrance pupil size and allows the eye to assume its most sensitive light gathering configuration. The light sensitive detecting system of the eye consists of the light sensitive receptors distributed over the back of the eyeball. This light sensitive membrane is called the retina. There are two kinds of photosensitive cells present in the retina. These cells are rods and cones, so named because of their shape. The rods are more sensitive to light and distinguish light from dark when the light level is very low. Vision at low levels of light intensity is called scotopic vision. The cones are less sensitive to light, but



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are capable of resolving light into components giving details of images and information involving the color of the received light. The rods and cones are distributed with densities greater than 100,000/mm2 on the retina, the greatest concentration of cones occurring in the region called the fovea. 20.3 The Defective Eye The most common defects of human vision involve the light refracting system of the eye. Normal vision provides clear vision for letters that subtendthree minutes of arc at a distance of 6.10 m (20 ft) from the observer. At this distance, defective eyes may produce images that are too blurred to read. If a person requires letters at 6.10 m (20 ft) that normal vision reads at 12.2m (40 ft) the person is said to have 20/40 vision. The numerator expresses the distance between observer and the letters; the denominator is the distance at which the normal eye can read these letters. Normal vision is expressed as 20/20 vision. EXAMPLE Find the size of the letters the normal eye can read at 6.10 m (20 ft), if the criterion is that they subtend 3 minutes of arc at the eye. Using this information what size object is required at 10 m? angle (rad) = object size/distance 1 min = 2.91 x 10-4 rad 3 min = 8.73 x 10-4 = x/6.10 m x = 0.533 cm At 10 m, x = 10 m x 8.73 x 10-4 = 8.73 x 10-3 = 0.87 cm. 20.4 Correction of Defective Sight Many of the defects of the human eye are correctable. Myopia is the term for nearsightedness. Myopia results when parallel light is focused in front of the retina. This means that there exists a farthest point of clear vision, usually only a few meters from the eye. Only objects closer to the eye than the farthest distance of clear vision can be properly focused on the retina. This defect is most frequently associated with an elongated eyeball. The correction for myopia is the use of a diverging lens with a focal length equal to the far point distance, the distance of farthest clear vision. This negative spectacle lens makes incoming parallel light appear to diverge from the far point (Figure 20.2).



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Farsightedness (hypermetropia) due to a short eyeball, results in parallel light being focused behind the retina Figure 20.3. The same result caused by the hardening of the lens with aging is called presbyopia. These two conditions mean that the near-point distance for a person is increased beyond 25 cm. The correction for each of these defects is the use of a converging lens, which produces for an object at the normal near point (25 cm) a virtual image at the actual near point of the eye. This means that the focal length of the correcting lens is given by the equation 1/f = 1/25 - 1/x, where x is the actual nearpoint of the eye (all distances in cm). The defect of astigmatism is the formation of a distorted image on the retina. The usual cause of this defect is the nonuniform curvature of the cornea. This defect can be corrected by using a lens that is ground to compensate for the cornea defect. Often this correction is obtained through the use of an appropriately shaped cylindrical lens.

EXAMPLES 1. A man has a far point of 2 m. Find the focal length of the corrective lens for seeing distant objects, and find the near point of this eye with corrective lens if the man has the normal near point without glasses. The lens needed must have a focal length of -2 m so that an object at 2 meters will have an image at infinity. The near point with this lens is x, where 1/x - 1/25 = 1/f (in cm). Thus x equals 200/7 cm. 2. A person has a near point of 50 cm. Find the corrective lens required to enable this person to read at the normal distance of 25 cm. Using the equation 1/f = 1/25 - 1/50, we find that the corrective lens should have a focal length of +50 cm. 20.5 The Receptive Eye As stated earlier research has provided evidence that individual receptor cells on the retina can be activated by single packets (photons) of light. Fortunately, such



