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Chapter 36

Image Formation

CHAPTE R OUTL I N E

36.1 Images Formed by FlatMirrors

36.2 Images Formed by SphericalMirrors

36.3 Images Formed byRefraction

36.4 Thin Lenses

36.5 Lens Aberrations

36.6 The Camera

36.7 The Eye

36.8 The Simple Magnifier

36.9 The Compound Microscope

36.10 The Telescope

1126

! The light rays coming from the leaves in the background of this scene did not form afocused image on the film of the camera that took this photograph. Consequently, the back-ground appears very blurry. Light rays passing though the raindrop, however, have beenaltered so as to form a focused image of the background leaves on the film. In this chapter,we investigate the formation of images as light rays reflect from mirrors and refract throughlenses. (Don Hammond/CORBIS)

1127

This chapter is concerned with the images that result when light rays encounter flatand curved surfaces. We find that images can be formed either by reflection or byrefraction and that we can design mirrors and lenses to form images with desired char-acteristics. We continue to use the ray approximation and to assume that light travelsin straight lines. Both of these steps lead to valid predictions in the field called geometricoptics. In subsequent chapters, we shall concern ourselves with interference and diffrac-tion effects—the objects of study in the field of wave optics.

36.1 Images Formed by Flat Mirrors

We begin by considering the simplest possible mirror, the flat mirror. Consider apoint source of light placed at O in Figure 36.1, a distance p in front of a flat mirror.The distance p is called the object distance. Light rays leave the source and arereflected from the mirror. Upon reflection, the rays continue to diverge (spreadapart). The dashed lines in Figure 36.1 are extensions of the diverging rays back to apoint of intersection at I. The diverging rays appear to the viewer to come from thepoint I behind the mirror. Point I is called the image of the object at O. Regardlessof the system under study, we always locate images by extending diverging rays backto a point at which they intersect. Images are located either at a point fromwhich rays of light actually diverge or at a point from which they appear todiverge. Because the rays in Figure 36.1 appear to originate at I, which is a distanceq behind the mirror, this is the location of the image. The distance q is called theimage distance.

Images are classified as real or virtual. A real image is formed when light rayspass through and diverge from the image point; a virtual image is formed whenthe light rays do not pass through the image point but only appear to divergefrom that point. The image formed by the mirror in Figure 36.1 is virtual. The imageof an object seen in a flat mirror is always virtual. Real images can be displayed on a

Mirror

O I

qp

Figure 36.1 An image formed by reflection from a flatmirror. The image point I is located behind the mirror aperpendicular distance q from the mirror (the imagedistance). The image distance has the same magnitude asthe object distance p.

screen (as at a movie), but virtual images cannot be displayed on a screen. We shall seean example of a real image in Section 36.2.

We can use the simple geometry in Figure 36.2 to examine the properties of theimages of extended objects formed by flat mirrors. Even though there are an infinitenumber of choices of direction in which light rays could leave each point on the object,we need to choose only two rays to determine where an image is formed. One of thoserays starts at P, follows a horizontal path to the mirror, and reflects back on itself. Thesecond ray follows the oblique path PR and reflects as shown, according to the law ofreflection. An observer in front of the mirror would trace the two reflected rays back tothe point at which they appear to have originated, which is point P! behind the mirror. Acontinuation of this process for points other than P on the object would result in a virtualimage (represented by a yellow arrow) behind the mirror. Because triangles PQR andP !QR are congruent, PQ " P !Q. We conclude that the image formed by an objectplaced in front of a flat mirror is as far behind the mirror as the object is in frontof the mirror.

Geometry also reveals that the object height h equals the image height h!. Let usdefine lateral magnification M of an image as follows:

(36.1)

This is a general definition of the lateral magnification for an image from any type ofmirror. (This equation is also valid for images formed by lenses, which we study inSection 36.4.) For a flat mirror, M " 1 for any image because h! " h.

Finally, note that a flat mirror produces an image that has an apparent left–rightreversal. You can see this reversal by standing in front of a mirror and raising your righthand, as shown in Figure 36.3. The image you see raises its left hand. Likewise, yourhair appears to be parted on the side opposite your real part, and a mole on your rightcheek appears to be on your left cheek.

This reversal is not actually a left–right reversal. Imagine, for example, lying on yourleft side on the floor, with your body parallel to the mirror surface. Now your head is onthe left and your feet are on the right. If you shake your feet, the image does not shake itshead! If you raise your right hand, however, the image again raises its left hand. Thus, themirror again appears to produce a left–right reversal but in the up–down direction!

The reversal is actually a front–back reversal, caused by the light rays going forwardtoward the mirror and then reflecting back from it. An interesting exercise is to standin front of a mirror while holding an overhead transparency in front of you so that youcan read the writing on the transparency. You will also be able to read the writing onthe image of the transparency. You may have had a similar experience if you haveattached a transparent decal with words on it to the rear window of your car. If the

M ! Image heightObject height

"h!

h

1128 C H A P T E R 3 6 • Image Formation

! PITFALL PREVENTION 36.1 Magnification Does

Not NecessarilyImply Enlargement

For optical elements other thanflat mirrors, the magnificationdefined in Equation 36.1 canresult in a number with magni-tude larger or smaller than 1.Thus, despite the cultural usageof the word magnification to meanenlargement, the image could besmaller than the object.

Lateral magnification

Object θθh R

QP P ′

Image

p q

h′

Active Figure 36.2 A geometricconstruction that is used to locatethe image of an object placed infront of a flat mirror. Because thetriangles PQR and P!QR are con-gruent, and h " h!.

At the Active Figures linkat http://www.pse6.com, youcan move the object and seethe effect on the image.

" p " " " q "

Figure 36.3 The image in the mirror of a person’s right hand is reversed front to back.This makes the right hand appear to be a left hand. Notice that the thumb is on the leftside of both real hands and on the left side of the image. That the thumb is not on theright side of the image indicates that there is no left-to-right reversal.

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decal can be read from outside the car, you can also read it when looking into yourrearview mirror from inside the car.

We conclude that the image that is formed by a flat mirror has the followingproperties.

S E C T I O N 3 6 . 1 • Images Formed by Flat Mirrors 1129

• The image is as far behind the mirror as the object is in front.

• The image is unmagnified, virtual, and upright. (By upright we mean that, if theobject arrow points upward as in Figure 36.2, so does the image arrow.)

• The image has front–back reversal.

Quick Quiz 36.1 In the overhead view of Figure 36.4, the image of the stoneseen by observer 1 is at C. At which of the five points A, B, C, D, or E does observer 2 seethe image?

Conceptual Example 36.1 Multiple Images Formed by Two Mirrors

Two flat mirrors are perpendicular to each other, as inFigure 36.5, and an object is placed at point O. In this situa-tion, multiple images are formed. Locate the positions ofthese images.

Solution The image of the object is at I1 in mirror 1 andat I2 in mirror 2. In addition, a third image is formed atI3. This third image is the image of I1 in mirror 2 or, equiva-lently, the image of I2 in mirror 1. That is, the image atI1 (or I2) serves as the object for I3. Note that to form thisimage at I3, the rays reflect twice after leaving the objectat O.

Mirror 2

Mirror 1

I1 I3

I2O

Figure 36.5 (Conceptual Example 36.1) When an object isplaced in front of two mutually perpendicular mirrors as shown,three images are formed.

Quick Quiz 36.2 You are standing about 2 m away from a mirror. Themirror has water spots on its surface. True or false: It is possible for you to see the waterspots and your image both in focus at the same time.

2 1

A B C D E

Figure 36.4 (Quick Quiz 36.1) Where does observer 2 see the image of the stone?


Conceptual Example 36.2 The Levitated Professor

The professor in the box shown in Figure 36.6 appears tobe balancing himself on a few fingers, with his feet offthe floor. He can maintain this position for a long time,and he appears to defy gravity. How was this illusioncreated?

Solution This is one of many magicians’ optical illusionsthat make use of a mirror. The box in which the professorstands is a cubical frame that contains a flat vertical mirrorpositioned in a diagonal plane of the frame. The professorstraddles the mirror so that one foot, which you see, is infront of the mirror, and the other foot, which you cannotsee, is behind the mirror. When he raises the foot in front ofthe mirror, the reflection of that foot also rises, so heappears to float in air.

Figure 36.6 (Conceptual Example 36.2) An optical illusion.

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Conceptual Example 36.3 The Tilting Rearview Mirror

Most rearview mirrors in cars have a day setting and anight setting. The night setting greatly diminishes theintensity of the image in order that lights from trailingvehicles do not blind the driver. How does such a mirrorwork?

Solution Figure 36.7 shows a cross-sectional view of arearview mirror for each setting. The unit consists of areflective coating on the back of a wedge of glass. In the daysetting (Fig. 36.7a), the light from an object behind the carstrikes the glass wedge at point 1. Most of the light entersthe wedge, refracting as it crosses the front surface, andreflects from the back surface to return to the front surface,where it is refracted again as it re-enters the air as ray B (forbright). In addition, a small portion of the light is reflectedat the front surface of the glass, as indicated by ray D (fordim).

B

D1

Daytime setting

Incidentlight

Reflectingside of mirror

(a)

Figure 36.7 (Conceptual Example 36.3) Cross-sectional views of a rearview mirror.(a) With the day setting, the silvered back surface of the mirror reflects a bright ray Binto the driver’s eyes. (b) With the night setting, the glass of the unsilvered frontsurface of the mirror reflects a dim ray D into the driver’s eyes.

B

D

Incidentlight

Nighttime setting

(b)

This dim reflected light is responsible for the image thatis observed when the mirror is in the night setting (Fig.36.7b). In this case, the wedge is rotated so that the pathfollowed by the bright light (ray B) does not lead to the eye.Instead, the dim light reflected from the front surface of thewedge travels to the eye, and the brightness of trailingheadlights does not become a hazard.

36.2 Images Formed by Spherical Mirrors

Concave Mirrors

A spherical mirror, as its name implies, has the shape of a section of a sphere. Thistype of mirror focuses incoming parallel rays to a point, as demonstrated by the col-ored light rays in Figure 36.8. Figure 36.9a shows a cross section of a spherical mirror,with its surface represented by the solid, curved black line. (The blue band representsthe structural support for the mirrored surface, such as a curved piece of glass onwhich the silvered surface is deposited.) Such a mirror, in which light is reflected fromthe inner, concave surface, is called a concave mirror. The mirror has a radius of cur-vature R, and its center of curvature is point C. Point V is the center of the sphericalsection, and a line through C and V is called the principal axis of the mirror.

Now consider a point source of light placed at point O in Figure 36.9b, where O isany point on the principal axis to the left of C. Two diverging rays that originate at Oare shown. After reflecting from the mirror, these rays converge and cross at the imagepoint I. They then continue to diverge from I as if an object were there. As a result, atpoint I we have a real image of the light source at O.

S E C T I O N 3 6 . 2 • Images Formed by Spherical Mirrors 1131

Figure 36.8 Red, blue, and green light raysare reflected by a curved mirror. Note thatthe three colored beams meet at a point.

Mirror

C V

(a)

Center ofcurvature R

Principalaxis

Mirror

O VI

(b)

C

Figure 36.9 (a) A concave mirror of radius R . The center of curvature C is located onthe principal axis. (b) A point object placed at O in front of a concave spherical mirrorof radius R, where O is any point on the principal axis farther than R from the mirrorsurface, forms a real image at I. If the rays diverge from O at small angles, they allreflect through the same image point.

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We shall consider in this section only rays that diverge from the object and make asmall angle with the principal axis. Such rays are called paraxial rays. All paraxial raysreflect through the image point, as shown in Figure 36.9b. Rays that are far from theprincipal axis, such as those shown in Figure 36.10, converge to other points on theprincipal axis, producing a blurred image. This effect, which is called spherical aberra-tion, is present to some extent for any spherical mirror and is discussed in Section 36.5.

We can use Figure 36.11 to calculate the image distance q from a knowledge of theobject distance p and radius of curvature R . By convention, these distances are mea-sured from point V. Figure 36.11 shows two rays leaving the tip of the object. One ofthese rays passes through the center of curvature C of the mirror, hitting the mirrorperpendicular to the mirror surface and reflecting back on itself. The second raystrikes the mirror at its center (point V ) and reflects as shown, obeying the law ofreflection. The image of the tip of the arrow is located at the point where these tworays intersect. From the gold right triangle in Figure 36.11, we see that tan# " h/p,and from the blue right triangle we see that tan# " $ h!/q. The negative sign is intro-duced because the image is inverted, so h! is taken to be negative. Thus, from Equation36.1 and these results, we find that the magnification of the image is

(36.2)

We also note from the two triangles in Figure 36.11 that have % as one angle that

from which we find that

(36.3)

If we compare Equations 36.2 and 36.3, we see that

Simple algebra reduces this to

(36.4)

This expression is called the mirror equation.If the object is very far from the mirror—that is, if p is so much greater than R

that p can be said to approach infinity—then 1/p # 0, and we see from Equation36.4 that q # R/2. That is, when the object is very far from the mirror, the imagepoint is halfway between the center of curvature and the center point on the mirror,as shown in Figure 36.12a. The incoming rays from the object are essentially parallel

1p

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R $ qp $ R

"qp

h!

h" $

R $ qp $ R

tan % "h

p $ R and tan % " $

h!

R $ q

M "h!

h" $

qp


Figure 36.10 Rays diverging fromthe object at large angles from theprincipal axis reflect from aspherical concave mirror tointersect the principal axis atdifferent points, resulting in ablurred image. This condition iscalled spherical aberration.

h

OPrincipal

axis

R

C

q

p

VIh′

αα

θθ

Figure 36.11 The imageformed by a spherical concavemirror when the object O liesoutside the center of curvatureC. This geometric constructionis used to derive Equation 36.4.

Mirror equation in terms ofradius of curvature

in this figure because the source is assumed to be very far from the mirror. We callthe image point in this special case the focal point F and the image distance thefocal length f, where

(36.5)

In Figure 36.8, the colored beams are traveling parallel to the principal axis and themirror reflects all three beams to the focal point. Notice that the point at which thethree beams intersect and the colors add is white.

Focal length is a parameter particular to a given mirror and therefore can be usedto compare one mirror with another. The mirror equation can be expressed in termsof the focal length:

(36.6)

Notice that the focal length of a mirror depends only on the curvature of the mirrorand not on the material from which the mirror is made. This is because the formationof the image results from rays reflected from the surface of the material. The situationis different for lenses; in that case the light actually passes through the material andthe focal length depends on the type of material from which the lens is made.

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Figure 36.12 (a) Light rays from a distant object (p : ') reflect from a concavemirror through the focal point F. In this case, the image distance q # R/2 " f, wheref is the focal length of the mirror. (b) Reflection of parallel rays from a concave mirror.

