Optical Lenses and Devices

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case of optical lenses, both human-made, for use in optical devices, and naturally ... allel to the optic axis will be bent by refraction at both surfaces of the lens ...

Optical Lenses and Devices


In the last chapter we learned the fundamental laws that tell us how light behaves when traveling through a transparent material. Here we apply those to the most important case of optical lenses, both human-made, for use in optical devices, and naturally occurring, as in the eye. After learning the basics of imaging using a single simple lens, we show how the human eye exquisitely functions to allow us to see in color at extremely high resolution. Two human-made optical devices, the magnifying glass and the compound microscope, are then examined. These can be fairly well understood with only the tools of geometric optics that we have learned. Understanding the huge arsenal of new optical microscopies studied in Chapter 23 requires knowledge of wave optics, presented in the next chapter. There we introduce the wave nature of light and some of its major consequences.

1. OPTICAL LENSES In the previous chapter, after introducing reflection and refraction we focused on the phenomena of imaging from reflections in spherical mirrors and total internal reflection. Here we turn to the portion of the light incident on a transparent glass surface that is transmitted. Lenses are perhaps the most important of optical devices. For a glass lens, we know that only about 4% of the incident light at near normal incidence (in the paraxial approximation, with light traveling nearly along the optic axis) will be reflected at each boundary with air and so most of the light will be transmitted after being refracted by the curved surfaces. Special optical antireflection coatings can even increase the transmitted light closer to 100%. Armed with the law of refraction, we can repeat an analysis similar to that of the last chapter for reflected light to trace the refracted rays of light and to describe the characteristics of the images formed by a lens. Lenses are usually made either of glass or clear plastic and are ground and polished to have spherical surfaces. Several varieties of two basic forms of lenses exist: converging lenses, those thicker at the center than at the edges, and diverging lenses, those thinner at the center than at the edges (Figure 21.1). Occasionally lenses are made with cylindrical or even other surface contours for special purposes; these are not discussed here. Many common lenses, especially those in cameras, are not simple single pieces of glass, but are compound lenses made by cementing many individual lenses together with a transparent glue that has a similar index of refraction to that of the glass. We discuss these later, focusing first on simple lenses, those that have negligible thickness compared to their diameter. These are known as thin lenses and the equations we introduce are limited to such lenses. Let’s first consider a double convex lens shown in Figure 21.2. Incident rays parallel to the optic axis will be bent by refraction at both surfaces of the lens toward the optic axis (Figure 21.2a). For a thin lens there is a common point, the focal point F, J. Newman, Physics of the Life Sciences, DOI: 10.1007/978-0-387-77259-2_21, © Springer Science+Business Media, LLC 2008



double convex

at which all these rays cross the optic axis, the straight line passing through the lens center and traveling perpendicular to both surfaces. The distance f from this point to the center of the lens is called the focal length of the lens and is the same distance to either side of the lens. That is, if the lens is rotated 180o about a vertical axis it will focus light at the same point. The focal length of a thin lens can be shown to be related to its radii of curvature R1 and R2 on each of its sides and its index of refraction n by the lens-maker’s equation


1 1 1  (n  1)a  b . f R1 R2 double concave


FIGURE 21.1 Converging (upper) and diverging (lower) lenses.


Note that this equation defines a single focal length for a lens, regardless of the side facing the incident light, even if the two radii of curvature are different. In this equation the radii are taken as positive if the surface is convex and negative if concave (discussed below). Note that a plane surface has an infinite radius of curvature.

Example 21.1 A plano-convex lens of refractive index 1.52 has a radius of curvature of 5 cm. First find its focal length. Suppose that a second plano-convex lens with the same index of refraction is cemented to the first along their planar faces. What radius of curvature is needed on this second lens to produce a net focal length 1/4 the value of the focal length of the first lens? Solution: According to Equation (21.1) the focal length of the first lens is [0.52/5]1  9.6 cm. The second lens must have a radius of curvature R, such that (9.6/4)  [0.52(1/5  1/R)]1. Solving for R, we find that R  1.7 cm.

FIGURE 21.2 (a) Refraction at convex lens surfaces tends to bend light toward the optic axis. (b) The focal point of a double convex thin lens.

For a double convex (or any converging) lens, if an object O is to be imaged in the lens, we can trace three characteristic rays to locate the image. For the object shown in Figure 21.3, those three rays from the object arrowhead are: (1) a ray parallel to the optic axis that will be focused through the focal point on the far side of the lens (in red); (2) a ray passing through the focal point on the same side of the lens that, on passing through the lens, will be refracted to lie parallel to the optic axis (in blue); and (3) a ray passing through the center of the lens that will continue undeviated (this occurs because both sides of the lens are essentially parallel; the negligible thickness of the lens eliminates any parallel displacement of the ray; shown in green). The image of the arrow tip representing the object can be determined by the common point at which these three rays cross (of course, any two of these will cross at the image point). Then the entire image I is known since the ray along the optic axis passes straight through and images the tail of the arrow. Note that in this case the image is upside down, or inverted, and smaller in size. In place of raytracing we can derive an equation that will allow us to find the image location as well as its lateral magnification. With the various distances defined in Figure 21.4, the derivation assumes paraxial optics (with incoming rays making small angles with the optic axis). There are two sets of similar right triangles. The first set (shown in green hatched lines) consists of one formed with the object height and distance as legs (with hypotenuse OC) and the other with the image height and distance as legs (and hypotenuse IC). From the similarity of these we have F





s h  , h¿ s¿


and a second set of similar triangles (one shown in red hatched lines and the other having F as a common vertex) yields




f h  . h¿ s¿  f





We can also invert Equation (21.2) to find an expression for the lateral magnification m m