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phenomena are not processed by the brain; the noise signals by such random activity would be most distracting. The research shows that of the order of 100 photons per second will excite a localized set of rods of about 100 in number causing visual perception to occur. The basis of this coincident detection system is still not clear, and it continues to be a current area of research. The wavelength, or color, sensitivity of the human eye is also a subject of present-day research. It has been established that the cone cells in the retina play the central role in color, or photopic, vision. The cones are most closely packed in the fovea region of the retina; so when you direct your visual attention to an object your eye muscles rotate your eyeballs so as to place the focused image of that object on the fovea of the retina. It has been found that there are four colors whose apparent color does not change as the intensity of the light focused upon the retina is increased above the threshold required to stimulate the cone cells. These "pure" colors are red, yellow, green, and blue, and for a typical observer these pure colors will have dominant wavelengths of 630 nm, 570 nm, 510 nm, and 470 nm, respectively. On the other hand, it is possible for a person to perceive yellow from a proper mixture of blue and red light, without any 570-nm light present. So a human being is capable of synthesizing colors in a way that is not clearly understood at the present time. In an attempt to refine our understanding of human vision, direct measurements of the response of the retinal cones to various exciting lights have been made. Three types of cone cells have been identified, each particularly sensitive to a different range of the spectrum. However, it is clear that these cones do not send nerve signals directly to the brain. The cone signals pass through a complex coding system before transmission to the brain where the visual messages are translated into a person's visual perception.

It has long been known that humans see objects differently in dim light and in bright light. We now know that the rod cells in the retina are most sensitive to low intensity of light, and these rod cells are most closely packed in the periphery of the retina, rather than in the fovea. To see objects in dim lighting you should look slightly to one side of the object, not directly at it, so that the light entering the eye falls not on the fovea but on the periphery of the retina where the rods are more concentrated. The rods and the cones also differ in their wavelength response to light. The rods are more sensitive to blue light than are the cones. The relative luminosity curves for scotopic (dim light) and photopic vision are shown in Figure 20.4. Relative luminosity is defined as the reciprocal of the normalized threshold intensities, obtained by dividing each wavelength threshold by the minimum threshold for the entire system. (Can you suggest an analogous analysis for the human ear?)



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The range of wavelengths that make up the visible spectrum (380 - 770 nm) is determined by the absorption of ultraviolet by the lens and by the insensitivity of the receptors in the eye to wavelengths greater than 770 nm. There are people who cannot discriminate colors. The degree of color- blindness varies among the population. There is evidence that some color blind persons have only two of the necessary pigments (thus they are called dichromates) for full color vision. Those with no color vision are called monochromates. 20.6 The Perceptive Eye The visual acuity of the eye is defined as the minimum angular separation between two objects that can be resolved by the eye. Several factors can set this limit, the spacing of receptor cells with independent nerve paths, diffraction effects (Rayleigh criterion), and lens system defects. The closest spacing of receptors in the human eye is that of cones in the fovea, where the center to center distance of adjacent cones is from 2 to 5 microns. For an eye with 2 cm between cornea and retina this corresponds to about 10-4 rad for the angle subtended by two stimulated cells separated by one unstimulated cell. This criterion determines the resolving power of the human eye. EXAMPLE Find the distance from the eye for two objects 1 cm apart that subtend an angle of 10- 4 rad s/r = θ 1 cm/r = 10-4 r = 100 m Using the Rayleigh criterion and appropriate values for wavelength and for the aperture of the eye, find the limit for visual acuity, and compare it with the 10-4 rad value calculated from receptor separation. Does this comparison support the idea that the visual acuity of the eye is essentially independent of color?

20.7 Binocular Vision Depth perception in human vision is primarily a result of binocular vision. Each eye produces a slightly different image on their respective retina due to the different angle the object subtends at each eyeball. The brain learns to interpret these slightly different images in terms of distance from the eye. The experimental data suggest that depth perception can be improved with practice. Experience is used to correlate the actual size of the object with the size of the image on the retina and the actual distance to the object. The following cues are important in depth perception: 1. viewing angle differences for each eyeball (Figure 20.5) 2. muscle tension for convergence upon a point of fixation 3. angle subtended by known object at a known distance 4. color contrast interpreted in terms of light scattering which relates to distance



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5. relation of light and shadows 6. overlapping contours 7. parallax (the apparent movement of objects in the visual field with observer motion) 8. context of object and environment