! PITFALL PREVENTION 36.2 The Focal Point Is

Not the Focus PointThe focal point is usually not thepoint at which the light rays focusto form an image. The focalpoint is determined solely by thecurvature of the mirror—it doesnot depend on the location ofthe object at all. In general, animage forms at a point differentfrom the focal point of a mirror(or a lens). The only exception iswhen the object is located infi-nitely far away from the mirror.

A satellite-dish antenna is a concave reflector for televi-sion signals from a satellite in orbit around the Earth. Thesignals are carried by microwaves that, because thesatellite is so far away, are parallel when they arrive at thedish. These waves reflect from the dish and are focusedon the receiver at the focal point of the dish.

Focal length

Mirror equation in terms of focallength

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Convex Mirrors

Figure 36.13 shows the formation of an image by a convex mirror—that is, onesilvered so that light is reflected from the outer, convex surface. This is sometimescalled a diverging mirror because the rays from any point on an object diverge afterreflection as though they were coming from some point behind the mirror. The imagein Figure 36.13 is virtual because the reflected rays only appear to originate at theimage point, as indicated by the dashed lines. Furthermore, the image is always uprightand smaller than the object. This type of mirror is often used in stores to foilshoplifters. A single mirror can be used to survey a large field of view because it forms asmaller image of the interior of the store.

We do not derive any equations for convex spherical mirrors because we can useEquations 36.2, 36.4, and 36.6 for either concave or convex mirrors if we adhere tothe following procedure. Let us refer to the region in which light rays move toward themirror as the front side of the mirror, and the other side as the back side. For example, inFigures 36.11 and 36.13, the side to the left of the mirrors is the front side, and theside to the right of the mirrors is the back side. Figure 36.14 states the sign conventionsfor object and image distances, and Table 36.1 summarizes the sign conventions for allquantities.

Ray Diagrams for Mirrors

The positions and sizes of images formed by mirrors can be conveniently deter-mined with ray diagrams. These graphical constructions reveal the nature of theimage and can be used to check results calculated from the mirror and magnifica-tion equations. To draw a ray diagram, we need to know the position of the objectand the locations of the mirror’s focal point and center of curvature. We then drawthree principal rays to locate the image, as shown by the examples in Figure 36.15.


FrontBack

O I F C

p q

Figure 36.13 Formation of an image by a spherical convex mirror. The image formedby the real object is virtual and upright.

Front, orreal, side

Reflected light

Back, orvirtual, side

p and q negative

No light

p and q positive

Incident light

Convex orconcave mirror

Figure 36.14 Signs of p and q forconvex and concave mirrors.

! PITFALL PREVENTION 36.3 Watch Your SignsSuccess in working mirrorproblems (as well as problemsinvolving refracting surfaces andthin lenses) is largely determinedby proper sign choices whensubstituting into the equations.The best way to become adept atthis is to work a multitude ofproblems on your own. Watchingyour instructor or reading the example problems is nosubstitute for practice.

Quantity Positive When Negative When

Object location (p) Object is in front of Object is in backmirror (real object) of mirror (virtual object)

Image location (q) Image is in front of Image is in back ofmirror (real image) mirror (virtual image)

Image height (h!) Image is upright Image is invertedFocal length (f ) Mirror is concave Mirror is convex

and radius (R)Magnification (M) Image is upright Image is inverted

Sign Conventions for Mirrors

Table 36.1


(a)

1

2

3

C FO

Front Back

I

Principal axis

(b)

1

2

3C F O I

Front Back

(c)

CFO I

1

2

3

Front Back

Active Figure 36.15 Ray diagrams for spherical mirrors, along with correspondingphotographs of the images of candles. (a) When the object is located so that the centerof curvature lies between the object and a concave mirror surface, the image is real,inverted, and reduced in size. (b) When the object is located between the focal pointand a concave mirror surface, the image is virtual, upright, and enlarged. (c) When theobject is in front of a convex mirror, the image is virtual, upright, and reduced in size.

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At the Active Figures linkat http://www.pse6.com, youcan move the objects andchange the focal length of themirrors to see the effect on theimages.

These rays all start from the same object point and are drawn as follows. We maychoose any point on the object; here, we choose the top of the object for simplicity.For concave mirrors (see Figs. 36.15a and 36.15b), we draw the following threeprincipal rays:


• Ray 1 is drawn from the top of the object parallel to the principal axis and isreflected through the focal point F.

• Ray 2 is drawn from the top of the object through the focal point and is reflectedparallel to the principal axis.

• Ray 3 is drawn from the top of the object through the center of curvature C and isreflected back on itself.

• Ray 1 is drawn from the top of the object parallel to the principal axis and isreflected away from the focal point F.

• Ray 2 is drawn from the top of the object toward the focal point on the back sideof the mirror and is reflected parallel to the principal axis.

• Ray 3 is drawn from the top of the object toward the center of curvature C on theback side of the mirror and is reflected back on itself.

The intersection of any two of these rays locates the image. The third ray serves asa check of the construction. The image point obtained in this fashion must alwaysagree with the value of q calculated from the mirror equation. With concave mirrors,note what happens as the object is moved closer to the mirror. The real, invertedimage in Figure 36.15a moves to the left as the object approaches the focal point.When the object is at the focal point, the image is infinitely far to the left. However,when the object lies between the focal point and the mirror surface, as shown inFigure 36.15b, the image is virtual, upright, and enlarged. This latter situation applieswhen you use a shaving mirror or a makeup mirror, both of which are concave. Yourface is closer to the mirror than the focal point, and you see an upright, enlargedimage of your face.

For convex mirrors (see Fig. 36.15c), we draw the following three principal rays:

Quick Quiz 36.3 You wish to reflect sunlight from a mirror onto somepaper under a pile of wood in order to start a fire. Which would be the best choice forthe type of mirror? (a) flat (b) concave (c) convex.

Quick Quiz 36.4 Consider the image in the mirror in Figure 36.16. Basedon the appearance of this image, you would conclude that (a) the mirror is concaveand the image is real. (b) the mirror is concave and the image is virtual. (c) themirror is convex and the image is real. (d) the mirror is convex and the image isvirtual.Figure 36.16 (Quick Quiz 36.4)

What type of mirror is this?

In a convex mirror, the image of an object is always virtual, upright, and reduced insize as shown in Figure 36.15c. In this case, as the object distance decreases, the virtualimage increases in size and moves away from the focal point toward the mirror as theobject approaches the mirror. You should construct other diagrams to verify how imageposition varies with object position.

! PITFALL PREVENTION 36.4 We Are Choosing a

Small Number ofRays

A huge number of light rays leaveeach point on an object (andpass through each point on animage). In a principal-ray dia-gram, which displays the charac-teristics of the image, we chooseonly a few rays that follow simplystated rules. Locating the imageby calculation complements thediagram.

NASA


Example 36.4 The Image formed by a Concave Mirror

image. In this case, the mirror equation gives

The image is virtual because it is located behind the mirror,as expected. The magnification of the image is

The image is twice as large as the object, and the positive signfor M indicates that the image is upright (see Fig. 36.15b).

What If? Suppose you set up the candle and mirror appara-tus illustrated in Figure 36.15a and described in part (A) ofthe example. While adjusting the apparatus, you accidentallystrike the candle with your elbow so that it begins to slidetoward the mirror at velocity vp. How fast does the image of thecandle move?

Answer We solve the mirror equation, Equation 36.6, for q :

Differentiating this equation with respect to time gives usthe velocity of the image vq " dq/dt :

For the object position of 25.0 cm in part (A), the velocity ofthe image is

Thus, the speed of the image is less than that of the objectin this case.

We can see two interesting behaviors of this function forvq . First, note that the velocity is negative regardless of thevalue of p or f. Thus, if the object moves toward the mirror, theimage moves toward the left in Figure 36.15 without regardfor the side of the focal point at which the object is located orwhether the mirror is concave or convex. Second, in the limitof p : 0, the velocity vq approaches $ vp . As the object movesvery close to the mirror, the mirror looks like a plane mirror,the image is as far behind the mirror as the object is in front,and both the object and the image move with the same speed.

vq " $f 2 vp

(p $ f )2 " $(10.0 cm)2

vp

(25.0 cm $ 10.0 cm)2 " $0.444 vp

vq "dqdt

"ddt

$ f pp $ f % " $

f 2

(p $ f )2 dpdt

" $f

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p $ f

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Assume that a certain spherical mirror has a focal length of& 10.0 cm. Locate and describe the image for objectdistances of

(A) 25.0 cm,

(B) 10.0 cm, and

(C) 5.00 cm.

Solution Because the focal length is positive, we know thatthis is a concave mirror (see Table 36.1).

(A) This situation is analogous to that in Figure 36.15a;hence, we expect the image to be real. We find the imagedistance by using Equation 36.6:

The magnification of the image is given by Equation 36.2:

The fact that the absolute value of M is less than unity tells usthat the image is smaller than the object, and the negative signfor M tells us that the image is inverted. Because q is positive,the image is located on the front side of the mirror and is real.

(B) When the object distance is 10.0 cm, the object islocated at the focal point. Now we find that

which means that rays originating from an object positioned atthe focal point of a mirror are reflected so that the image isformed at an infinite distance from the mirror; that is, the raystravel parallel to one another after reflection. This is the situa-tion in a flashlight, where the bulb filament is placed at thefocal point of a reflector, producing a parallel beam of light.

(C) When the object is at p " 5.00 cm, it lies halfwaybetween the focal point and the mirror surface, as shown inFigure 36.15b. Thus, we expect a magnified, virtual, upright

'q "

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" $16.7 cm25.0 cm

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Investigate the image formed for various object positions and mirror focal lengths at the Interactive Worked Example linkat http://www.pse6.com.

Interactive

Example 36.5 The Image from a Convex Mirror

(B) the magnification of the image.

Solution (A) This situation is depicted in Figure 36.15c.We should expect to find an upright, reduced, virtual image.To find the image position, we use Equation 36.6:

An anti-shoplifting mirror, as shown in Figure 36.17, showsan image of a woman who is located 3.0 m from the mirror.The focal length of the mirror is $ 0.25 m. Find

(A) the position of her image and

Interactive

36.3 Images Formed by Refraction

In this section we describe how images are formed when light rays are refracted at theboundary between two transparent materials. Consider two transparent media havingindices of refraction n1 and n2, where the boundary between the two media is a spheri-cal surface of radius R (Fig. 36.18). We assume that the object at O is in the mediumfor which the index of refraction is n1. Let us consider the paraxial rays leaving O. Aswe shall see, all such rays are refracted at the spherical surface and focus at a singlepoint I, the image point.

Figure 36.19 shows a single ray leaving point O and refracting to point I. Snell’s lawof refraction applied to this ray gives

Because #1 and #2 are assumed to be small, we can use the small-angle approximationsin # # # (with angles in radians) and say that

Now we use the fact that an exterior angle of any triangle equals the sum of the twoopposite interior angles. Applying this rule to triangles OPC and PIC in Figure 36.19 gives

( " # 2 & )

# 1 " % & (

n1# 1 " n2# 2

n1 sin # 1 " n2 sin # 2


The negative value of q indicates that her image is virtual, orbehind the mirror, as shown in Figure 36.15c.

(B) The magnification of the image is

The image is much smaller than the woman, and it isupright because M is positive.

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"1

$0.25 m

Figure 36.17 (Example 36.5) Convex mirrors, often used for security in department stores, provide wide-angle viewing.

Investigate the image formed for various object positions and mirror focal lengths at the Interactive Worked Example linkat http://www.pse6.com.

n1 < n2

O I

p q

n2n1R

Figure 36.18 An image formed byrefraction at a spherical surface. Raysmaking small angles with the principal axisdiverge from a point object at O and arerefracted through the image point I.

©19

90 P

aul S

ilver

man

/Fun

dam

enta

l Pho

togr

aphs

If we combine all three expressions and eliminate #1 and #2, we find that

(36.7)

From Figure 36.19, we see three right triangles that have a common vertical leg oflength d. For paraxial rays (unlike the relatively large-angle ray shown in Fig. 36.19),the horizontal legs of these triangles are approximately p for the triangle containingangle %, R for the triangle containing angle (, and q for the triangle containing angle). In the small-angle approximation, tan# # #, so we can write the approximate rela-tionships from these triangles as follows:

We substitute these expressions into Equation 36.7 and divide through by d to give

(36.8)

For a fixed object distance p, the image distance q is independent of the angle that theray makes with the axis. This result tells us that all paraxial rays focus at the same point I.

As with mirrors, we must use a sign convention if we are to apply this equation to avariety of cases. We define the side of the surface in which light rays originate as thefront side. The other side is called the back side. Real images are formed by refractionin back of the surface, in contrast with mirrors, where real images are formed in frontof the reflecting surface. Because of the difference in location of real images, therefraction sign conventions for q and R are opposite the reflection sign conventions.For example, q and R are both positive in Figure 36.19. The sign conventions for spher-ical refracting surfaces are summarized in Table 36.2.

We derived Equation 36.8 from an assumption that n1 * n2 in Figure 36.19. Thisassumption is not necessary, however. Equation 36.8 is valid regardless of which indexof refraction is greater.

n1

p&

n2

q"

n2 $ n1

R

tan % # % #dp tan ( # ( #

dR tan ) # ) #

dq

n1% & n2) " (n2 $ n1)(

S E C T I O N 3 6 . 3 • Images Formed by Refraction 1139

Relation between object andimage distance for a refractingsurface

O

P

R

C

n1 n2

d

p q

γα βI

θ1 θ2

Figure 36.19 Geometry used to derive Equation 36.8, assuming that n1 * n2.


Object location (p) Object is in front of Object is in back of surface (real object) surface (virtual object)

Image location (q) Image is in back of Image is in front of surface (real image) surface (virtual image)

Image height (h!) Image is upright Image is invertedRadius (R) Center of curvature Center of curvature

is in back of surface is in front of surface

Sign Conventions for Refracting Surfaces

Table 36.2

Flat Refracting Surfaces

If a refracting surface is flat, then R is infinite and Equation 36.8 reduces to

(36.9)

From this expression we see that the sign of q is opposite that of p. Thus, accordingto Table 36.2, the image formed by a flat refracting surface is on the sameside of the surface as the object. This is illustrated in Figure 36.20 for thesituation in which the object is in the medium of index n1 and n1 is greater than n 2.In this case, a virtual image is formed between the object and the surface. If n1 isless than n 2, the rays in the back side diverge from each other at lesser angles thanthose in Figure 36.20. As a result, the virtual image is formed to the left of theobject.

q " $n2

n1 p

n1

p" $

n2

q


Example 36.7 Gaze into the Crystal Ball

Solution Because n1 + n 2, where n2 " 1.00 is the index ofrefraction for air, the rays originating from the coin arerefracted away from the normal at the surface and divergeoutward. Hence, the image is formed inside the paper-weight and is virtual. Applying Equation 36.8 and noting

A set of coins is embedded in a spherical plastic paper-weight having a radius of 3.0 cm. The index of refraction ofthe plastic is n1 " 1.50. One coin is located 2.0 cm from theedge of the sphere (Fig. 36.21). Find the position of theimage of the coin.