Combining the two equations, we have that s/s  f/(s f ), and after cross-multiplying and dividing both sides by the product (s s f ) and simplifying the result (try it!), we find the lens equation 1 1 1   . s s¿ f



s¿ h¿  , s h


FIGURE 21.3 Raytracing with a thin converging lens. The image lies at the intersection of the three numbered rays (see text for explanation of the construction).


where the minus sign was introduced so that an inverted image has a negative magnification, and an erect image has a positive magnification. In the case worked out above using ray diagrams, the object distance is greater than the focal length and we can see from Equation (21.4) that then because 1/s  1/f, the image distance s is positive, indicating that the image is real and that the image will be inverted (because m  0 in that case). If s  s the image will be magnified, whereas if s  s the image will be smaller than the object. It is easy to see that the dividing line occurs when s  2f, because then s  2f as well and the image will be “life-size,” but still inverted. If s  2f, then f  s  2f and the image will be smaller, whereas if f  s  2f, the image will be magnified. The reciprocal of the focal length for a lens is called its power P, where 1 P . f


Lens power is measured in reciprocal meters which are called diopters (D). Thus, the shorter the focal length of a lens is, the stronger its power. The diopter unit is mainly used in coding eyeglass lenses, a topic we return to in the next section. At this point, if we generalize Equations (21.4) and (21.5) using a set of sign rules, these equations are then valid for all thin lenses no matter what the configuration. These rules are given in Table 21.1. We illustrate their use with a few examples.

Table 21.1 Sign Conventions for Thin Lenses Quantity



 If object in front* of lens  If object behind lens  If image behind lens  If image in front of lens  If erect  If inverted  If surface is convex  If surface is concave  If converging  If diverging

s h, h R1, R2 f

FIGURE 21.4 Geometry to derive lens equation and magnification (Equations (21.4) and (21.5)). s-f O


h h

C I s



Front and back are with respect to incident light; that is, the front of the lens faces the incident light.








FIGURE 21.5 Raytracing when the object O is within one focal length of a converging lens. The image I is virtual, erect, and magnified.

First consider the situation in Figure 21.5 where the object is closer to the lens than one focal length. In this case, Equation (21.4) predicts that s will be negative because 1/s  1/f. What does this mean in terms of the image? The figure shows raytracing in this case. It is clear that the F focused rays do not converge and that therefore there is no real image, no place at which a screen can be put to see an image. On the other hand, a viewer on the far side of the lens from the object looking back through the lens will see the rays appear to emanate from a (virtual) image behind the lens, to be larger than the object, and to be erect. That the image is larger and erect follows from Equation (21.5) because s is greater than s and is negative, making m  1 (note the agreement with our sign conventions in Table 21.1). As the object approaches the focal point, the virtual image recedes to larger distances and is magnified to ever greater size. You may recognize this application of a lens as a magnifying glass (Figure 21.6). We discuss this situation further after we take a look at the eye in Section 3. As a second application of the lens equation consider the diverging lens shown in Figure 21.7. According to the lens-maker equation, because the radii of curvature are both taken as negative, so is the focal length. Therefore, no matter where an object is placed on the left of the lens in the figure, according to the lens equation, the image will always be virtual with the object appearing smaller and erect. This is so because with 1/s  0, when subtracted from 1/| f |, we have that 1 1 1   s¿ ƒfƒ s

FIGURE 21.6 A happy magnifying glass.

so that we must always have s  0 and |s|  s. Figure 21.7 shows the raytracing diagram for one such situation. As a final case, we examine the problem of where to put a converging lens in order to image an object on a screen when the total object to screen distance is fixed at a distance L. In that case (s  s)  L, and we must find the possible lens locations, or the possible individual s and s values. Writing the lens equation as s  s¿ 1 1 1  , or ss¿  fL,   s s¿ ss¿ f we need to solve for possible s and s values. Substituting for s  (L  s) into the above, we have s(L  s)  fL or s2  sL  fL  0. Solving the quadratic equation, there are two possible solutions given by s

L ;

1L2  4fL 2

or s 

4f L L ; 1 , 2 2A L

as long as f  L/4 or L  4f. To each of these values of s, let’s call them s and s, both of which are positive, there corresponds a value of s (s  L  s) and a magnification m  (s/s). Because the two values s and s add up to L, it is clear that the two solutions are s  s and s  s on the one hand and s  s and s  s on the other. In the first case, the lens is closer to the screen than to the object and the magnification is less than 1. The image will be inverted and reduced in size and, of






FIGURE 21.7 Raytracing for a diverging lens to form the image I of an object O. Ray 1 is parallel to the optic axis and diverges as if it originated at the focal point; ray 2 is aimed at the focal point on the other side of the lens and emerges parallel to the axis; ray 3 passes undeviated through the lens center. The virtual image is found at the extrapolated location where these rays cross.

1 2 3 I





s s'

course, real because it is actually formed on a screen. In the second case, the lens is closer to the object and the real image is enlarged but still inverted (Figure 21.8). With a handheld magnifying glass this is an easy and interesting experiment to try.

Example 21.2 A 0.2 cm tall object lies 10 cm from a 25 cm focal length magnifying glass. Find the location, magnification, and size of the image. Is it erect or inverted? Real or virtual? Solution: By direct substitution into the lens Equation (21.4), using s  10 cm and f  25 cm, we find that s  16.7 cm. Because s  0, we know that the image is on the same side of the lens as the object and therefore a virtual image. The magnification is given by m  s/s  1.7, so that the object appears to be (0.2)(1.7)  0.34 cm tall and erect.

As was mentioned above, many lenses are compound or thick lenses composed of multiple lenses cemented together, whereas other optical systems may consist of multiple individual thin lenses. Situations in which there are multiple thin lenses can be handled using the formalism of this section. One begins by finding the image of the object in the first lens (closest to object) and then simply treats this image as the object to be imaged by the second lens, and so on. Using the same consistent sign convention given in Table 21.1, such problems can be analyzed without any new concepts. For example, if the two (thin) lenses are in contact, then we can derive a simple formula for the overall focal length of the combination as follows. Writing the lens equation for the first lens we have that

FIGURE 21.8 The two solutions to imaging an object on a screen a fixed distance away from an object.