ENRICHMENT 20.8 Derivation of the Power of Lenses Used Close to the Eye To derive the power of a lens to be used close to the eye, let the eye lens be replaced by a thin convex lens of fixed focal length f e in meters. Then the power of the eye will be given by Pe where Pe = 1/fe (20.4) Next take another lens of focal length f1 in meters with a power of P1 = 1/f1 and place it close to the first lens. If an object is placed several meters away from the lens we can find the image distance q1 by applying the thin-lens equation (Equation 18.13), 1/q1 = 1/f1 - 1/p =(p - f1)/pf1 Then q1 = pf1 /(p - f1 ) (20.5) This image distance q1 becomes the virtual (negative) object distance for the eye lens so that the final location of the image in the eye is given by qe: 1/qe = 1/fe -pe = 1/fe +q1 where pe(=-q1) is the object distance to the eye lens. But the power of the two lens system, Ps = 1/fs = 1/p + 1/qe = 1/f1 -1/q1 + 1/fe + 1/q1 Ps = 1/f1 + 1/fe =P1 + Pe (20.6) Hence the power of a lens-eye system is the sum of the glass lens power plus the eye lens power.



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EXAMPLE A man has a near point of 50 cm from his eyes and a far point at infinity. What is the useful accommodation power of his eyes? Consider the lens of the eye as a simple lens. 1/fe = 1/p + 1/q For the far point 1/f∞ = 1/∞ + 1/q∞ = 1/q∞ Now 1/f = P∞ = 1/q∞ Pn = 1/0.50 + P∞ = 2.0 + 1/q∞ = 1/fn accommodation power = P - P∞ = 2.0 diopters What lens should be prescribed for this individual? For the normal eye the near point should be 25 cm. This means that the image formed by the lens must be at least 50 cm from the eye lens. 1/p + 1/q = 1/f 1/25 - 1/50 = 1/f 1/f - (2 - 1)/50 = 1 / 50 cm P = 1/0.50 m = 2.0 diopters A general rule for an auxiliary lens for the eye is that the image formed by this lens must be between the near and far point of the unaided eye. One then uses the principles of a multiple lens system to prescribe the corrective lens.

SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of this summary with the number of the section where you can find related content material. Human Vision 1. For each of the following optical properties write a statement to relate it to human

vision: light intensity, light wavelength, lens power, and resolving power. Visual Defect 2. A person who suffers from hypermetropia is unable to clearly see objects that are

______ the eye lens, has the ______ point distance ______ in size, and can wear corrective eyeglasses of _______ power.

3. A person who suffers from myopia is unable to clearly see objects that are ______ the eye lens, has the ______ point distance in size, and can wear corrective eyeglasses of ______ power.



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Characteristics of Vision Match each term below with the lettered items. 4. visual acuity 5. scotopic vision 6. photopic vision 7. visual accommodation 8. dichromat 9. binocular vision

a. looking at photographs b. two-colored rug c. contact-lens holder d. perspicacity e. depth perception f. color perception g. low intensity h. carrots i. color blindness j. blue sensitive k. red sensitive l. fovea region

m. 0.01 rad n. reciprocal of normalized intensities o. lens power p. iris aperture size q. diopters r. cones s. rods t. 20/20 u. red, yellow, blue, green v. Rayleigh criterion w. eye muscle tension x. perspiration

Corrective Lenses 10. A young person with a near point of 20 cm, is not able to see an object clearly if it is more than 5 m from his eye. What kind of lens should you prescribe to correct his vision?

a. +5 diopters b. -5 diopters c. +0.2 diopters

d. -0.2 diopters e. -2.0 diopters

Answers 1. Section 20.5, Section 20.2, Section 20.6 2. near, near, increased, positive (Section 20.4) 3. far from far, decreased, negative (Section 20.4) 4. d, l, m, p, t, v (Section 20.6) 5. g, j, s (Section 20.5)

6. f, i, k, l, r, u (Section 20.5) 7. o, q (Section 20.2) 8. f, i (Section 20.5) 9. e, w (Section 20,7) 10. d (Section 20.2)