O

I

q

p

n1 > n2

n1 n2

Active Figure 36.20 The imageformed by a flat refracting surfaceis virtual and on the same side ofthe surface as the object. All raysare assumed to be paraxial.

At the Active Figures linkat http://www.pse6.com, youcan move the object to see theeffect on the location of theimage. Quick Quiz 36.5 In Figure 36.18, what happens to the image point I as the

object point O is moved to the right from very far away to very close to the refractingsurface? (a) It is always to the right of the surface. (b) It is always to the left of the surface. (c) It starts off to the left and at some position of O, I moves to the rightof the surface. (d) It starts off to the right and at some position of O, I moves to the leftof the surface.

Quick Quiz 36.6 In Figure 36.20, what happens to the image point I asthe object point O moves toward the right-hand surface of the material of index ofrefraction n1? (a) It always remains between O and the surface, arriving at thesurface just as O does. (b) It moves toward the surface more slowly than O so thateventually O passes I. (c) It approaches the surface and then moves to the right ofthe surface.

Conceptual Example 36.6 Let’s Go Scuba Diving!

when a person under water views objects with the nakedeye. In this case, light rays from an object focus behindthe retina, resulting in a blurred image. When a mask isused, the air space between the eye and the mask surfaceprovides the normal amount of refraction at the eye–airinterface, and the light from the object focuses on theretina.

It is well known that objects viewed under water with thenaked eye appear blurred and out of focus. However, ascuba diver using a mask has a clear view of underwaterobjects. Explain how this works, using the facts that theindices of refraction of the cornea, water, and air are 1.376,1.333, and 1.00029, respectively.

Solution Because the cornea and water have almost iden-tical indices of refraction, very little refraction occurs

36.4 Thin Lenses

Lenses are commonly used to form images by refraction in optical instruments, suchas cameras, telescopes, and microscopes. We can use what we just learned aboutimages formed by refracting surfaces to help us locate the image formed by a lens.We recognize that light passing through a lens experiences refraction at twosurfaces. The development we shall follow is based on the notion that the image

S E C T I O N 3 6 . 4 • Thin Lenses 1141

3.0 cm

2.0 cm

q

n2n1

n1 > n2

Figure 36.21 (Example 36.7) Light rays from a coin embeddedin a plastic sphere form a virtual image between the surfaceof the object and the sphere surface. Because the object is insidethe sphere, the front of the refracting surface is the interior ofthe sphere.

from Table 36.2 that R is negative, we obtain

The negative sign for q indicates that the image is in front ofthe surface—in other words, in the same medium as theobject, as shown in Figure 36.21. Being in the same mediumas the object, the image must be virtual. (See Table 36.2.)The coin appears to be closer to the paperweight surfacethan it actually is.

$1.7 cmq "

1.50

2.0 cm&

1q

"1.00 $ 1.50

$3.0 cm

n1

p&

n2

q"

n2 $ n1

R

Example 36.8 The One That Got Away

The apparent height h! of the fish is

and the fish appears to be approximately three-fourths itsactual height.

" 0.752h

h! " q top $ q bottom " $0.752d $ [$0.752(d & h)]

A small fish is swimming at a depth d below the surface of apond (Fig. 36.22). What is the apparent depth of the fish, asviewed from directly overhead?

Solution Because the refracting surface is flat, R is infinite.Hence, we can use Equation 36.9 to determine the locationof the image with p " d. Using the indices of refractiongiven in Figure 36.22, we obtain

Because q is negative, the image is virtual, as indicated bythe dashed lines in Figure 36.22. The apparent depth isapproximately three-fourths the actual depth.

What If? What if you look more carefully at the fish andmeasure its apparent height, from its upper fin to its lowerfin? Is the apparent height h! of the fish different from theactual height h?

Answer Because all points on the fish appear to be frac-tionally closer to the observer, we would predict that theheight would be smaller. If we let the distance d in Figure36.22 be measured to the top fin and the distance to thebottom fin be d & h, then the images of the top and bottomof the fish are located at

q bottom " $0.752(d & h)

q top " $0.752d

$0.752dq " $n2

n1 p " $

1.001.33

d "

d

q

n2 = 1.00

n1 = 1.33

Figure 36.22 (Example 36.8) The apparent depth q of thefish is less than the true depth d. All rays are assumed to beparaxial.

formed by one refracting surface serves as the object for the second surface.We shall analyze a thick lens first and then let the thickness of the lens be approxi-mately zero.

Consider a lens having an index of refraction n and two spherical surfaces withradii of curvature R1 and R2, as in Figure 36.23. (Note that R1 is the radius of curva-ture of the lens surface that the light from the object reaches first and that R2 is theradius of curvature of the other surface of the lens.) An object is placed at point O at adistance p1 in front of surface 1.

Let us begin with the image formed by surface 1. Using Equation 36.8 and assum-ing that n1 " 1 because the lens is surrounded by air, we find that the image I1 formedby surface 1 satisfies the equation

(36.10)

where q 1 is the position of the image due to surface 1. If the image due tosurface 1 is virtual (Fig. 36.23a), q 1 is negative, and it is positive if the image is real(Fig. 36.23b).

Now we apply Equation 36.8 to surface 2, taking n1 " n and n 2 " 1. (We make thisswitch in index because the light rays approaching surface 2 are in the material of thelens, and this material has index n.) Taking p 2 as the object distance for surface 2 andq2 as the image distance gives

(36.11)

We now introduce mathematically the fact that the image formed by the firstsurface acts as the object for the second surface. We do this by noting from Figure36.23 that p2, measured from surface 2, is related to q1 as follows:

Virtual image from surface 1 (Fig. 36.23a): p2 " $ q1 & t (q1 is negative)

Real image from surface 1 (Fig. 36.23b): p2 " $ q1 & t (q1 is positive)

np 2

&1q 2

"1 $ n

R 2

1p 1

&nq 1

"n $ 1

R 1


tp1

q1

p2

O

I1

C1

Surface 1

R1

nSurface 2

R2

n1 = 1

tp1

q1

O

C1

Surface 1

R1

nSurface 2

R2

n1 = 1

p2

I1

(a)

(b)

Figure 36.23 To locate the image formedby a lens, we use the virtual image at I1formed by surface 1 as the object for theimage formed by surface 2. The point C1 isthe center of curvature of surface 1. (a) Theimage due to surface 1 is virtual so that I1 isto the left of the surface. (b) The image dueto surface 1 is real so that I1 is to the right ofthe surface.

where t is the thickness of the lens. For a thin lens (one whose thickness is smallcompared to the radii of curvature), we can neglect t. In this approximation, we seethat p 2 " $ q 1 for either type of image from surface 1. (If the image from surface 1is real, the image acts as a virtual object, so p 2 is negative.) Hence, Equation 36.11becomes

(36.12)

Adding Equations 36.10 and 36.12, we find that

(36.13)

For a thin lens, we can omit the subscripts on p1 and q2 in Equation 36.13 and call theobject distance p and the image distance q, as in Figure 36.24. Hence, we can writeEquation 36.13 in the form

(36.14)

This expression relates the image distance q of the image formed by a thin lens to theobject distance p and to the lens properties (index of refraction and radii of curva-ture). It is valid only for paraxial rays and only when the lens thickness is much lessthan R1 and R2.

The focal length f of a thin lens is the image distance that corresponds to an infi-nite object distance, just as with mirrors. Letting p approach ' and q approach f inEquation 36.14, we see that the inverse of the focal length for a thin lens is

(36.15)

This relationship is called the lens makers’ equation because it can be used to deter-mine the values of R1 and R2 that are needed for a given index of refraction and adesired focal length f. Conversely, if the index of refraction and the radii of curvatureof a lens are given, this equation enables a calculation of the focal length. If the lens isimmersed in something other than air, this same equation can be used, with n inter-preted as the ratio of the index of refraction of the lens material to that of the sur-rounding fluid.

Using Equation 36.15, we can write Equation 36.14 in a form identical to Equation36.6 for mirrors:

(36.16)1p

&1q

"1f

1f

" (n $ 1) $ 1R 1

$1

R 2%

1p

&1q

" (n $ 1) $ 1R 1

$1

R 2%

1p 1

&1q 2

" (n $ 1) $ 1R 1

$1

R 2%

$nq 1

&1q 2

"1 $ n

R 2


C1C2

O

p q

I

R2R1

Figure 36.24 Simplified geometry for a thin lens.

Lens makers’ equation

Thin lens equation


f f

f f

(a)

(b)

F1 F2F1 F2

F1F2F1 F2

Figure 36.25 (Left) Effects of a converging lens (top) and a diverging lens (bottom)on parallel rays. (Right) Parallel light rays pass through (a) a converging lens and (b) a diverging lens. The focal length is the same for light rays passing through a givenlens in either direction. Both focal points F1 and F2 are the same distance from the lens.

Henr

y Lea

p an

d Ji

m Le

hman

Front

p positiveq negative

Incident light

Back

p negativeq positive

Refracted light

Figure 36.26 A diagram forobtaining the signs of p and q for athin lens. (This diagram alsoapplies to a refracting surface.)


Object location (p) Object is in front of Object is in back oflens (real object) lens (virtual object)

Image location (q) Image is in back of Image is in front oflens (real image) lens (virtual image)

Image height (h!) Image is upright Image is invertedR1 and R2 Center of curvature Center of curvature

is in back of lens is in front of lensFocal length ( f ) Converging lens Diverging lens

Sign Conventions for Thin Lenses

Table 36.3

This equation, called the thin lens equation, can be used to relate the image distanceand object distance for a thin lens.

Because light can travel in either direction through a lens, each lens has two focalpoints, one for light rays passing through in one direction and one for rays passingthrough in the other direction. This is illustrated in Figure 36.25 for a biconvex lens(two convex surfaces, resulting in a converging lens) and a biconcave lens (two con-cave surfaces, resulting in a diverging lens).

Figure 36.26 is useful for obtaining the signs of p and q, and Table 36.3 gives thesign conventions for thin lenses. Note that these sign conventions are the same asthose for refracting surfaces (see Table 36.2). Applying these rules to a biconvex lens,we see that when p + f , the quantities p, q, and R1 are positive, and R2 is negative.Therefore, p, q, and f are all positive when a converging lens forms a real image of anobject. For a biconcave lens, p and R2 are positive and q and R1 are negative, with theresult that f is negative.

! PITFALL PREVENTION 36.5 A Lens Has Two

Focal Points butOnly One FocalLength

A lens has a focal point on eachside, front and back. However,there is only one focal length—each of the two focal points islocated the same distance fromthe lens (Fig, 36.25). This can beseen mathematically by inter-changing R1 and R2 in Equation36.15 (and changing the signs ofthe radii because back and fronthave been interchanged). As aresult, the lens forms an image ofan object at the same point if itis turned around. In practice thismight not happen, because reallenses are not infinitesimally thin.

Various lens shapes are shown in Figure 36.27. Note that a converging lens isthicker at the center than at the edge, whereas a diverging lens is thinner at the centerthan at the edge.

Magnification of Images

Consider a thin lens through which light rays from an object pass. As with mirrors (Eq.36.2), we could analyze a geometric construction to show that the lateral magnificationof the image is

From this expression, it follows that when M is positive, the image is upright and onthe same side of the lens as the object. When M is negative, the image is inverted andon the side of the lens opposite the object.

Ray Diagrams for Thin Lenses

Ray diagrams are convenient for locating the images formed by thin lenses or systemsof lenses. They also help clarify our sign conventions. Figure 36.28 shows such dia-grams for three single-lens situations.

To locate the image of a converging lens (Fig. 36.28a and b), the following three raysare drawn from the top of the object:

M "h!

h" $

qp


(a)

(b)

Convex–concave

Convex–concave

Plano–concave

Plano–convex

Biconvex

Biconcave

Figure 36.27 Various lens shapes.(a) Converging lenses have a posi-tive focal length and are thickest atthe middle. (b) Diverging lenseshave a negative focal length andare thickest at the edges.

O

(a)

F1

Front

F2

Back

I

1

23

I

(b)

F1

Front

F2

Back

O

1

2

3

O

(c)

F1

Front

F2

Back

I

1

2

3

Active Figure 36.28 Ray diagrams for locating the image formedby a thin lens. (a) When the object is in front of and outside thefocal point of a converging lens, the image is real, inverted, and onthe back side of the lens. (b) When the object is between the focalpoint and a converging lens, the image is virtual, upright, largerthan the object, and on the front side of the lens. (c) When anobject is anywhere in front of a diverging lens, the image is virtual,upright, smaller than the object, and on the front side of the lens.

At the Active Figures link at http://www.pse6.com, youcan move the objects and change the focal length of thelenses to see the effect on the images.

• Ray 1 is drawn parallel to the principal axis. After being refracted by the lens, thisray passes through the focal point on the back side of the lens.

• Ray 2 is drawn through the center of the lens and continues in a straight line.

• Ray 3 is drawn through the focal point on the front side of the lens (or as ifcoming from the focal point if p * f ) and emerges from the lens parallel to theprincipal axis.

To locate the image of a diverging lens (Fig. 36.28c), the following three rays aredrawn from the top of the object:


• Ray 1 is drawn parallel to the principal axis. After being refracted by the lens,this ray emerges directed away from the focal point on the front side of thelens.

• Ray 2 is drawn through the center of the lens and continues in a straight line.

• Ray 3 is drawn in the direction toward the focal point on the back side of the lensand emerges from the lens parallel to the principal axis.

For the converging lens in Figure 36.28a, where the object is to the left of the focalpoint (p + f ), the image is real and inverted. When the object is between the focalpoint and the lens (p * f ), as in Figure 36.28b, the image is virtual and upright. For adiverging lens (see Fig. 36.28c), the image is always virtual and upright, regardless ofwhere the object is placed. These geometric constructions are reasonably accurate onlyif the distance between the rays and the principal axis is much less than the radii of thelens surfaces.