1 1 1   , s1 s1¿ f1



with a similar equation for the second lens, 1 1 1   . s2 s2¿ f2 But, using the first image as the object for the second lens means that s2  s1¿, so that on adding the two equations together we find that FIGURE 21.9 Spherical aberration. Parallel rays far off-axis will focus at different distances along the optic axis (horizontal double-headed arrow). The image in the paraxial focal plane, shown by the vertical line, will therefore be blurred laterally (vertical double-headed arrow).

1 1 1 1    . s1 s2¿ f1 f2 This equation can be interpreted as treating the two lenses in contact with each other as a single lens with a combined focal length f given by 1 1 1 +  . f1 f2 f


Most lenses in optical devices are, in fact, compound lenses designed to compensate for aberrations and so Equation (21.7) tells how to find the net focal length of the compound lens. We use this idea to analyze the compound microscope in the next section. But what is the purpose of cementing multiple lenses together? All lenses suffer from various defects in the quality of the image they produce. Collectively these are termed lens aberrations. We can distinguish two classes of aberrations: monochromatic, those involving a single color, and chromatic, due to the dispersion of the lens material, refracting different wavelengths (colors) differently due to a variation in index of refraction. There are five major monochromatic aberrations, all of which distort the imaging of a single point of the object to a single point of the image. One of these, spherical aberration, is simply due to the spherical lens curvature giving rise to distortion when incident rays are far from the optic axis. In particular, as shown in Figure 21.9, parallel rays from a distant point source arriving at different distances from the optic axis will be imaged at slightly different points on the axis, resulting in a blurred image of the point object. Limiting the accepted rays to paraxial rays close to the optic axis with a stop, or aperture, can reduce spherical aberration. The four other monochromatic aberrations have to do with off-axis imaging and examples of their images are shown in Figure 21.10. 1. Coma with the comet Hyakutake showing its shape.


C’ B’




FIGURE 21.10 Continued





2. Distortion with an example of barrel distortion from a wide-angle camera shot of a brick wall.



Image (pincushion distortion)

Image (barrel distortion)

3. Curvature of field with the painting “Anna’s Bedroom” by Scott Kahn illustrating this aberration.


object Curved image plane

Ideal image

4. Astigmatism with a painting illustrating this aberration.



Imageradial lines in focus

Imagetangential lines in focus

FIGURE 21.10 The four other monochromatic aberrations, other than spherical aberration, with schematics and example photos.

Chromatic aberrations due to the dispersion of glass result in different focal points for each color (Figure 21.11). The shorter wavelength light (violet end of the visible spectrum) experiences a larger index of refraction and is therefore refracted more and brought to a closer focal point. Compound lenses made from a converging and a diverging lens of different index of refraction glasses are designed to minimize chromatic aberration and are known as achromatic lenses (Figure 21.12). The longer optical path



FIGURE 21.11 Chromatic aberration, showing the effect of dispersion on the focusing of different colored light in a white light beam. Shorter wavelengths experience greater refraction due to the higher index of refraction.


of the longer wavelength light in the diverging lens seen in the figure compensates for the smaller index of refraction at these wavelengths and brings the various colors to a common focus. All good quality cameras are made with achromatic lenses, the better ones having very thick multiple lenses to minimize other distortions as well.


FIGURE 21.12 Using an achromatic lens (compound lens corrected for chromatic aberration) there is much less chromatic blur. ACHROMATIC LENS


Eyes are our visual window to the world. They are complex structures that are capable of very high resolution, are extremely sensitive to light, capable even of detecting single photons, can give color and depth perception, and adjust to focus on objects from as close as 10 cm, in young eyes, to “infinity.” In this section we first describe the structure of the eye with the aim of relating its anatomy to its functioning, as well as to some diseases. Then we focus on the retina and the visual pigment to describe in some detail the actual transduction of photon energy to an electrical response in the optic nerve. A schematic cross-section of the human eye is shown in Figure 21.13. Starting from the outside, the external covering of the eye consists of three layers. Most of the outermost layer is the sclera, the white of the eye, a tough fibrous layer containing nerve endings but no blood vessels. The sclera covers about 85% of the eyeball, roughly a 2.5 cm sphere, but the front portion consists of a transparent 12 mm diameter cornea with an index of refraction of about 1.38. The cornea is the most refracting surface in the eye, with the largest index transition from n  1 in air. The next two inner layers are the choroid, filled with pigments and blood vessels, and the retina, the site of photon detection. Neither of these layers extends into the cornea region (see Figure 21.13). At the front end of the eye behind the cornea is a liquid-filled chamber with the aqueous humor, which is continually drained and replaced, bounded also by the lens and the iris. A buildup of pressure in this region can produce a condition known as glaucoma, which can lead to blindness. The lens is a double convex lens made from a crystalline array of 25% protein and 10% lipids, having an index of refraction of about 1.42. It is one of the few parts of our bodies that are preserved without any turnover of their cells. With age, or disease, the lens loses its perfect crystallinity and develops defects that scatter light. These are known as cataracts and, when sufficiently large, can adversely affect vision by “clouding” the eye, or scattering light just as if you tried to view the world through a thin layer of milk. The shape of the lens is controlled by ciliary muscles that can change its focusing ability in a process known as accommodation. Normally, without any shape change of the lens, we can focus on objects from about 20 feet to infinity. This ability is due to the finite thickness of the photon detection region allowing light