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ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve single concept problems. Equations 1/p + 1/q = 1/ f (18.13) P = 1/f (m) (20.1) P∞ = 1/l (m) (20.1) Pn = 1/near point + 1/l (20.3) Problems 1. At what distance should a person with 20/20 vision be able to read a letter 1 cm high? 2. What is the power of the lens for each example in Section 20.4, Correction of

Defective Sight? Answers 1. 11.5 m 2. -0.5 diopters, + 2.0 diopters EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. When appropriate, the numerical answer is given in brackets at the end of the exercise. Section 20.2 1. A girl can adjust the power of her eye's lens-cornea system between limits of +60.0

diopters and +65.0 diopters. With the lens relaxed she can see distant objects such as stars clearly.

a. Find the girl's near point. b. How far is her retina from her eye lens? [20.0 cm, 1.67 cm] Section 20.4 2. A myopic student wears contact lenses of -8.00 diopter power in order to see distant

objects. The distance from the eye lens to the retina is 2.00 cm. a. Find the power of the student's eye when relaxed. b. If the student can give an extra +4.00 diopters to his eye lens with ciliary muscles,

find the near point of his eyes without glasses. [58.0 diopters, 8.33 cm]



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3. A nearsighted person wears glasses of f = -50.0 cm. Through these glasses he has perfect distant vision, but he cannot focus clearly on the objects closer than 25.0 cm. How close can he focus if he removes his glasses? What is his far point? [16.7 cm, 50.0 cm]

4. A pair of bifocal lenses have components with focal lengths of 50 cm and -250 cm. What are the near and far points for the wearer of these bifocal lenses? [50 cm and 250 cm]

PROBLEMS Each of the following problems may involve more than one physical concept. When appropriate, the answer is given in brackets at the end of the problem. 5. Typical vision data for a female subject is as follows:

Near Point Far Point

without glasses 13 cm 39 cm

with glasses 20 cm ∞

What is the subject's power of accommodation without glasses? With glasses, what is the power of accommodation of the optical system? Explain these results. What is the power of the glasses lens? Does this person suffer from myopia, hyperopia, or presbyopia? [5.1 diopters, 5 diopters -2.6 diopters, myopia]

6. An automobile is approaching you on a straight road on a pitch black night. The lights on the car are 1.5 m apart. Assume the aperture of your eye is 1 cm and the effective wavelength of the light is 550 nm.

a. Using the Rayleigh criterion, how far will the automobile be from you when you can just be sure that you are seeing an automobile and not a motorcycle?

b. Using the spacing of receptor cells in the eye, how far will the automobile be from you when you can just be sure that you are seeing an automobile and not a motorcycle?

c. According to NASA sources the minimum separation acuity for the average astronaut is 0.4 minutes of arc. How far will the automobile be from an astronaut before he can be sure that it is not a motorcycle? [2.2 x 104 m, 1.5 x 104 m, 1.3 x 104 m]

7. A photon of visible light has energy of about 3.6 x 10-19 J. The absolute luminance threshold for the dark-adapted human eye is 10-5 mL (milli-Lambert= 1.47 x 10-6 watts/cm2). How many photons per second per cm2 are required for minimum human vision? [4.1 x 107 photons/cm2-sec]

8. In the fovea portion of the eye there are 136 thousand cones per square millimeter, and the lower absolute threshold of illumination is 10-3 mL. Using the average energy of a visible photon of light as 3.6 x 10-19 J, how many photons per second are required to provoke a visual response from a cone? [300 photons/cone - sec]



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9. On a horizontal angle of 20ø from the fovea there are 158 thousand rods per square millimeter, and the lower absolute threshold of illumination is 10-5 mL. Using an average energy of visible light photons as 3.6 x 10-19 J, how many photons per second are required to provoke a visual response from a rod? [2.58 photons/rod - sec]

10. Given the cone and rod average distribution per unit area in problems 8 and 9, what is the minimum acuity angle for vision using only cones? Using only rods? The distance from the cornea to the retina is 25 mm. [2.16 x 10-4 rad, 2 x 10-4 rad]

11. Design an experiment to measure relative luminosity for the eye as shown in Figure 20.4.

12. An older person has a near point of 75 cm. Find the correct lenses to prescribe for this person. [bifocals, clear glass and +2.67 diopters]

chapter 20 human vision - doane...

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