Note that refraction occurs only at the surfaces of the lens. A certain lens designtakes advantage of this fact to produce the Fresnel lens, a powerful lens without greatthickness. Because only the surface curvature is important in the refracting qualities ofthe lens, material in the middle of a Fresnel lens is removed, as shown in the crosssections of lenses in Figure 36.29. Because the edges of the curved segments causesome distortion, Fresnel lenses are usually used only in situations in which imagequality is less important than reduction of weight. A classroom overhead projectoroften uses a Fresnel lens; the circular edges between segments of the lens can be seenby looking closely at the light projected onto a screen.

Figure 36.29 The Fresnel lens onthe left has the same focal length asthe thick lens on the right but ismade of much less glass.

Quick Quiz 36.7 What is the focal length of a pane of window glass?(a) zero (b) infinity (c) the thickness of the glass (d) impossible to determine

Quick Quiz 36.8 Diving masks often have a lens built into the glass fordivers who do not have perfect vision. This allows the individual to dive without thenecessity for glasses, because the lenses in the faceplate perform the necessaryrefraction to provide clear vision. The proper design allows the diver to see clearly withthe mask on both under water and in the open air. Normal eyeglasses have lenses thatare curved on both the front and back surfaces. The lenses in a diving mask should becurved (a) only on the front surface (b) only on the back surface (c) on both the frontand back surfaces.

Example 36.9 Images Formed by a Converging Lens

Solution

(A) First we construct a ray diagram as shown in Figure36.30a. The diagram shows that we should expect a real,inverted, smaller image to be formed on the back side of thelens. The thin lens equation, Equation 36.16, can be used tofind the image distance:

1p

&1q

"1f

A converging lens of focal length 10.0 cm forms images ofobjects placed

(A) 30.0 cm,

(B) 10.0 cm, and

(C) 5.00 cm from the lens.

In each case, construct a ray diagram, find the imagedistance and describe the image.

Interactive


The positive sign for the image distance tells us that theimage is indeed real and on the back side of the lens. Themagnification of the image is

Thus, the image is reduced in height by one half, and thenegative sign for M tells us that the image is inverted.

(B) No calculation is necessary for this case becausewe know that, when the object is placed at the focalpoint, the image is formed at infinity. This is readilyverified by substituting p " 10.0 cm into the thin lensequation.

(C) We now move inside the focal point. The ray diagramin Figure 36.30b shows that in this case the lens acts as amagnifying glass; that is, the image is magnified, upright,on the same side of the lens as the object, and virtual.Because the object distance is 5.00 cm, the thin lens equa-tion gives

$10.0 cmq "

1

5.00 cm&

1q

"1

10.0 cm

$0.500M " $qp

" $15.0 cm30.0 cm

"

&15.0 cmq "

1

30.0 cm&

1q

"1

10.0 cm

and the magnification of the image is

The negative image distance tells us that the image is virtualand formed on the side of the lens from which the light isincident, the front side. The image is enlarged, and the posi-tive sign for M tells us that the image is upright.

What If? What if the object moves right up to the lenssurface, so that p B 0? Where is the image?

Answer In this case, because p ** R , where R is either of theradii of the surfaces of the lens, the curvature of the lens canbe ignored and it should appear to have the same effect as aplane piece of material. This would suggest that the image isjust on the front side of the lens, at q " 0. We can verify thismathematically by rearranging the thin lens equation:

If we let p : 0, the second term on the right becomes verylarge compared to the first and we can neglect 1/f. Theequation becomes

Thus, q is on the front side of the lens (because it has theopposite sign as p), and just at the lens surface.

q " $p " 0

1q

" $1p

1q

"1f

$1p

&2.00M " $qp

" $$ $10.0 cm5.00 cm % "

(b)

O F2I, F1

10.0 cm5.00 cm

10.0 cm

(a)

O F1

F2 I

15.0 cm

30.0 cm

10.0 cm

Figure 36.30 (Example 36.9) An image is formed by a converging lens. (a) The objectis farther from the lens than the focal point. (b) The object is closer to the lens thanthe focal point.

Investigate the image formed for various object positions and lens focal lengths at the Interactive Worked Example link athttp://www.pse6.com.


Example 36.10 The Case of a Diverging Lens

This result confirms that the image is virtual, smaller thanthe object, and upright.

(B) When the object is at the focal point, the ray diagramappears as in Figure 36.31b. In the thin lens equation, usingp " 10.0 cm, we have

The magnification of the image is

Notice the difference between this situation and that for aconverging lens. For a diverging lens, an object at the focalpoint does not produce an image infinitely far away.

(C) When the object is inside the focal point, at p " 5.00 cm,the ray diagram in Figure 36.31c shows that we expect avirtual image that is smaller than the object and upright. In

&0.500M " $qp

" $$ $5.00 cm10.0 cm % "

$5.00 cmq "

1

10.0 cm&

1q

"1

$10.0 cm

Repeat Example 36.9 for a diverging lens of focal length10.0 cm.

Solution(A) We begin by constructing a ray diagram as in Figure36.31a taking the object distance to be 30.0 cm. The dia-gram shows that we should expect an image that is virtual,smaller than the object, and upright. Let us now apply thethin lens equation with p " 30.0 cm:

The magnification of the image is

&0.250M " $qp

" $$ $7.50 cm30.0 cm % "

$7.50 cmq "

1

30.0 cm&

1q

"1

$10.0 cm

1p

&1q

"1f

IO

(c)

F1

5.00 cm

3.33 cm

10.0 cm

F2

IO

(a)

F1

30.0 cm

10.0 cm7.50 cm

F2 IO, F1

(b)

5.00 cm

10.0 cm

F2

Figure 36.31 (Example 36.10) An image is formed by a diverging lens. (a) The objectis farther from the lens than the focal point. (b) The object is at the focal point.(c) The object is closer to the lens than the focal point.

Interactive

Combination of Thin Lenses

If two thin lenses are used to form an image, the system can be treated in the followingmanner. First, the image formed by the first lens is located as if the second lens were notpresent. Then a ray diagram is drawn for the second lens, with the image formed by thefirst lens now serving as the object for the second lens. The second image formed is thefinal image of the system. If the image formed by the first lens lies on the back side of thesecond lens, then that image is treated as a virtual object for the second lens (that is, inthe thin lens equation, p is negative). The same procedure can be extended to a systemof three or more lenses. Because the magnification due to the second lens is performedon the magnified image due to the first lens, the overall magnification of the image dueto the combination of lenses is the product of the individual magnifications.

Let us consider the special case of a system of two lenses of focal lengths f1 and f2in contact with each other. If p1 " p is the object distance for the combination, applica-tion of the thin lens equation (Eq. 36.16) to the first lens gives

1p

&1q 1

"1f 1


this case, the thin lens equation gives

$3.33 cmq "

1

5.00 cm&

1q

"1

$10.0 cm

and the magnification of the image is

This confirms that the image is virtual, smaller than theobject, and upright.

&0.667M " $$ $3.33 cm5.00 cm % "

Example 36.11 A Lens Under Water

equation by the second gives

Because f air " 40.0 cm, we find that

f water " 3.71f air " 3.71(40.0 cm) "

The focal length of any lens is increased by a factor(n $ 1)/(n! $ 1) when the lens is immersed in a fluid,where n! is the ratio of the index of refraction n of the lensmaterial to that of the fluid.

148 cm

f water

f air"

n $ 1n! $ 1

"1.52 $ 11.14 $ 1

" 3.71

A converging glass lens (n " 1.52) has a focal length of40.0 cm in air. Find its focal length when it is immersed inwater, which has an index of refraction of 1.33.

Solution We can use the lens makers’ equation (Eq. 36.15)in both cases, noting that R1 and R 2 remain the same in airand water:

where n! is the ratio of the index of refraction of glassto that of water: n! " 1.52/1.33 " 1.14. Dividing the first

1

f water" (n! $ 1) $ 1

R1$

1R 2

%

1

f air" (n $ 1) $ 1

R1$

1R 2

%

Investigate the image formed for various object positions and lens focal lengths at the Interactive Worked Example link athttp://www.pse6.com.

Light from a distant object is brought into focusby two converging lenses.He

nry L

eap

and

Jim

Lehm

an

where q1 is the image distance for the first lens. Treating this image as the object forthe second lens, we see that the object distance for the second lens must be p2 " $ q1.(The distances are the same because the lenses are in contact and assumed to be infini-tesimally thin. The object distance is negative because the object is virtual.) Therefore,for the second lens,

where q " q2 is the final image distance from the second lens, which is the image distancefor the combination. Adding the equations for the two lenses eliminates q1 and gives

If we consider replacing the combination with a single lens that will form an image atthe same location, we see that its focal length is related to the individual focal lengthsby

(36.17)

Therefore, two thin lenses in contact with each other are equivalent to a singlethin lens having a focal length given by Equation 36.17.

1f

"1f 1

&1f 2

1p

&1q

"1f 1

&1f 2

$1q 1

&1q

"1f 2

1p 2

&1q 2

"1f 2


Focal length for a combinationof two thin lenses in contact

Example 36.12 Where Is the Final Image?

The magnification of this image is

The image formed by this lens acts as the object for thesecond lens. Thus, the object distance for the second lens is20.0 cm $ 15.0 cm " 5.00 cm. We again apply the thin lensequation to find the location of the final image:

The magnification of the second image is

Thus, the overall magnification of the system is

To finalize the problem, note that the negative sign on theoverall magnification indicates that the final image isinverted with respect to the initial object. The fact that theabsolute value of the magnification is less than one tells usthat the final image is smaller than the object. The fact thatq2 is negative tells us that the final image is on the front, orleft, side of lens 2. All of these conclusions are consistentwith the ray diagram in Figure 36.32b.

$0.667M " M1M2 " ($0.500)(1.33) "

&1.33M2 " $q 2

p 2" $

($6.67 cm)5.00 cm

"

$6.67 cmq 2 "

1

5.00 cm&

1q 2

"1

20.0 cm

$0.500M1 " $q 1

p 1" $

15.0 cm30.0 cm

"

Two thin converging lenses of focal lengths f1 " 10.0 cmand f2 " 20.0 cm are separated by 20.0 cm, as illustrated inFigure 36.32a. An object is placed 30.0 cm to the left oflens 1. Find the position and the magnification of the finalimage.

Solution Conceptualize by imagining light rays passingthrough the first lens and forming a real image (becausep + f ) in the absence of the second lens. Figure 36.32bshows these light rays forming the inverted image I1. Oncethe light rays converge to the image point, they do notstop. They continue through the image point and interactwith the second lens. The rays leaving the image pointbehave in the same way as the rays leaving an object. Thus,the image of the first lens serves as the object of the secondlens. We categorize this problem as one in which we applythe thin lens equation, but in stepwise fashion to the twolenses.

To analyze the problem, we first draw a ray diagram(Figure 36.32b) showing where the image from the first lensfalls and how it acts as the object for the second lens. Thelocation of the image formed by lens 1 is found from thethin lens equation:

&15.0 cmq 1 "

1

30.0 cm&

1q 1

"1

10.0 cm

1p 1

&1q 1

"1f

Interactive


30.0 cm 20.0 cm

Object

f1 = 10.0 cm f2 = 20.0 cm(a)

(b)

I2 I1

Lens 1 Lens 220.0 cm

6.67 cm

15.0 cm10.0 cm

O1

30.0 cm

What If? Suppose we want to create an upright image withthis system of two lenses. How must the second lens bemoved in order to achieve this?

Answer Because the object is farther from the first lensthan the focal length of that lens, we know that the firstimage is inverted. Consequently, we need the second lens to

invert the image once again so that the final image isupright. An inverted image is only formed by a converginglens if the object is outside the focal point. Thus, the imagedue to the first lens must be to the left of the focal point ofthe second lens in Figure 36.32b. To make this happen, wemust move the second lens at least as far away from the firstlens as the sum q1 & f2 " 15.0 cm & 20.0 cm " 35.0 cm.

Figure 36.32 (Example 36.12) (a) A combination of two converging lenses. (b) Theray diagram showing the location of the final image due to the combination of lenses.The black dots are the focal points of lens 1 while the red dots are the focal points oflens 2.

Investigate the image formed by a combination of two lenses at the Interactive Worked Example link at http://www.pse6.com.

Conceptual Example 36.13 Watch Your p’s and q’s!

related according to the same equation—the thin lensequation.

The curve in the upper right portion of the f " & 10 cmgraph corresponds to an object located on the front side of alens, which we have drawn as the left side of the lens in ourprevious diagrams. When the object is at positive infinity, areal image forms at the focal point on the back side (the posi-tive side) of the lens, q " f. (The incoming rays are parallel inthis case.) As the object moves closer to the lens, the image

Use a spreadsheet or a similar tool to create two graphs ofimage distance as a function of object distance—one for alens for which the focal length is 10 cm and one for a lensfor which the focal length is $ 10 cm.

Solution The graphs are shown in Figure 36.33. In eachgraph, a gap occurs where p " f, which we shalldiscuss. Note the similarity in the shapes—a result of thefact that image and object distances for both lenses are

36.5 Lens Aberrations

Our analysis of mirrors and lenses assumes that rays make small angles with the princi-pal axis and that the lenses are thin. In this simple model, all rays leaving a pointsource focus at a single point, producing a sharp image. Clearly, this is not always true.When the approximations used in this analysis do not hold, imperfect images areformed.

A precise analysis of image formation requires tracing each ray, using Snell’s law ateach refracting surface and the law of reflection at each reflecting surface. This proce-dure shows that the rays from a point object do not focus at a single point, with theresult that the image is blurred. The departures of actual images from the ideal pre-dicted by our simplified model are called aberrations.

Spherical Aberrations

Spherical aberrations occur because the focal points of rays far from the principal axis ofa spherical lens (or mirror) are different from the focal points of rays of the samewavelength passing near the axis. Figure 36.34 illustrates spherical aberration for parallelrays passing through a converging lens. Rays passing through points near the center of


–50 –40 –30 –20 –10 0 10 20 30 40 50–50

–40

–30

–20

–10

0

10

20

30

40

50

Virtual object Real object

Real image

Virtual image

q(cm)

p(cm)

f = 10 cm

(a)

–50 –40 –30 –20 –10 0 10 20 30 40 50–50

–40

–30

–20

–10

0

10

20

30

40

50

Virtual object Real object

Real image

Virtual image

q(cm)

p(cm)

f = –10 cm

(b)

Figure 36.33 (Conceptual Example 36.13) (a) Image position as a function of objectposition for a lens having a focal length of & 10 cm. (b) Image position as a function ofobject position for a lens having a focal length of $ 10 cm.

moves farther from the lens, corresponding to the upwardpath of the curve. This continues until the object is located atthe focal point on the near side of the lens. At this point, therays leaving the lens are parallel, making the image infinitelyfar away. This is described in the graph by the asymptoticapproach of the curve to the line p " f " 10 cm.