to be focused at slightly different distances (see below). To see objects closer than that distance, the eye cannot remain relaxed, but the lens must change shape, thickening to give a tighter focus. The iris serves as an adjustable aperture and is pigmented, giving the eye its characteristic coloring. The central opening, known as the pupil, is the photon entrance path. Filling the eyeball is a gel-like material, the vitreous humor, which is more or less permanent. Six pairs of muscles control the movement of the eyeball in its socket, allowing us to focus images of interest on the highest acuity region of the retina, the fovea. This region of the retina, also known as the macula, has only cones, the photon receptor cells responsible for color vision, with each cone having a direct connection to a different optic neuron, or nerve cell. The macula therefore has the highest spatial resolution on the retina; outside the fovea there are roughly 10 cones per neuron connection or 125 rods, the other type of photon receptor, per neuron. These neurons collect in the optic disc, creating a blind spot with no visual pigment, and lead to the optic nerve bundle. Before we consider the photon detection process on the retina in more detail, consider the overall optical arrangement of the eye. As shown in Figure 21.14, parallel rays of light (from an object at “infinity”) are focused by the relaxed lens to a point on the retina some 2 cm behind the lens, in fact at the central fovea which lies on the visual axis. An object at a large but finite distance is focused onto the retina as an inverted image. Our brain interprets this inverted image as erect. The total equivalent lens of the eye is a thick lens system, composed of the cornea, aqueous humor, and lens, subject to all of the aberrations mentioned in the last section. One function of the iris is to reduce the aperture size to limit incoming rays to be paraxial, thus reducing aberrations. Often the eyeball is either elongated along the visual axis (myopia, or nearsightedness) or shortened in that direction (hyperopia, or farsightedness). A third defect in which the cornea is not spherical, but oval in shape, having different focusing properties along two different orthogonal directions, is known as astigmatism. All three of these defects can be corrected for by placing lenses in front of the eye (either as eyeglasses or contact lenses). Figure 21.15 shows ray diagrams for each of these, together with their corrections through the use of a lens. In myopia (top), parallel rays are brought to a focus in front of the retina (hence the name nearsightedness), blurring the image on the retina. By using a diverging lens, the image can be formed on the retina. The worse the myopia, the greater the power of the corrective lens needed. In hyperopia (middle), parallel rays would be focused behind the retina (far-sightedness) and so have not yet converged to form a clear image on the retina. A converging lens will move the focal point onto the retina. Again, the worse the eye’s vision is, the stronger power corrective lenses needed. Inexpensive “reading glasses,” simply matched converging plastic lenses, are sold in various diopter ratings for several dollars and are usually adequate for reading purposes. Astigmatism (bottom) produces a distortion in imaging so that two perpendicular lines cannot be both brought into focus. A cylindrical lens that focuses light along only one axis can correct this defect.

Example 21.3 Suppose the focal length of a person’s eye is 3.0 cm when fully relaxed (looking at a distant object). If the person’s retina is 3.3 cm behind the eye lens (a nearsighted eye compared to the normal distance of 3.0 cm), what must be the focal length of the corrective lenses so that this person can see “objects at infinity?”


FIGURE 21.13 The human eye in cross-section.

FIGURE 21.14 A distant object imaged on the retina.


Solution: Using the thin lens equation with s   and s¿  3.3 cm, we find an effective focal length of 3.3 cm needed. Because the effective focal length of such a two-lens system (the lens of the eye and the corrective lenses) is given, from Equation (21.7), by 1 feffective

1 flens

1 feye


we can find the focal length of the needed lens to be flens  a

1 feffective

1 feye




1 1 1  b   33.0 cm 3.3 3.0

(or in diopters, (1/.33 m)  3 D).

There are some interesting points to be made about nearsightedness. For example, if the near point of the normal eye is 25 cm (smin) and the distance to the retina is 3 cm (s), then the minimum focal length of the normal eye is 2.7 cm (from 1 1 1   b. s s¿ f Normal eyes produce an inverted real image on the retina that is m  s¿/s  3 cm/25 cm  0.12 times as large as the object. Let’s assume the nearsighted person in Example 21.3 has the same minimum focal length. Then, without corrective lenses, at closest focus 1 1 1 = + , smin 2.7 3.3

FIGURE 21.15 Three common focusing problems with the eye and their correction with eyeglasses.

which leads to smin  14.9 cm. Nearsighted people can see clearly closer to their eyes than normal-sighted persons; they have a smaller near point, which is the closest distance an object can be brought into focus, without corrective glasses. The size of the real image for this case is 3.3 cm/14.9 cm  0.22, almost twice as large as that of the normal-sighted person! Nearsighted people who are stamp or coin collectors have a great advantage over normal-sighted persons; they often don’t need magnifying glasses to see fine details. Unfortunately for the nearsighted, this close vision advantage vanishes when they put on corrective lenses. Suppose the person above is wearing the 3 D corrective lenses designed for distance vision and is looking at an object 25 cm away. The 3 D lens turns out to create a virtual image 14 cm in front of the lens that serves as the object for the eye’s lens, using 1 1 1 1 1 b   0.14 m. s¿  a  b  a  3  s f 0.25

Near-sighted eye


Near-sighted eye, corrected

Far-sighted eye

Far-sighted eye, corrected

Astigmatic eye

Astigmatic eye, corrected

But, that is approximately the closest the bare eye lens can focus, therefore with corrective lenses on, the person can no longer see as close as without them. (An object 14 cm away would form a virtual image only 10 cm in front of the glasses, too close to be in focus.) Note also that the virtual image formed by the corrective lenses is smaller than the object by a factor of 14 cm/25 cm  0.56. This reduction cancels the magnification advantage the nearsighted person enjoyed, too. In fact, because the corrective lens is diverging, it always forms a reduced size virtual image in front of




the real object. The eye uses this image as its object and so nearsighted people with corrective lenses always perceive objects to be both closer to them and smaller, compared with what a normal person sees. As the eye ages, the lens shape becomes more resistant to change by the ciliary muscles and so the eye cannot accommodate as well to bring close objects into focus. This weakening of accommodation is known as presbyopia. In the young eye, the near point can be as short as 7 cm. With age, the near point moves farther away due to presbyopia, having a mean of about 1 m for a 60 year old individual. Producing a similar effect as hyperopia, this can be corrected with a converging lens. People with myopia will often see an improvement in their vision with age due to presbyopia and will often require bifocal lenses with the lower portion made converging for reading or close vision and the upper portion made diverging for distance vision. The structure of the retina is shown in cross-section in Figure 21.16. Note the striking fact that incident light must travel through the network of nerve cells (retinal ganglion cells) before reaching the photoreceptors, lying partly immersed in a layer of pigmented cells. Fortunately these cells are transparent, but only about 50% of the light that strikes the cornea gets to the retina, and only about 20% of that gets to the light detecting cells. These cells, the rods and cones, permanent and not replaced over time, are, however, 100% efficient. Light that is not absorbed by the photoreceptors