As the object moves inside the focal point, the imagebecomes virtual and located near q " $ '. We are nowfollowing the curve in the lower left portion of Figure36.33a. As the object moves closer to the lens, the virtualimage also moves closer to the lens. As p : 0, the image dis-tance q also approaches 0. Now imagine that we bring theobject to the back side of the lens, where p * 0. The objectis now a virtual object, so it must have been formed by someother lens. For all locations of the virtual object, the image

distance is positive and less than the focal length. The finalimage is real, and its position approaches the focal point asp becomes more and more negative.

The f " $ 10 cm graph shows that a distant real objectforms an image at the focal point on the front side of thelens. As the object approaches the lens, the image remainsvirtual and moves closer to the lens. But as we continuetoward the left end of the p axis, the object becomes virtual.As the position of this virtual object approaches the focalpoint, the image recedes toward infinity. As we pass thefocal point, the image shifts from a location at positiveinfinity to one at negative infinity. Finally, as the virtualobject continues moving away from the lens, the imageis virtual, starts moving in from negative infinity, andapproaches the focal point.

Figure 36.34 Spherical aberrationcaused by a converging lens. Doesa diverging lens cause sphericalaberration?

the lens are imaged farther from the lens than rays passing through points near the edges.Figure 36.10 earlier in the chapter showed a similar situation for a spherical mirror.

Many cameras have an adjustable aperture to control light intensity and reducespherical aberration. (An aperture is an opening that controls the amount of lightpassing through the lens.) Sharper images are produced as the aperture size isreduced because with a small aperture only the central portion of the lens is exposedto the light; as a result, a greater percentage of the rays are paraxial. At the same time,however, less light passes through the lens. To compensate for this lower light intensity,a longer exposure time is used.

In the case of mirrors, spherical aberration can be minimized through the use of aparabolic reflecting surface rather than a spherical surface. Parabolic surfaces are notused often, however, because those with high-quality optics are very expensive to make.Parallel light rays incident on a parabolic surface focus at a common point, regardlessof their distance from the principal axis. Parabolic reflecting surfaces are used in manyastronomical telescopes to enhance image quality.

Chromatic Aberrations

The fact that different wavelengths of light refracted by a lens focus at different pointsgives rise to chromatic aberrations. In Chapter 35 we described how the index of refrac-tion of a material varies with wavelength. For instance, when white light passes through alens, violet rays are refracted more than red rays (Fig. 36.35). From this we see that thefocal length of a lens is greater for red light than for violet light. Other wavelengths (notshown in Fig. 36.35) have focal points intermediate between those of red and violet.

Chromatic aberration for a diverging lens also results in a shorter focal length forviolet light than for red light, but on the front side of the lens. Chromatic aberrationcan be greatly reduced by combining a converging lens made of one type of glass and adiverging lens made of another type of glass.

S E C T I O N 3 6 . 6 • The Camera 1153

VioletRed

RedViolet

F V

FR

Figure 36.35 Chromatic aberra-tion caused by a converging lens.Rays of different wavelengths focusat different points.

Quick Quiz 36.9 A curved mirrored surface can have (a) spherical aberra-tion but not chromatic aberration (b) chromatic aberration but not spherical aberra-tion (c) both spherical aberration and chromatic aberration.

36.6 The Camera

The photographic camera is a simple optical instrument whose essential features areshown in Figure 36.36. It consists of a light-tight chamber, a converging lens thatproduces a real image, and a film behind the lens to receive the image. One focusesthe camera by varying the distance between lens and film. This is accomplished with anadjustable bellows in antique cameras and with some other mechanical arrangement incontemporary models. For proper focusing—which is necessary for the formation ofsharp images—the lens-to-film distance depends on the object distance as well as onthe focal length of the lens.

The shutter, positioned behind the lens, is a mechanical device that is opened forselected time intervals, called exposure times. One can photograph moving objects byusing short exposure times or photograph dark scenes (with low light levels) by usinglong exposure times. If this adjustment were not available, it would be impossible totake stop-action photographs. For example, a rapidly moving vehicle could moveenough in the time interval during which the shutter is open to produce a blurredimage. Another major cause of blurred images is the movement of the camera whilethe shutter is open. To prevent such movement, either short exposure times or atripod should be used, even for stationary objects. Typical shutter speeds (that is,exposure times) are (1/30)s, (1/60)s, (1/125)s, and (1/250)s. For handheld cameras,

ShutterLens

Aperture

Film

Image

qp

Figure 36.36 Cross-sectional viewof a simple camera. Note that inreality, p ++ q.

the use of slower speeds can result in blurred images (due to movement), but the useof faster speeds reduces the gathered light intensity. In practice, stationary objects arenormally shot with an intermediate shutter speed of (1/60)s.

More expensive cameras have an aperture of adjustable diameter to further controlthe intensity of the light reaching the film. As noted earlier, when an aperture of smalldiameter is used, only light from the central portion of the lens reaches the film; inthis way spherical aberration is reduced.

The intensity I of the light reaching the film is proportional to the area of the lens.Because this area is proportional to the square of the diameter D, we conclude that I isalso proportional to D2. Light intensity is a measure of the rate at which energy isreceived by the film per unit area of the image. Because the area of the image is propor-tional to q2 and q # f (when p ++ f , so p can be approximated as infinite), we concludethat the intensity is also proportional to 1/f 2, and thus I , D2/f 2. The brightness of theimage formed on the film depends on the light intensity, so we see that the imagebrightness depends on both the focal length and the diameter of the lens.

The ratio f/D is called the f -number of a lens:

(36.18)

Hence, the intensity of light incident on the film varies according to the followingproportionality:

(36.19)

The f -number is often given as a description of the lens “speed.” The lower the f -number,the wider the aperture and the higher the rate at which energy from the lightexposes the film—thus, a lens with a low f -number is a “fast” lens. The conventionalnotation for an f -number is “f/” followed by the actual number. For example, “f/4”means an f -number of 4—it does not mean to divide f by 4! Extremely fast lenses, whichhave f -numbers as low as approximately f/1.2, are expensive because it is very difficult tokeep aberrations acceptably small with light rays passing through a large area of the lens.Camera lens systems (that is, combinations of lenses with adjustable apertures) are oftenmarked with multiple f -numbers, usually f/2.8, f/4, f/5.6, f/8, f/11, and f/16. Any one ofthese settings can be selected by adjusting the aperture, which changes the value of D.Increasing the setting from one f -number to the next higher value (for example, fromf/2.8 to f/4) decreases the area of the aperture by a factor of two. The lowest f -numbersetting on a camera lens corresponds to a wide-open aperture and the use of themaximum possible lens area.

Simple cameras usually have a fixed focal length and a fixed aperture size, with anf -number of about f/11. This high value for the f -number allows for a large depth offield, meaning that objects at a wide range of distances from the lens form reasonablysharp images on the film. In other words, the camera does not have to be focused.

Digital cameras are similar to the cameras we have described here except that thelight does not form an image on photographic film. The image in a digital camerais formed on a charge-coupled device (CCD), which digitizes the image, turning it intobinary code, as we discussed for sound in Section 17.5. (The CCD is described inSection 40.2.) The digital information is then stored on a memory chip for playbackon the screen of the camera, or it can be downloaded to a computer and sent to afriend or relative through the Internet.

I , 1

( f/D)2 , 1

( f -number)2

f -number ! fD


Quick Quiz 36.10 A camera can be modeled as a simple converging lensthat focuses an image on the film, acting as the screen. A camera is initially focused ona distant object. To focus the image of an object close to the camera, the lens must be(a) moved away from the film. (b) left where it is. (c) moved toward the film.

36.7 The Eye

Like a camera, a normal eye focuses light and produces a sharp image. However, themechanisms by which the eye controls the amount of light admitted and adjusts toproduce correctly focused images are far more complex, intricate, and effective thanthose in even the most sophisticated camera. In all respects, the eye is a physiologicalwonder.

Figure 36.37 shows the basic parts of the human eye. Light entering the eye passesthrough a transparent structure called the cornea (Fig. 36.38), behind which are a clearliquid (the aqueous humor), a variable aperture (the pupil, which is an opening in theiris), and the crystalline lens. Most of the refraction occurs at the outer surface of theeye, where the cornea is covered with a film of tears. Relatively little refraction occursin the crystalline lens because the aqueous humor in contact with the lens has an aver-age index of refraction close to that of the lens. The iris, which is the colored portion

S E C T I O N 3 6 . 7 • The Eye 1155

Example 36.14 Finding the Correct Exposure Time

time. If I is the light intensity reaching the film, then ina time interval -t the energy per unit area received bythe film is proportional to I -t. Comparing the two situa-tions, we require that I1 -t1 " I2 -t2, where -t1 is thecorrect exposure time for f/1.8 and -t2 is the correctexposure time for f/4. Using this result together withEquation 36.19, we find that

As the aperture size is reduced, exposure time mustincrease.

1100

s-t 2 " $ f 2 -numberf 1 -number %

2 -t 1 " $ 4

1.8 %2 $ 1

500 s% #

-t 1

( f1 -number)2 " -t 2

( f 2 -number)2

The lens of a certain 35-mm camera (where 35 mm is thewidth of the film strip) has a focal length of 55 mm and aspeed (an f -number) of f/1.8. The correct exposure time forthis speed under certain conditions is known to be (1/500) s.

(A) Determine the diameter of the lens.

Solution From Equation 36.18, we find that

(B) Calculate the correct exposure time if the f -number ischanged to f/4 under the same lighting conditions.

Solution The total light energy hitting the film is propor-tional to the product of the intensity and the exposure

31 mmD "f

f -number"

55 mm1.8

"

Pupil

Cornea

Crystallinelens

Ciliarymuscle

Retinalarteries

and veins

Retina

Vitreoushumor

IrisOpticnerve

Aqueoushumor

Figure 36.37 Important parts of the eye.Figure 36.38 Close-up photograph of thecornea of the human eye.

From

Lenn

art N

ilsso

n, in

colla

bora

tion

with

Jan

Lind

berg

, Beh

old

Man

:A

Phot

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phic

Jour

ney o

f Disc

over

y Ins

ide

the

Body

, Bos

ton,

Littl

e,Br

own

& Co

., 19

74

of the eye, is a muscular diaphragm that controls pupil size. The iris regulates theamount of light entering the eye by dilating the pupil in low-light conditions and con-tracting the pupil in high-light conditions. The f -number range of the eye is fromabout f/2.8 to f/16.

The cornea–lens system focuses light onto the back surface of the eye, the retina,which consists of millions of sensitive receptors called rods and cones. When stimulatedby light, these receptors send impulses via the optic nerve to the brain, where an imageis perceived. By this process, a distinct image of an object is observed when the imagefalls on the retina.

The eye focuses on an object by varying the shape of the pliable crystalline lensthrough an amazing process called accommodation. An important component ofaccommodation is the ciliary muscle, which is situated in a circle around the rim of thelens. Thin filaments, called zonules, run from this muscle to the edge of the lens. Whenthe eye is focused on a distant object, the ciliary muscle is relaxed, tightening thezonules that attach the muscle to the edge of the lens. The force of the zonules causesthe lens to flatten, increasing its focal length. For an object distance of infinity, thefocal length of the eye is equal to the fixed distance between lens and retina, about1.7 cm. The eye focuses on nearby objects by tensing the ciliary muscle, which relaxesthe zonules. This action allows the lens to bulge a bit, and its focal length decreases,resulting in the image being focused on the retina. All these lens adjustments takeplace so swiftly that we are not even aware of the change.

Accommodation is limited in that objects very close to the eye produce blurredimages. The near point is the closest distance for which the lens can accommodateto focus light on the retina. This distance usually increases with age and has anaverage value of 25 cm. Typically, at age 10 the near point of the eye is about 18 cm.It increases to about 25 cm at age 20, to 50 cm at age 40, and to 500 cm or greaterat age 60. The far point of the eye represents the greatest distance for which thelens of the relaxed eye can focus light on the retina. A person with normal visioncan see very distant objects and thus has a far point that can be approximated asinfinity.

Recall that the light leaving the mirror in Figure 36.8 becomes white where itcomes together but then diverges into separate colors again. Because nothing but airexists at the point where the rays cross (and hence nothing exists to cause the colors toseparate again), seeing white light as a result of a combination of colors must be a vi-sual illusion. In fact, this is the case. Only three types of color-sensitive cells are presentin the retina; they are called red, green, and blue cones because of the peaks of thecolor ranges to which they respond (Fig. 36.39). If the red and green cones are stimu-lated simultaneously (as would be the case if yellow light were shining on them), thebrain interprets what we see as yellow. If all three types of cones are stimulated by theseparate colors red, blue, and green, as in Figure 36.8, we see white. If all three types ofcones are stimulated by light that contains all colors, such as sunlight, we again seewhite light.

Color televisions take advantage of this visual illusion by having only red, green,and blue dots on the screen. With specific combinations of brightness in these three


Relativesensitivity

420 nm 534 nm 564 nm

Wavelength

Figure 36.39 Approximate color sensitivity of the three types of cones in the retina.

primary colors, our eyes can be made to see any color in the rainbow. Thus, theyellow lemon you see in a television commercial is not really yellow, it is redand green! The paper on which this page is printed is made of tiny, matted,translucent fibers that scatter light in all directions; the resultant mixture of colorsappears white to the eye. Snow, clouds, and white hair are not really white. In fact,there is no such thing as a white pigment. The appearance of these things is aconsequence of the scattering of light containing all colors, which we interpret aswhite.

Conditions of the Eye

When the eye suffers a mismatch between the focusing range of the lens–corneasystem and the length of the eye, with the result that light rays from a near object reachthe retina before they converge to form an image, as shown in Figure 36.40a, the con-dition is known as farsightedness (or hyperopia). A farsighted person can usually seefaraway objects clearly but not nearby objects. Although the near point of a normal eyeis approximately 25 cm, the near point of a farsighted person is much farther away.The refracting power in the cornea and lens is insufficient to focus the light from allbut distant objects satisfactorily. The condition can be corrected by placing a converg-ing lens in front of the eye, as shown in Figure 36.40b. The lens refracts the incomingrays more toward the principal axis before entering the eye, allowing them to convergeand focus on the retina.

A person with nearsightedness (or myopia), another mismatch condition, canfocus on nearby objects but not on faraway objects. The far point of the nearsightedeye is not infinity and may be less than 1 m. The maximum focal length of the near-sighted eye is insufficient to produce a sharp image on the retina, and rays from adistant object converge to a focus in front of the retina. They then continue pastthat point, diverging before they finally reach the retina and causing blurred vision

S E C T I O N 3 6 . 7 • The Eye 1157

Converginglens

Contractedmuscle

Imagebehind retina

Imageat retina

(a)

(b)

ObjectNearpoint

Nearpoint Object

Relaxed lens

Figure 36.40 (a) When a farsighted eye looks at an object located between the nearpoint and the eye, the image point is behind the retina, resulting in blurred vision. Theeye muscle contracts to try to bring the object into focus. (b) Farsightedness is cor-rected with a converging lens.