FIGURE 21.16 The human retina. (top) Light passes through a network of nerve cells from the bottom of this drawing before being detected by the rods and cones. (bottom) Fluorescence microscopy image of the retina.



FIGURE 21.17 Hexagonally packed cone cells at the fovea section of the retina in a living human eye. The image is taken at a location about 300 m from the foveal center (which is equal to about 1 degree of visual angle) and it is about 150 m across. The small spots are single cone photoreceptors which, at this location are separated by about 5 m. The dark shadow is that of a blood vessel which runs above the photoreceptor layer.

FIGURE 21.18 Cone cells of the human retina.


is subsequently absorbed by a layer of pigmented cells to prevent stray reflections of light within the retina. There are about 125 million rods and 7 million cones on the retina distributed such that only cones are at the central fovea, where vision is most acute (Figure 21.17). The rods and cones are named by their shapes, but are somewhat similar in overall structure (Figure 21.18). There is an inner segment, filled with mitochondria to supply the energy needed for the light transduction, through which the light must also pass. The retina consumes the greatest amount of oxygen per unit weight of any tissue in the body. Inner and outer segments are connected through a thin cilialike portion containing microtubules. Light transduction takes place in the outer segments, the rod outer segments having been studied much more thoroughly than those of the cone. They are each 20 m long and 2 m in diameter and contain stacks of rod discs that are membranes containing the visual pigment rhodopsin, with about 105 rhodopsin molecules per m2 of surface area. Rhodopsin consists of two parts: a protein portion with 348 amino acids, known as opsin, and a smaller hydrocarbon part C20H28O, a derivative of vitamin A known as retinal, the light-absorbing portion (Figure 21.19). The structure of retinal is shown in Figure 21.20. There are two possible sterioisomers, or conformations, of retinal: 11-cis-retinal found in the dark and all-trans-retinal that nearly instantaneously forms after the absorption of a single photon. With the advent of femtosecond (1015 s) laser pulses, this first step in the vision process, the isomerization of retinal, has been found to occur within about 500 femtoseconds. Subsequent to this initial conformational change there is a sequence of conformational steps, discovered using pulsed laser spectroscopy, and other events that lead to an eventual electrical signal at the neuron. Each photon absorbed by a rhodopsin leads to the hydrolysis of over 100,000 molecules of cyclic GMP, the crucial signaling molecule in the subsequent transduction. The reduction in cyclic GMP, needed to keep Na channels open in the rod membrane, causes the eventual polarization of the membrane and electrical signal. The electrical signal that is sent from the retina to the brain over the optic nerve is not simply the sum of all rod and cone firings. Somehow the activity of the many rod and cone interconnections “preprocesses” information about the light falling on them so that a significant part of “seeing” occurs prior to what goes on in the visual cortex of the brain. It takes time to perform the preprocessing of visual information in the rod and cone networks, perhaps 0.1–0.2 s, a time that matches the response time of our nervous system. We do not see the instantaneous values of the electric fields of the EM light waves, which vary about 1015 times per second. Rather we see the effects of electric fields that have been averaged over many cycles of oscillation. In fact, the signal sent from the eye to the brain is not directly related to E fields, but rather to the average intensity, proportional to E2 averaged over many cycles. Now, rods and cones may detect EM intensity, but not equally well at all frequencies. The retinal molecule acts like a damped oscillator. When driven by the oscillating E field of the light, these molecules vibrate resonantly at different driving frequencies. There are three kinds of cone




FIGURE 21.19 A rod cell on left with a model of the seven-helix rhodopsin with the front two helices of opsin cut away to show the retinal molecule at the active site.

cells (with three different resonant frequencies close to “red,” “green,” and “blue”) and one kind of rod cell, with a resonant peak response between “green” and “blue.” The absorption spectrum of rhodopsin shown in Figure 21.21 indicates that darkadapted rods are most sensitive (have their strongest absorption) in the green-blue at 500 nm. In strong light this shifts slightly to 550 nm, but with a single absorption spectrum rods are not able to distinguish all of the different colors we can see in bright light. Cones are much less sensitive to light (the figure does not show this because the absorption peaks are all normalized); in fact the cones effectively “turn off” in dim light. In dim light there is little distinction between colors; everything appears gray. In bright light, the cones take over. They have three different types of visual pigments, each having a maximum absorption at a different visible wavelength, corresponding to a different color. It has long been known that (almost) any light color can be represented as a sum of three “primary” light colors: red (R), green (G), and blue (B). The RGB system is the basis for color TV, for example. On a color TV screen, each pixel is divided into an R, a G, and a B subpixel. Three electron beams sweep rapidly across the screen lighting each pixel with a certain amount of R, of G, and of B. Clearly it is tempting to explain the RGB color system in terms of the three different cone cells. Undoubtedly, there is some connection between the two, but the connection must be fairly subtle, and, as yet, is still not worked out. One reason for this situation is that the “R” cone cells actually have their response maximum at a frequency that is closer to yellow than to red. A second reason is

FIGURE 21.20 The chemical structure of retinal, the light-sensitive portion of rhodopsin.