(Fig. 36.41a). Nearsightedness can be corrected with a diverging lens, as shown inFigure 36.41b. The lens refracts the rays away from the principal axis before they enterthe eye, allowing them to focus on the retina.

Beginning in middle age, most people lose some of their accommodation ability asthe ciliary muscle weakens and the lens hardens. Unlike farsightedness, which is amismatch between focusing power and eye length, presbyopia (literally, “old-agevision”) is due to a reduction in accommodation ability. The cornea and lens do nothave sufficient focusing power to bring nearby objects into focus on the retina. Thesymptoms are the same as those of farsightedness, and the condition can be correctedwith converging lenses.

In the eye defect known as astigmatism, light from a point source produces a lineimage on the retina. This condition arises when either the cornea or the lens or bothare not perfectly symmetric. Astigmatism can be corrected with lenses that have differ-ent curvatures in two mutually perpendicular directions.

Optometrists and ophthalmologists usually prescribe lenses1 measured in diopters:the power P of a lens in diopters equals the inverse of the focal length in meters: P " 1/f.For example, a converging lens of focal length & 20 cm has a power of & 5.0 diopters, anda diverging lens of focal length $ 40 cm has a power of $ 2.5 diopters.


(a)

(b)

Diverginglens

Relaxed lens

Image infront of retina

Imageat retina

Far point

Far point

Object

Object

1 The word lens comes from lentil, the name of an Italian legume. (You may have eaten lentil soup.)Early eyeglasses were called “glass lentils” because the biconvex shape of their lenses resembled theshape of a lentil. The first lenses for farsightedness and presbyopia appeared around 1280; concaveeyeglasses for correcting nearsightedness did not appear for more than 100 years after that.

Figure 36.41 (a) When a nearsighted eye looks at an object that lies beyond the eye’sfar point, the image is formed in front of the retina, resulting in blurred vision.(b) Nearsightedness is corrected with a diverging lens.

Quick Quiz 36.11 Two campers wish to start a fire during the day. Onecamper is nearsighted and one is farsighted. Whose glasses should be used to focus theSun’s rays onto some paper to start the fire? (a) either camper (b) the nearsightedcamper (c) the farsighted camper.

36.8 The Simple Magnifier

The simple magnifier consists of a single converging lens. As the name implies, thisdevice increases the apparent size of an object.

Suppose an object is viewed at some distance p from the eye, as illustrated inFigure 36.42. The size of the image formed at the retina depends on the angle # sub-tended by the object at the eye. As the object moves closer to the eye, # increases anda larger image is observed. However, an average normal eye cannot focus on an objectcloser than about 25 cm, the near point (Fig. 36.43a). Therefore, # is maximum at thenear point.

To further increase the apparent angular size of an object, a converging lens canbe placed in front of the eye as in Figure 36.43b, with the object located at point O,just inside the focal point of the lens. At this location, the lens forms a virtual,upright, enlarged image. We define angular magnification m as the ratio of theangle subtended by an object with a lens in use (angle # in Fig. 36.43b) to the anglesubtended by the object placed at the near point with no lens in use (angle #0 in Fig.36.43a):

(36.20)

The angular magnification is a maximum when the image is at the near point of theeye—that is, when q " $ 25 cm. The object distance corresponding to this image

m ! #

#0

S E C T I O N 3 6 . 8 • The Simple Magnifier 1159

Example 36.15 A Case of Nearsightedness

We use a negative sign for the image distance because theimage is virtual and in front of the eye. As you should havesuspected, the lens must be a diverging lens (one with a neg-ative focal length) to correct nearsightedness.

$2.5 mf "

1p

&1q

"1'

&1

$2.5 m"

1f

A particular nearsighted person is unable to see objectsclearly when they are beyond 2.5 m away (the far point ofthis particular eye). What should the focal length be in alens prescribed to correct this problem?

Solution The purpose of the lens in this instance is to“move” an object from infinity to a distance where it can beseen clearly. This is accomplished by having the lensproduce an image at the far point. From the thin lensequation, we have

θ

p

Figure 36.42 The size of theimage formed on the retinadepends on the angle # subtendedat the eye.

θθ

25 cm

(a)

h

25 cm

h'

I

h

pF O θ

(b)

θ0θ

Figure 36.43 (a) An objectplaced at the near point of theeye (p " 25 cm) subtends anangle #0 # h/25 at the eye.(b) An object placed near thefocal point of a converginglens produces a magnified imagethat subtends an angle # # h!/25 at the eye.

distance can be calculated from the thin lens equation:

where f is the focal length of the magnifier in centimeters. If we make the small-angleapproximations

(36.21)

Equation 36.20 becomes

(36.22)

Although the eye can focus on an image formed anywhere between the near pointand infinity, it is most relaxed when the image is at infinity. For the image formed bythe magnifying lens to appear at infinity, the object has to be at the focal point of thelens. In this case, Equations 36.21 become

and the magnification is

(36.23)

With a single lens, it is possible to obtain angular magnifications up to about 4 withoutserious aberrations. Magnifications up to about 20 can be achieved by using one or twoadditional lenses to correct for aberrations.

mmin "#

# 0"

25 cmf

# 0 #h

25 and # #

hf

mmax " 1 &25 cm

f

mmax "#

# 0"

h/ph/25

"25p

"25

25f/(25 & f )

tan # 0 # # 0 #h

25 and tan # # # #

hp

p "25f

25 & f

1p

&1

$25 cm"

1f


A simple magnifier, also called amagnifying glass, is used to view anenlarged image of a portion of amap.

Example 36.16 Maximum Magnification of a Lens

When the eye is relaxed, the image is at infinity. In this case,we use Equation 36.23:

2.5mmin "25 cm

f"

25 cm10 cm

"

What is the maximum magnification that is possible with alens having a focal length of 10 cm, and what is the magnifi-cation of this lens when the eye is relaxed?

Solution The maximum magnification occurs when theimage is located at the near point of the eye. Under thesecircumstances, Equation 36.22 gives

3.5" 1 &25 cm10 cm

"mmax " 1 &25 cm

f

36.9 The Compound Microscope

A simple magnifier provides only limited assistance in inspecting minute details ofan object. Greater magnification can be achieved by combining two lenses in adevice called a compound microscope, a schematic diagram of which is shown inFigure 36.44a. It consists of one lens, the objective, that has a very short focallength fo * 1 cm and a second lens, the eyepiece, that has a focal length fe of a few

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centimeters. The two lenses are separated by a distance L that is much greater thaneither fo or fe . The object, which is placed just outside the focal point of the objec-tive, forms a real, inverted image at I1, and this image is located at or close to thefocal point of the eyepiece. The eyepiece, which serves as a simple magnifier,produces at I2 a virtual, enlarged image of I1. The lateral magnification M1 of thefirst image is $ q1/p1. Note from Figure 36.44a that q1 is approximately equal to Land that the object is very close to the focal point of the objective: p1 # fo. Thus, thelateral magnification by the objective is

The angular magnification by the eyepiece for an object (corresponding to the imageat I1) placed at the focal point of the eyepiece is, from Equation 36.23,

me "25 cm

fe

Mo # $Lfo

S E C T I O N 3 6 . 9 • The Compound Microscope 1161

Objective Eyepiece

L

(a)

I2

O

Fo

fo

p1 q1

Fe I l

fe

Active Figure 36.44 (a) Diagram of a compound microscope, which consists of anobjective lens and an eyepiece lens. (b) A compound microscope. The three-objectiveturret allows the user to choose from several powers of magnification. Combinations ofeyepieces with different focal lengths and different objectives can produce a wide rangeof magnifications.

At the Active Figures linkat http://www.pse6.com, youcan adjust the focal lengths ofthe objective and eyepiecelenses to see the effect on thefinal image.

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The overall magnification of the image formed by a compound microscope is definedas the product of the lateral and angular magnifications:

(36.24)

The negative sign indicates that the image is inverted.The microscope has extended human vision to the point where we can view previ-

ously unknown details of incredibly small objects. The capabilities of this instrumenthave steadily increased with improved techniques for precision grinding of lenses. Aquestion often asked about microscopes is: “If one were extremely patient and careful,would it be possible to construct a microscope that would enable the human eye to seean atom?” The answer is no, as long as light is used to illuminate the object. Thereason is that, for an object under an optical microscope (one that uses visible light) tobe seen, the object must be at least as large as a wavelength of light. Because thediameter of any atom is many times smaller than the wavelengths of visible light, themysteries of the atom must be probed using other types of “microscopes.”

The ability to use other types of waves to “see” objects also depends on wavelength.We can illustrate this with water waves in a bathtub. Suppose you vibrate your hand inthe water until waves having a wavelength of about 15 cm are moving along thesurface. If you hold a small object, such as a toothpick, so that it lies in the path of thewaves, it does not appreciably disturb the waves; they continue along their path “oblivi-ous” to it. Now suppose you hold a larger object, such as a toy sailboat, in the path ofthe 15-cm waves. In this case, the waves are considerably disturbed by the object.Because the toothpick is smaller than the wavelength of the waves, the waves do not“see” it. (The intensity of the scattered waves is low.) Because it is about the same sizeas the wavelength of the waves, however, the boat creates a disturbance. In other words,the object acts as the source of scattered waves that appear to come from it.

Light waves behave in this same general way. The ability of an optical microscope toview an object depends on the size of the object relative to the wavelength of the lightused to observe it. Hence, we can never observe atoms with an optical microscope2

because their dimensions are small (* 0.1 nm) relative to the wavelength of the light(* 500 nm).

36.10 The Telescope

Two fundamentally different types of telescopes exist; both are designed to aid inviewing distant objects, such as the planets in our Solar System. The refracting tele-scope uses a combination of lenses to form an image, and the reflecting telescopeuses a curved mirror and a lens.

The lens combination shown in Figure 36.45a is that of a refracting telescope.Like the compound microscope, this telescope has an objective and an eyepiece. Thetwo lenses are arranged so that the objective forms a real, inverted image of a distantobject very near the focal point of the eyepiece. Because the object is essentially atinfinity, this point at which I1 forms is the focal point of the objective. The eyepiecethen forms, at I2, an enlarged, inverted image of the image at I1. In order to providethe largest possible magnification, the image distance for the eyepiece is infinite. Thismeans that the light rays exit the eyepiece lens parallel to the principal axis, andthe image of the objective lens must form at the focal point of the eyepiece. Hence,the two lenses are separated by a distance fo & fe , which corresponds to the length ofthe telescope tube.

M " Mome " $Lfo

$ 25 cmfe %


2 Single-molecule near-field optic studies are routinely performed with visible light having wave-lengths of about 500 nm. The technique uses very small apertures to produce images having resolu-tions as small as 10 nm.

The angular magnification of the telescope is given by #/#o , where #o is the anglesubtended by the object at the objective and # is the angle subtended by the finalimage at the viewer’s eye. Consider Figure 36.45a, in which the object is a very greatdistance to the left of the figure. The angle #o (to the left of the objective) subtended bythe object at the objective is the same as the angle (to the right of the objective)subtended by the first image at the objective. Thus,

where the negative sign indicates that the image is inverted.

tan #o # #o # $h!

fo

S E C T I O N 3 6 . 10 • The Telescope 1163

Objective lens

Eyepiece lens

fo

fe fe

Fe Fo

I1

Fe'

θh'

(a)

I2

oθ

oθ

At the Active Figures linkat http://www.pse6.com, youcan adjust the focal lengths ofthe objective and eyepiecelenses to see the effect on thefinal image.

Active Figure 36.45 (a) Lens arrangement in a refracting telescope, with the object atinfinity. (b) A refracting telescope.

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The angle # subtended by the final image at the eye is the same as the angle that aray coming from the tip of I1 and traveling parallel to the principal axis makes with theprincipal axis after it passes through the lens. Thus,

We have not used a negative sign in this equation because the final image is notinverted; the object creating this final image I2 is I1, and both it and I2 point in thesame direction. Hence, the angular magnification of the telescope can be expressed as

(36.25)

and we see that the angular magnification of a telescope equals the ratio of the objec-tive focal length to the eyepiece focal length. The negative sign indicates that theimage is inverted.

When we look through a telescope at such relatively nearby objects as the Moonand the planets, magnification is important. However, individual stars in our galaxy areso far away that they always appear as small points of light no matter how great themagnification. A large research telescope that is used to study very distant objects musthave a great diameter to gather as much light as possible. It is difficult and expensiveto manufacture large lenses for refracting telescopes. Another difficulty with largelenses is that their weight leads to sagging, which is an additional source of aberration.These problems can be partially overcome by replacing the objective with a concavemirror, which results in a reflecting telescope. Because light is reflected from themirror and does not pass through a lens, the mirror can have rigid supports on theback side. Such supports eliminate the problem of sagging.

Figure 36.46a shows the design for a typical reflecting telescope. Incoming lightrays pass down the barrel of the telescope and are reflected by a parabolic mirror atthe base. These rays converge toward point A in the figure, where an image would beformed. However, before this image is formed, a small, flat mirror M reflects the lighttoward an opening in the side of the tube that passes into an eyepiece. This particulardesign is said to have a Newtonian focus because Newton developed it. Figure 36.46b

m "#

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Eyepiece

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Figure 36.46 (a) A Newtonian-focus reflecting telescope. (b) A reflecting telescope.This type of telescope is shorter than that in Figure 36.45b.

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shows such a telescope. Note that in the reflecting telescope the light never passesthrough glass (except through the small eyepiece). As a result, problems associatedwith chromatic aberration are virtually eliminated. The reflecting telescope can bemade even shorter by orienting the flat mirror so that it reflects the light back towardthe objective mirror and the light enters an eyepiece in a hole in the middle of themirror.

The largest reflecting telescopes in the world are at the Keck Observatory on MaunaKea, Hawaii. The site includes two telescopes with diameters of 10 m, each containing36 hexagonally shaped, computer-controlled mirrors that work together to form a largereflecting surface. In contrast, the largest refracting telescope in the world, at the YerkesObservatory in Williams Bay, Wisconsin, has a diameter of only 1 m.

Summary 1165

The lateral magnification M of the image due to a mirror or lens is defined as theratio of the image height h! to the object height h and is equal to the negative of theratio of the image distance q to the object distance p :

(36.1, 36.2)

In the paraxial ray approximation, the object distance p and image distance q for aspherical mirror of radius R are related by the mirror equation:

(36.4, 36.6)

where f " R/2 is the focal length of the mirror.An image can be formed by refraction from a spherical surface of radius R . The

object and image distances for refraction from such a surface are related by

(36.8)

where the light is incident in the medium for which the index of refraction is n1 and isrefracted in the medium for which the index of refraction is n2.

The inverse of the focal length f of a thin lens surrounded by air is given by thelens makers’ equation:

(36.15)

Converging lenses have positive focal lengths, and diverging lenses have negativefocal lengths.

For a thin lens, and in the paraxial ray approximation, the object and imagedistances are related by the thin lens equation:

(36.16)

The ratio of the focal length of a camera lens to the diameter of the lens is calledthe f -number of the lens:

(36.18)

The intensity of light incident on the film in the camera varies according to:

(36.19)I , 1

( f/D)2 , 1

( f -number)2

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S U M M A RY

Take a practice test forthis chapter by clicking onthe Practice Test link athttp://www.pse6.com.


The maximum magnification of a single lens of focal length f used as a simplemagnifier is

(36.22)

The overall magnification of the image formed by a compound microscope is:

(36.24)

where fo and fe are the focal lengths of the objective and eyepiece lenses, respectively,and L is the distance between the lenses.

The angular magnification of a refracting telescope can be expressed as

(36.25)

where fo and fe are the focal lengths of the objective and eyepiece lenses, respectively.The angular magnification of a reflecting telescope is given by the same expressionwhere fo is the focal length of the objective mirror.

m " $fofe

M " $Lfo

$ 25 cmfe %

mmax " 1 &25 cm

f

1. What is wrong with the caption of the cartoon shown inFigure Q36.1?

5. Why does a clear stream, such as a creek, always appear tobe shallower than it actually is? By how much is its depthapparently reduced?

6. Consider the image formed by a thin converging lens. Under what conditions is the image (a) inverted, (b) upright, (c) real, (d) virtual, (e) larger than the object,and (f) smaller than the object?

7. Repeat Question 6 for a thin diverging lens.

8. Use the lens makers’ equation to verify the sign of thefocal length of each of the lenses in Figure 36.27.

9. If a solid cylinder of glass or clear plastic is placed abovethe words LEAD OXIDE and viewed from above as shown

Q U E S T I O N S

Figure Q36.1 “Most mirrors reverse left and right. This onereverses top and bottom.”

2. Consider a concave spherical mirror with a real object. Isthe image always inverted? Is the image always real? Giveconditions for your answers.

3. Repeat the preceding question for a convex spherical mirror.

4. Do the equations 1/p & 1/q " 1/f or M " $ q/p apply tothe image formed by a flat mirror? Explain your answer. Figure Q36.9

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Questions 1167

in Figure Q36.9, the LEAD appears inverted but theOXIDE does not. Explain.

10. In Figure 36.28a, assume that the blue object arrow isreplaced by one that is much taller than the lens. Howmany rays from the object will strike the lens? How manyprincipal rays can be drawn in a ray diagram?

11. A zip-lock plastic sandwich bag filled with water can act asa crude converging lens in air. If the bag is filled with airand placed under water, is the effective lens converging ordiverging?

12. Explain why a mirror cannot give rise to chromaticaberration.

Why do some automobile mirrors have printed on themthe statement “Objects in mirror are closer than theyappear”? (See Fig. P36.19.)

14. Can a converging lens be made to diverge light if it isplaced into a liquid? What If? How about a convergingmirror?

Explain why a fish in a spherical goldfish bowl appearslarger than it really is.

16. Why do some emergency vehicles have the symbolwritten on the front?

17. A lens forms an image of an object on a screen. What happens to the image if you cover the top half of the lenswith paper?

Lenses used in eyeglasses, whether converging or diverg-ing, are always designed so that the middle of the lenscurves away from the eye, like the center lenses of Figure36.27a and b. Why?

19. Which glasses in Figure Q36.19 correct nearsightednessand which correct farsightedness?

18.

AMBULANCE

15.

13.

22. In a Jules Verne novel, a piece of ice is shaped to form a mag-nifying lens to focus sunlight to start a fire. Is this possible?

23. The f -number of a camera is the focal length of the lensdivided by its aperture (or diameter). How can the f -numberof the lens be changed? How does changing this numberaffect the required exposure time?

24. A solar furnace can be constructed by using a concavemirror to reflect and focus sunlight into a furnace enclo-sure. What factors in the design of the reflecting mirrorwould guarantee very high temperatures?

25. One method for determining the position of an image,either real or virtual, is by means of parallax. If a finger orother object is placed at the position of the image, asshown in Figure Q36.25, and the finger and image areviewed simultaneously (the image is viewed through thelens if it is virtual), the finger and image have the sameparallax; that is, if they are viewed from different positions,the image will appear to move along with the finger. Usethis method to locate the image formed by a lens. Explainwhy the method works.

Figure Q36.19

Finger

Image

Figure Q36.25

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20. A child tries on either his hyperopic grandfather’s or hismyopic brother’s glasses and complains that “everythinglooks blurry.” Why do the eyes of a person wearing glassesnot look blurry? (See Figure Q36.19.)

21. Consider a spherical concave mirror, with the objectlocated to the left of the mirror beyond the focal point.Using ray diagrams, show that the image moves to the leftas the object approaches the focal point.

26. Figure Q36.26 shows a lithograph by M. C. Escher titledHand with Reflection Sphere (Self-Portrait in Spherical Mirror).Escher had this to say about the work: “The picture

Figure Q36.26

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shows a spherical mirror, resting on a left hand. But as aprint is the reverse of the original drawing on stone, itwas my right hand that you see depicted. (Being left-handed, I needed my left hand to make the drawing.)Such a globe reflection collects almost one’s wholesurroundings in one disk-shaped image. The wholeroom, four walls, the floor, and the ceiling, everything,albeit distorted, is compressed into that one small circle.Your own head, or more exactly the point between youreyes, is the absolute center. No matter how you turn ortwist yourself, you can’t get out of that central point.You are immovably the focus, the unshakable core, ofyour world.” Comment on the accuracy of Escher’sdescription.

27. You can make a corner reflector by placing three flatmirrors in the corner of a room where the ceiling meetsthe walls. Show that no matter where you are in theroom, you can see yourself reflected in the mirrors—upside down.

28. A converging lens of short focal length can take light di-verging from a small source and refract it into a beam ofparallel rays. A Fresnel lens, as shown in Figure 36.29, isused for this purpose in a lighthouse. A concave mirrorcan take light diverging from a small source and reflect itinto a beam of parallel rays. Is it possible to make a Fresnelmirror? Is this an original idea, or has it already beendone? Suggestion: Look at the walls and ceiling of an audi-torium.

Section 36.1 Images Formed by Flat Mirrors

1. Does your bathroom mirror show you older or youngerthan you actually are? Compute an order-of-magnitudeestimate for the age difference, based on data that youspecify.

2. In a church choir loft, two parallel walls are 5.30 m apart.The singers stand against the north wall. The organistfaces the south wall, sitting 0.800 m away from it. Toenable her to see the choir, a flat mirror 0.600 m wide ismounted on the south wall, straight in front of her. Whatwidth of the north wall can she see? Suggestion: Draw a top-view diagram to justify your answer.

Determine the minimum height of a vertical flat mirror inwhich a person 5!10. in height can see his or her fullimage. (A ray diagram would be helpful.)

4. Two flat mirrors have their reflecting surfaces facingeach other, with the edge of one mirror in contact withan edge of the other, so that the angle between themirrors is %. When an object is placed between themirrors, a number of images are formed. In general, ifthe angle % is such that n% " 360°, where n is an integer,the number of images formed is n $ 1. Graphically, findall the image positions for the case n " 6 when a pointobject is between the mirrors (but not on the anglebisector).

5. A person walks into a room with two flat mirrors on oppo-site walls, which produce multiple images. When theperson is located 5.00 ft from the mirror on the left walland 10.0 ft from the mirror on the right wall, find thedistance from the person to the first three images seen inthe mirror on the left.

6. A periscope (Figure P36.6) is useful for viewing objectsthat cannot be seen directly. It finds use in submarinesand in watching golf matches or parades from behind acrowd of people. Suppose that the object is a distance

3.

p1 from the upper mirror and that the two flat mirrorsare separated by a distance h. (a) What is the distance ofthe final image from the lower mirror? (b) Is the finalimage real or virtual? (c) Is it upright or inverted?(d) What is its magnification? (e) Does it appear to beleft–right reversed?

1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide

= coached solution with hints available at http://www.pse6.com = computer useful in solving problem

= paired numerical and symbolic problems

P R O B L E M S

h

Figure P36.6

Section 36.2 Images Formed by Spherical Mirrors

7. A concave spherical mirror has a radius of curvature of20.0 cm. Find the location of the image for objectdistances of (a) 40.0 cm, (b) 20.0 cm, and (c) 10.0 cm. Foreach case, state whether the image is real or virtual andupright or inverted. Find the magnification in each case.

Problems 1169

8. At an intersection of hospital hallways, a convex mirror ismounted high on a wall to help people avoid collisions.The mirror has a radius of curvature of 0.550 m. Locateand describe the image of a patient 10.0 m from themirror. Determine the magnification.

A spherical convex mirror has a radius of curvaturewith a magnitude of 40.0 cm. Determine the position ofthe virtual image and the magnification for objectdistances of (a) 30.0 cm and (b) 60.0 cm. (c) Are theimages upright or inverted?

10. A large church has a niche in one wall. On the floor planit appears as a semicircular indentation of radius 2.50 m. Aworshiper stands on the center line of the niche, 2.00 mout from its deepest point, and whispers a prayer. Where isthe sound concentrated after reflection from the back wallof the niche?

A concave mirror has a radius of curvature of 60.0 cm.Calculate the image position and magnification of an objectplaced in front of the mirror at distances of (a) 90.0 cm and(b) 20.0 cm. (c) Draw ray diagrams to obtain the imagecharacteristics in each case.

12. A concave mirror has a focal length of 40.0 cm. Determinethe object position for which the resulting image is uprightand four times the size of the object.

13. A certain Christmas tree ornament is a silver sphere havinga diameter of 8.50 cm. Determine an object location forwhich the size of the reflected image is three-fourths thesize of the object. Use a principal-ray diagram to arrive at adescription of the image.

14. (a) A concave mirror forms an inverted image four timeslarger than the object. Find the focal length of the mirror,assuming the distance between object and image is 0.600 m.(b) A convex mirror forms a virtual image half the sizeof the object. Assuming the distance between image andobject is 20.0 cm, determine the radius of curvature of themirror.

15. To fit a contact lens to a patient’s eye, a keratometer can beused to measure the curvature of the front surface of theeye, the cornea. This instrument places an illuminatedobject of known size at a known distance p from thecornea. The cornea reflects some light from the object,forming an image of the object. The magnification M ofthe image is measured by using a small viewing telescopethat allows comparison of the image formed by thecornea with a second calibrated image projected into thefield of view by a prism arrangement. Determine the ra-dius of curvature of the cornea for the case p " 30.0 cmand M " 0.013 0.

16. An object 10.0 cm tall is placed at the zero mark of ameter stick. A spherical mirror located at some point onthe meter stick creates an image of the object that isupright, 4.00 cm tall, and located at the 42.0-cm markof the meter stick. (a) Is the mirror convex or concave?(b) Where is the mirror? (c) What is the mirror’s focallength?

A spherical mirror is to be used to form, on a screen located5.00 m from the object, an image five times the size of theobject. (a) Describe the type of mirror required. (b) Whereshould the mirror be positioned relative to the object?

17.

11.

9.

18. A dedicated sports car enthusiast polishes the inside andoutside surfaces of a hubcap that is a section of a sphere.When she looks into one side of the hubcap, she sees animage of her face 30.0 cm in back of the hubcap. She thenflips the hubcap over and sees another image of her face10.0 cm in back of the hubcap. (a) How far is her facefrom the hubcap? (b) What is the radius of curvature ofthe hubcap?

19. You unconsciously estimate the distance to an objectfrom the angle it subtends in your field of view. Thisangle # in radians is related to the linear height of theobject h and to the distance d by # " h/d. Assume thatyou are driving a car and another car, 1.50 m high, is24.0 m behind you. (a) Suppose your car has a flatpassenger-side rearview mirror, 1.55 m from your eyes.How far from your eyes is the image of the car followingyou? (b) What angle does the image subtend in yourfield of view? (c) What If? Suppose instead that your carhas a convex rearview mirror with a radius of curvatureof magnitude 2.00 m (Fig. P36.19). How far from youreyes is the image of the car behind you? (d) What angledoes the image subtend at your eyes? (e) Based on itsangular size, how far away does the following car appearto be?

Figure P36.19

20. Review Problem. A ball is dropped at t " 0 from rest 3.00 mdirectly above the vertex of a concave mirror that has aradius of curvature of 1.00 m and lies in a horizontal plane.(a) Describe the motion of the ball’s image in the mirror.(b) At what time do the ball and its image coincide?


Section 36.3 Images Formed by Refraction

A cubical block of ice 50.0 cm on a side is placed on a levelfloor over a speck of dust. Find the location of the imageof the speck as viewed from above. The index of refractionof ice is 1.309.

22. A flint glass plate (n " 1.66) rests on the bottom of anaquarium tank. The plate is 8.00 cm thick (vertical dimen-sion) and is covered with a layer of water (n " 1.33)12.0 cm deep. Calculate the apparent thickness of theplate as viewed from straight above the water.

A glass sphere (n " 1.50) with a radius of 15.0 cm has atiny air bubble 5.00 cm above its center. The sphere isviewed looking down along the extended radius contain-ing the bubble. What is the apparent depth of the bubblebelow the surface of the sphere?

24. A simple model of the human eye ignores its lens entirely.Most of what the eye does to light happens at the outersurface of the transparent cornea. Assume that this surfacehas a radius of curvature of 6.00 mm, and assume that theeyeball contains just one fluid with a refractive index of1.40. Prove that a very distant object will be imaged on theretina, 21.0 mm behind the cornea. Describe the image.

25. One end of a long glass rod (n " 1.50) is formed into aconvex surface with a radius of curvature of 6.00 cm. Anobject is located in air along the axis of the rod. Find theimage positions corresponding to object distances of(a) 20.0 cm, (b) 10.0 cm, and (c) 3.00 cm from the end ofthe rod.

26. A transparent sphere of unknown composition is observedto form an image of the Sun on the surface of the sphereopposite the Sun. What is the refractive index of thesphere material?

27. A goldfish is swimming at 2.00 cm/s toward the front wallof a rectangular aquarium. What is the apparent speed ofthe fish measured by an observer looking in from outsidethe front wall of the tank? The index of refraction of wateris 1.33.