FIGURE 21.21 The absorption spectrum of the “red,” “green,” and “blue” cones and of the rods (shown in black).




498 534 564

500 Wavelength (nm)


that some people only have two of the three cone cells, yet many of them seem to perceive color about as well as people with the normal distribution of cells. Somehow, their brains fill in the missing information. Finally, in people who are red–green color blind, all three cells appear to be present. So, the final word is not in yet.


FIGURE 21.22 (a) An object at the near point of the eye subtending an angle . (b) An object viewed through a magnifying glass when placed at its focal point, now subtending an angle . h

θ a

Image at ∞ h b


Optical devices abound in our technologically oriented world, from the supermarket optical scanner to the sophisticated digital video camera. Nearly all of these devices incorporate multiple lenses, except for the simple magnifying glass which we study just below. As we have seen, to analyze imaging problems with multiple lenses we straightforwardly treat the image from the first lens as the object for the second lens, and so on. Similarly the overall magnification is the product of the magnifications from each lens in the combination. We show an example of this in the optical microscope below. Recall that the near point is the closest distance that an object can be placed from the eye and remain in focus. If you want to see an object in more detail, you must bring it even closer to your eyes. In that way the image of the object on the retina will be enlarged. This allows more detail to be seen, because the image is then spread out over more detection sites increasing the spatial resolution on the retina, limited ultimately by the density of nerve cell connections. The unaided human eye thus has a limited ability to see detail because the near point limits our ability to bring objects as close as we might like to increase the size of a focused image on the retina. Figure 21.22a shows a small object at the near point and the image formed on the retina. If we take the near point to be 25 cm, a typical value, then the angle subtended by the object is equal to  h/25 cm as shown in the diagram. To increase the image size even further on the retina, but still have the image in focus, a magnifying glass (convex lens) is needed. The converging lens increases the focusing ability of the lens of our eye and allows us to bring the object closer to our eye while still keeping the image in focus. The angular magnification, or magnifying power, is defined in terms of the increased angle subtended by the object as compared to that at the near point of the unaided eye (see Figure 21.22)

θ f


u¿ . u





Using a magnifying glass with a relaxed eye focused at infinity and with the eye directly behind the magnifier, a virtual image at infinity is formed when the object is placed at the close focal point of the magnifier. Under these conditions   h/f as shown in the diagram and we can write the angular magnification as

fobj L


Image at ∞


h a b f a

h b 25 cm

25 cm . f


The smaller the focal length of the lens, the greater is the magnification seen by the eye. The exact position of the object relative to the focal point of the magnifier turns out to be unimportant, only changing the magnification by a small amount. The eye accommodates these small changes to make the image clear on the retina resulting in a small increase in magnification. The maximum magnification occurs when the image viewed through the magnifying glass occurs at the near point of the eye. To obtain higher magnification still, a compound microscope can be used. Figure 21.23 shows a schematic drawing of such a microscope with two lenses, an eyepiece that functions as a magnifying glass and an objective lens that further magnifies the object. The overall magnification is the product of that produced by each lens. The object is placed just outside the focal point of the objective, s ~ fobj, so that an inverted real image is formed with a lateral magnification of mobj 

FIGURE 21.23 Optics of a simple compound microscope. The object is placed just outside the focal point of the objective lens so that an enlarged, inverted real image is formed just inside the focal point of the eyepiece lens. Its image is a further enlarged virtual image viewed at infinity by a relaxed eye.

s¿ s¿ ,  s fobj

from Equation (21.5). This image then acts as the object for the eyepiece, adjusted to place the final virtual image at infinity, so that the eye can be relaxed as it views the image. In this case, we can write that s¿  1L  feye2, where L is the distance between the lenses, as shown in the diagram. The overall magnification compared to that at the near point with the unaided eye is then m  mobj meye  a

L  feye fobj


25 cm 25 cm # L b L , feye fobj feye


FIGURE 21.24 A basic compound microscope.

Dual eyepieces

because feye is generally much smaller than L, where all distances are given in cm. Usually the short focal length lenses of a microscope are compound lenses designed to eliminate aberrations. Magnifications of over 1000 are readily obtained. Figure 21.24 shows a photo of a basic compound microscope with its three main features: the built-in light source or condenser, providing a uniform brightness with the use of lenses; a stage, designed to securely hold the sample and usually to move it about in the horizontal plane; and the barrel of the microscope, holding the lenses and usually allowing different objectives with different focal lengths to be used. Microscopy has developed substantially in the recent past. The use of high-sensitivity video cameras and electronic image-processing techniques, as well as the development of several new types of microscopy discussed in Chapter 24, have broadened the versatility of the microscope in studying fundamental processes in biology.

O P T I C A L D E V I C E S : T H E M AG N I F Y I N G G L A S S



Objective lenses on rotating mount x–y moveable stage

Light source


CHAPTER SUMMARY Thin optical lenses have a focal length that is given by the lens-maker formula 1 1 1  (n  1)a  b , f R1 R2


where n is the index of refraction of the lens material and R1 and R2 are the radii of curvature of the two faces of the lens. An object located a distance s from the lens will produce an image through a thin lens of focal length f at a distance s given by the lens equation,

If two thin lenses are placed in contact, the overall focal length of the pair is given by 1 1 1 + = . f1 f2 f

The structure, optics, and detection properties of the eye are discussed in Section 2 of this chapter with a discussion of some common vision disorders and their correction with lenses. A simple magnifying glass of focal length f will produce a virtual image with a magnification given by mu 

1 1 1   . s s¿ f

h¿ s¿   . s h


where 25 cm is taken as the near point of the eye. Similarly, the overall magnification of a compound microscope, made from an eyepiece and an objective lens, is given by (21.5)

These three equations work under a wide variety of conditions if one uses the sign conventions of Table 21.1.

QUESTIONS 1. Distinguish carefully between a converging and diverging lens, a thin and a thick lens, and a real and a virtual image. Which combinations do not exist? For example, is there a thin diverging lens that forms a real image? 2. Carefully define all the symbols in the lens-maker equation. In a thin double convex lens made from n  1.5 glass with both sides having 25 cm radii of curvature, what is the predicted focal length? 3. Review the raytracing algorithm for finding the image of a (real) object through a thin lens. Distinguish the four cases of a converging lens with object closer or farther than the focal length, and a diverging lens with the object closer or farther than a focal length. Make a sketch of an example of each case. Classify each case according to whether the image is (erect or inverted), (real or virtual), (magnified more or less than 1 ), and (closer or farther than a focal length from the lens). 4. Consider the object–convex lens–real image configuration. If one places one’s eye at the image location, one will not see the image; however, if one moves the eye farther back, away from both the lens and the image location, the image will become visible. Why is this?


25 cm , f


The lateral magnification of the image is m


m  mobj meye L

25 cm # L , fobj feye


where L is the lens separation distance.

5. Compare the (real) image of a person formed by a converging lens to the (virtual) image of that person in a plane mirror. In which is up/down reversed, left/right reversed, and the magnification possibly changed? 6. In applications where a large light source is to be focused, such as a lighthouse, stage lighting in a theater, or an overhead projector, both a large diameter and a thick lens are needed. Fresnel, realizing that the refraction occurs at the glass surface, designed a lens (now called a Fresnel lens) that keeps the large curvature and lens size, but collapses the lens down to a nearly planar lens by removing the glass interior. This overcomes the problem of weight, bulk, and cost of such a glass lens. Based on the figure below explain how this lens works.




7. What is the physical origin of chromatic aberration in a thin lens? Of spherical aberration? Which rays are brought to focus closer to the lens: blue or red? paraxial (those parallel to and close to the optic axis) or offaxis rays? 8. Why does closing down the iris reduce aberrations of the lens of the eye? 9. One way to correct the eye for myopia is with laser surgery called photorefractive keratectomy (PRK) in which the cornea is reshaped so that light from a distance object focuses on the retina instead of in front of the retina. What do you think the laser procedure does to the cornea? This surgery will not correct for the age-related presbyopia that leads to the need for reading glasses. 10. Discuss the origin of presbyopia and the need for reading glasses as one ages. 11. Which photoreceptors, rods or cones, give us our most acute vision? Our color vision? Our night vision? 12. Explain in basic terms why the magnification of a microscope (or any two-lens system) is equal to the product of the magnifications of the objective and eyepiece. MULTIPLE CHOICE QUESTIONS 1. A light-emitting object is 10 cm from a thin lens. An upright virtual image is formed 20 cm behind the object. Which of the following is true? The lens has a focal length of (a) 6.7 cm, (b) 15 cm, (c) 15 cm, (d) 20 cm. 2. The distance from the eye’s lens to the retina for a given person is 3.0 cm. This person clearly sees an object 27 cm in front of his eye. The focal length of the eye’s lens in this case is (a) 2.7 cm, (b) 3.0 cm, (c) 3.4 cm, (d) 3.4 cm. 3. A 10 cm tall object 25 cm from a converging lens has its real image 50 cm from the lens. The object appears to be (a) 20 cm tall and erect, (b) 5 cm tall and inverted, (c) 5 cm tall and erect, (d) 20 cm tall and inverted. 4. A light source with an arrow pointing up is placed at the zero mark on an optical bench. A convex lens of unknown focal length is placed with its center at the 30 cm mark on the bench. A focused image appears on a collector when placed at the 60 cm mark on the bench and nowhere else. What must be true about the image? It is (a) real and inverted, (b) real and upright, (c) virtual and inverted, (d) virtual and upright. 5. The focal length of the lens in the previous question must be (a) 15 cm, (b) 15 cm, (c) 30 cm, (d) 60 cm. 6. Suppose a concave lens is inserted at the 15 cm mark on the bench in question 4. What would you have to do to the collector to find a focused image now? (a) Leave it at 60 cm. (b) Move it to some position











between the 30 cm mark and the 60 cm mark. (c) Move it to a position farther out than the 60 cm mark. (d) You can’t move the collector anywhere to get a focused image because no real image will be formed by this arrangement of lenses. In drawing a ray diagram for a converging lens with an object farther away than a focal length which of the following is not a correct ray to draw: (a) a ray parallel to the optic axis deflects at the lens to appear to come from the near focal point, (b) a ray parallel to the optic axis deflects at the lens to go through the focal point on the far side of the lens, (c) a ray going through the near focal point deflects at the lens and emerges from the lens parallel to the optic axis, (d) a ray straight through the lens center. A plano-convex lens of crown glass with a radius of curvature of 50 cm has a focal length of (a) 1.0 m, (b) 0.5 m, (c) 2.0 m, (d) 1.0 m. A double concave lens of crown glass with radii of curvature magnitudes of 25 cm and 50 cm has a focal length of (a) 0.33 m, (b) 3 m, (c) 0.33 m, (d) 33 m. Chromatic aberration in lenses is due to (a) dispersion, (b) interference, (c) total internal reflection, (d) varying degrees of absorption of different colors of light. Which of the following best describes nearsightedness? The lens in the eye produces an image of an object (a) 5 m away that would form behind the retina, (b) 5 m away that forms in front of the retina, (c) 10 cm away that would form behind the retina, (d) 10 cm away that forms in front of the retina. Without corrective lenses a woman can see an object clearly no closer than 0.5 m from her face. With corrective lenses she can see the object clearly as close as 0.1 m from her face. When the object is at 0.1 m her corrective lenses must form a (a) real image 0.5 m in front of her face, (b) real image 0.1 m in front of her face, (c) virtual image 0.5 m in front of her face, (d) virtual image 0.1 m in front of her face. A prescription for corrective lenses reads 5 D for each lens. These corrective lenses are (a) diverging with focal length equal to 5 m, (b) diverging with focal length equal to 0.2 m, (c) converging with focal length equal to 5 m, (d) converging with focal length equal to 0.2 m. It is essentially impossible for humans to react to a visual stimulus in less than 0.1 s. This is because that is about how long it takes (a) light to pass from the lens of the eye to the retina, (b) light to make one complete cycle of oscillation, (c) molecules in the cones to complete one cycle of vibration when they are excited by light, (d) firing from the retinal cells to be processed before being sent to the brain. In very dim light you can see an object better by looking at it out of the corner of your eye than straight on. That is because (a) cones are more concentrated on


the lens x, in order that the camera be able to take sharp photographs of objects positioned anywhere from 50 cm to infinity, measured from the front surface of the camera body.

the periphery of the retina than rods and cones function better in dim light, (b) cones are more concentrated on the periphery of the retina than rods and rods function better in dim light, (c) rods are more concentrated on the periphery of the retina than cones and cones function better in dim light, (d) rods are more concentrated on the periphery of the retina than cones and rods function better in dim light. 16. Color vision is due to (a) the varying sensitivity of different rhodopsins to different wavelengths of light, (b) three different kinds of rod cells, (c) the dispersion of the lens of the eye, (d) the pigmented epithelial cells. 17. A magnifying glass produces an image that is (a) upright and real, (b) inverted and real, (c) upright and virtual, (d) inverted and virtual. 18. Light entering the eye passes through the following layers in the order (a) lens, cornea, ganglion cells, retina, (b) cornea, lens, ganglion cells, retina, (c) lens, cornea, retina, ganglion cells, (d) cornea, lens, retina, ganglion cells. PROBLEMS 1. Sunlight can be focused on the ground by a lens when it is held 24 cm above the surface. What is the power of the lens? 2. A 5.0 diopter lens is used to magnify an insect when held 12 cm away. Describe the type of image, its position, and lateral magnification. Draw a ray diagram sketch. 3. A pinhole can function as a “lens”. Consider a box with a very small hole in one side. The hole admits light from any and all points outside to the inside but from any one point outside only the ray directed at the hole can enter the box. Show with a diagram the image of the object obtained by the light admitted by the pinhole. What is the magnification? Such boxes can be used as cameras, provided they are mounted on a rock-solid surface. Film exposures are typically many seconds or minutes and the image quality can be superb. 4. An old-fashioned box camera has a fixed lens and a depth of 15 cm. Suppose the camera lens is designed to be optimal for taking photographs of objects 3 m away. What is the focal length of the lens? camera



15 cm

5. A camera has a lens with adjustable position. The camera depth d  4 cm. Determine the focal length of the lens and the necessary allowable extension of


d Subject at minimum object distance

x 50 cm

6. How far from its infinity setting must a 35 mm lens be moved so that it produces a sharp image of an object 3 m away? 7. Suppose the image of a creature swimming in a dish of water is projected onto a screen. The distance from the dish to the lens is 36 cm and the screen is 4.5 m away from the mirror and lens. (The mirror simply redirects the light from the lens to the screen. Treat the distance between the lens and mirror as small enough so that it can be neglected in the specification of the image distance. Thus, object distance sv is 36 cm and image distance sh is 4.5 m.)



(a) What is the focal length of the projection lens? (b) If the creature swims at 1 cm per second in the dish, how fast does the image move on the screen? 8. A 35 mm film slide projector has a projection lens with a focal length of 135 mm. (a) Where should the slide be placed if the projection screen is 3 m away from the projection lens? (b) What is the magnification? (c) Now for the hard part (judging by the difficulty that even well-educated lecturers have with this one in practice). If we want to get a true (upright, nonreversed) image of the slide on the screen, how should the slide be placed into the projector? Should it be flipped upside down? Should it be reversed left and right? Should the slide be inserted backwards? (What does “backwards” mean here?) 9. A quick and easy way to get an approximate determination of the focal length of a convex lens is to measure the distance from the lens to an image of a light or other bright source some distance away. Suppose one has a lens that in fact has a focal length of 10 cm.




A fluorescent ceiling light with a grill cover is 1.5 m above the lens and a student is able to see an image of the lighting fixture grill on the back of his hand. He declares that the focal length of the lens is equal to the distance between lens and hand. Calculate the actual hand–lens separation distance for a sharp image and show that the error in the value of the focal length determined this way is less than 10%. Error  [(focal length  distance measured)/focal length] 100%.

10. Consider a convex lens of focal length 20 cm. Calculate the image distance for each of the following object distances: , 4 m, 2 m, 1 m, 80 cm, 60 cm, 40 cm, 20 cm. 11. It is often necessary to convey or relay the image of an object while keeping the image size unchanged


(unit magnification). Consider an object for which an image of the same size is desired, exactly 1 m away. (a) What is the focal length and the location of the single lens that will accomplish this? (b) The image of the object, using one lens, will be inverted. A relay system using two identical convex lenses will invert the inversion, yielding an upright image. Design such a system for the same initial situation as in the previous part. 12. If a certain microscope has an eyepiece with 12 magnification and it is desired to view a specimen with an overall magnification of 60 , what is the power of the objective that must be used? 13. The magnification of a compound microscope can be slightly improved if the final image is not at infinity but rather at the near point of the eye. Derive the formula for the magnification under this condition. This arrangement tends to produce eye strain because the iris must be under tension to have the lens of the eye constantly focusing at the near point. 14. Tom Cruise catches a reporter shooting pictures of his daughter at his home. He claims the reporter was trespassing. To prove his point, he gives as evidence the film the police took from the reporter. His daughter’s height of 0.62 m is 2.89 mm high on the film, and the focal length of the camera lens that the police seized was 210 mm. How far away from the baby was the reporter standing? Could the reporter be trespassing?