Section 36.4 Thin Lenses

28. A contact lens is made of plastic with an index of refrac-tion of 1.50. The lens has an outer radius of curvature of& 2.00 cm and an inner radius of curvature of & 2.50 cm.What is the focal length of the lens?

The left face of a biconvex lens has a radius of curva-ture of magnitude 12.0 cm, and the right face has a radiusof curvature of magnitude 18.0 cm. The index of refrac-tion of the glass is 1.44. (a) Calculate the focal length ofthe lens. (b) What If? Calculate the focal length the lenshas after is turned around to interchange the radii of cur-vature of the two faces.

30. A converging lens has a focal length of 20.0 cm. Locate theimage for object distances of (a) 40.0 cm, (b) 20.0 cm, and(c) 10.0 cm. For each case, state whether the image is realor virtual and upright or inverted. Find the magnificationin each case.

31. A thin lens has a focal length of 25.0 cm. Locate anddescribe the image when the object is placed (a) 26.0 cmand (b) 24.0 cm in front of the lens.

29.

23.

21.

32. An object located 32.0 cm in front of a lens forms animage on a screen 8.00 cm behind the lens. (a) Find thefocal length of the lens. (b) Determine the magnification.(c) Is the lens converging or diverging?

The nickel’s image in Figure P36.33 has twice thediameter of the nickel and is 2.84 cm from the lens. Deter-mine the focal length of the lens.

33.

Figure P36.33

34. A person looks at a gem with a jeweler’s loupe—aconverging lens that has a focal length of 12.5 cm. Theloupe forms a virtual image 30.0 cm from the lens.(a) Determine the magnification. Is the image uprightor inverted? (b) Construct a ray diagram for thisarrangement.

35. Suppose an object has thickness dp so that it extends fromobject distance p to p & dp. Prove that the thickness dq ofits image is given by ($ q2/p2)dp, so that the longitudinalmagnification dq/dp " $ M 2, where M is the lateralmagnification.

36. The projection lens in a certain slide projector is a singlethin lens. A slide 24.0 mm high is to be projected so thatits image fills a screen 1.80 m high. The slide-to-screendistance is 3.00 m. (a) Determine the focal length of theprojection lens. (b) How far from the slide should the lensof the projector be placed in order to form the image onthe screen?

An object is located 20.0 cm to the left of a diverging lenshaving a focal length f " $ 32.0 cm. Determine (a) thelocation and (b) the magnification of the image. (c) Construct a ray diagram for this arrangement.

38. An antelope is at a distance of 20.0 m from a converginglens of focal length 30.0 cm. The lens forms an image ofthe animal. If the antelope runs away from the lens at aspeed of 5.00 m/s, how fast does the image move? Doesthe image move toward or away from the lens?

39. In some types of optical spectroscopy, such as photolumi-nescence and Raman spectroscopy, a laser beam exits froma pupil and is focused on a sample to stimulate electro-magnetic radiation from the sample. The focusing lensusually has an antireflective coating preventing any lightloss. Assume a 100-mW laser is located 4.80 m from thelens, which has a focal length of 7.00 cm. (a) How far fromthe lens should the sample be located so that an image ofthe laser exit pupil is formed on the surface of the sample?(b) If the diameter of the laser exit pupil is 5.00 mm, what

37.

Problems 1171

is the diameter of the light spot on the sample? (c) What isthe light intensity at the spot?

40. Figure P36.40 shows a thin glass (n " 1.50) converginglens for which the radii of curvature are R1 " 15.0 cm andR2 " $ 12.0 cm. To the left of the lens is a cube having aface area of 100 cm2. The base of the cube is on the axis ofthe lens, and the right face is 20.0 cm to the left of thelens. (a) Determine the focal length of the lens. (b) Drawthe image of the square face formed by the lens. What typeof geometric figure is this? (c) Determine the area of theimage.

45. Two rays traveling parallel to the principal axis strike alarge plano-convex lens having a refractive index of 1.60(Fig. P36.45). If the convex face is spherical, a ray near theedge does not pass through the focal point (sphericalaberration occurs). Assume this face has a radius ofcurvature of 20.0 cm and the two rays are at distancesh1 " 0.500 cm and h2 " 12.0 cm from the principal axis.Find the difference -x in the positions where each crossesthe principal axis.

20.0 cm

F F

Figure P36.40

C

R ∆x

Figure P36.45

41. An object is at a distance d to the left of a flat screen. Aconverging lens with focal length f * d/4 is placedbetween object and screen. (a) Show that two lens posi-tions exist that form an image on the screen, and deter-mine how far these positions are from the object. (b) Howdo the two images differ from each other?

42. Figure 36.36 diagrams a cross section of a camera. It has asingle lens of focal length 65.0 mm, which is to form animage on the film at the back of the camera. Suppose theposition of the lens has been adjusted to focus the imageof a distant object. How far and in what direction must thelens be moved to form a sharp image of an object that is2.00 m away?

43. The South American capybara is the largest rodent onEarth; its body can be 1.20 m long. The smallest rodent isthe pygmy mouse found in Texas, with an average bodylength of 3.60 cm. Assume that a pygmy mouse is observedby looking through a lens placed 20.0 cm from the mouse.The whole image of the mouse is the size of a capybara.Then the lens is moved a certain distance along its axis,and the image of the mouse is the same size as before!How far was the lens moved?

Section 36.5 Lens Aberrations

44. The magnitudes of the radii of curvature are 32.5 cm and42.5 cm for the two faces of a biconcave lens. The glass hasindex of refraction 1.53 for violet light and 1.51 for redlight. For a very distant object, locate and describe (a) theimage formed by violet light, and (b) the image formed byred light.

Section 36.6 The Camera

46. A camera is being used with a correct exposure at f/4 anda shutter speed of (1/16) s. In order to photograph arapidly moving subject, the shutter speed is changed to(1/128) s. Find the new f -number setting needed tomaintain satisfactory exposure.

Section 36.7 The Eye

A nearsighted person cannot see objects clearly beyond25.0 cm (her far point). If she has no astigmatism andcontact lenses are prescribed for her, what power and typeof lens are required to correct her vision?

48. The accommodation limits for Nearsighted Nick’s eyes are18.0 cm and 80.0 cm. When he wears his glasses, he cansee faraway objects clearly. At what minimum distance is heable to see objects clearly?

49. A person sees clearly when he wears eyeglasses that have apower of $ 4.00 diopters and sit 2.00 cm in front of hiseyes. If the person wants to switch to contact lenses, whichare placed directly on the eyes, what lens power should beprescribed?

Section 36.8 The Simple Magnifier

Section 36.9 The Compound Microscope

Section 36.10 The Telescope

50. A lens that has a focal length of 5.00 cm is used as amagnifying glass. (a) To obtain maximum magnification,where should the object be placed? (b) What is themagnification?

47.


51. The distance between eyepiece and objective lens in acertain compound microscope is 23.0 cm. The focal lengthof the eyepiece is 2.50 cm, and that of the objectiveis 0.400 cm. What is the overall magnification of themicroscope?

52. The desired overall magnification of a compound micro-scope is 140/. The objective alone produces a lateral mag-nification of 12.0/. Determine the required focal lengthof the eyepiece.

The Yerkes refracting telescope has a 1.00-m diameterobjective lens of focal length 20.0 m. Assume it is used withan eyepiece of focal length 2.50 cm. (a) Determine themagnification of the planet Mars as seen through this tele-scope. (b) Are the Martian polar caps right side up orupside down?

54. Astronomers often take photographs with the objectivelens or mirror of a telescope alone, without an eyepiece.(a) Show that the image size h! for this telescope is givenby h! " f h/( f $ p) where h is the object size, f is the objec-tive focal length, and p is the object distance. (b) What If?Simplify the expression in part (a) for the case in whichthe object distance is much greater than objective focallength. (c) The “wingspan” of the International SpaceStation is 108.6 m, the overall width of its solar panel con-figuration. Find the width of the image formed by a tele-scope objective of focal length 4.00 m when the station isorbiting at an altitude of 407 km.

55. Galileo devised a simple terrestrial telescope that producesan upright image. It consists of a converging objective lensand a diverging eyepiece at opposite ends of the telescopetube. For distant objects, the tube length is equal to theobjective focal length minus the absolute value of theeyepiece focal length. (a) Does the user of the telescopesee a real or virtual image? (b) Where is the final image?(c) If a telescope is to be constructed with a tube of length10.0 cm and a magnification of 3.00, what are the focallengths of the objective and eyepiece?

56. A certain telescope has an objective mirror with an aper-ture diameter of 200 mm and a focal length of 2 000 mm.It captures the image of a nebula on photographic film atits prime focus with an exposure time of 1.50 min. Toproduce the same light energy per unit area on the film,what is the required exposure time to photograph thesame nebula with a smaller telescope, which has anobjective with a diameter of 60.0 mm and a focal length of900 mm?

Additional Problems

57. The distance between an object and its upright image is20.0 cm. If the magnification is 0.500, what is the focallength of the lens that is being used to form the image?

58. The distance between an object and its upright image is d.If the magnification is M, what is the focal length of thelens that is being used to form the image?

59. Your friend needs glasses with diverging lenses of focallength $ 65.0 cm for both eyes. You tell him he looks good

53.

when he doesn’t squint, but he is worried about how thickthe lenses will be. Assuming the radius of curvature of thefirst surface is R1 " 50.0 cm and the high-index plastic hasa refractive index of 1.66, (a) find the required radius ofcurvature of the second surface. (b) Assume the lens isground from a disk 4.00 cm in diameter and 0.100 cmthick at the center. Find the thickness of the plastic at theedge of the lens, measured parallel to the axis. Suggestion:Draw a large cross-sectional diagram.

60. A cylindrical rod of glass with index of refraction 1.50 isimmersed in water with index 1.33. The diameter of therod is 9.00 cm. The outer part of each end of the rod hasbeen ground away to form each end into a hemisphere ofradius 4.50 cm. The central portion of the rod withstraight sides is 75.0 cm long. An object is situated in thewater, on the axis of the rod, at a distance of 100 cm fromthe vertex of the nearer hemisphere. (a) Find the locationof the final image formed by refraction at both surfaces.(b) Is the final image real or virtual? Upright or inverted?Enlarged or diminished?

61. A zoom lens system is a combination of lenses that producesa variable magnification while maintaining fixed objectand image positions. The magnification is varied bymoving one or more lenses along the axis. While multiplelenses are used in practice to obtain high-quality images,the effect of zooming in on an object can be demonstratedwith a simple two-lens system. An object, two converginglenses, and a screen are mounted on an optical bench.The first lens, which is to the right of the object, has afocal length of 5.00 cm, and the second lens, which is tothe right of the first lens, has a focal length of 10.0 cm.The screen is to the right of the second lens. Initially, anobject is situated at a distance of 7.50 cm to the left of thefirst lens, and the image formed on the screen has a mag-nification of & 1.00. (a) Find the distance between theobject and the screen. (b) Both lenses are now movedalong their common axis, while the object and the screenmaintain fixed positions, until the image formed on thescreen has a magnification of & 3.00. Find the displace-ment of each lens from its initial position in (a). Can thelenses be displaced in more than one way?

62. The object in Figure P36.62 is midway between the lensand the mirror. The mirror’s radius of curvature is20.0 cm, and the lens has a focal length of $ 16.7 cm. Con-sidering only the light that leaves the object and travelsfirst toward the mirror, locate the final image formed bythis system. Is this image real or virtual? Is it upright orinverted? What is the overall magnification?

Lens ObjectMirror

25.0 cm

Figure P36.62

Problems 1173

63. An object placed 10.0 cm from a concave spherical mirrorproduces a real image 8.00 cm from the mirror. If the objectis moved to a new position 20.0 cm from the mirror, what isthe position of the image? Is the latter image real or virtual?

64. In many applications it is necessary to expand or todecrease the diameter of a beam of parallel rays of light.This change can be made by using a converging lens and adiverging lens in combination. Suppose you have a con-verging lens of focal length 21.0 cm and a diverging lens offocal length $ 12.0 cm. How can you arrange these lensesto increase the diameter of a beam of parallel rays? Bywhat factor will the diameter increase?

A parallel beam of light enters a glass hemisphereperpendicular to the flat face, as shown in Figure P36.65.The magnitude of the radius is 6.00 cm, and the index ofrefraction is 1.560. Determine the point at which the beamis focused. (Assume paraxial rays.)

65.

and the other is inverted. Both images are 1.50 times largerthan the object. The lens has a focal length of 10.0 cm. Thelens and mirror are separated by 40.0 cm. Determine thefocal length of the mirror. Do not assume that the figure isdrawn to scale.

The disk of the Sun subtends an angle of 0.533° at theEarth. What are the position and diameter of the solarimage formed by a concave spherical mirror with a radiusof curvature of 3.00 m?

70. Assume the intensity of sunlight is 1.00 kW/m2 at a particu-lar location. A highly reflecting concave mirror is to bepointed toward the Sun to produce a power of at least 350 Wat the image. (a) Find the required radius Ra of the circularface area of the mirror. (b) Now suppose the light intensity isto be at least 120 kW/m2 at the image. Find the requiredrelationship between Ra and the radius of curvature R of themirror. The disk of the Sun subtends an angle of 0.533° atthe Earth.

In a darkened room, a burning candle is placed 1.50 mfrom a white wall. A lens is placed between candle and wallat a location that causes a larger, inverted image to formon the wall. When the lens is moved 90.0 cm toward thewall, another image of the candle is formed. Find (a) thetwo object distances that produce the specified images and(b) the focal length of the lens. (c) Characterize thesecond image.

72. Figure P36.72 shows a thin converging lens for which theradii of curvature are R 1 " 9.00 cm and R 2 " $ 11.0 cm.The lens is in front of a concave spherical mirror withthe radius of curvature R " 8.00 cm. (a) Assume itsfocal points F1 and F2 are 5.00 cm from the center ofthe lens. Determine its index of refraction. (b) The lensand mirror are 20.0 cm apart, and an object is placed8.00 cm to the left of the lens. Determine the position ofthe final image and its magnification as seen by the eyein the figure. (c) Is the final image inverted or upright?Explain.

71.

69.

n

R

Air

I

q

Figure P36.65

66. Review problem. A spherical lightbulb of diameter 3.20 cmradiates light equally in all directions, with power 4.50 W.(a) Find the light intensity at the surface of the bulb.(b) Find the light intensity 7.20 m away from the center ofthe bulb. (c) At this 7.20-m distance a lens is set up with itsaxis pointing toward the bulb. The lens has a circular facewith a diameter 15.0 cm and has a focal length of 35.0 cm.Find the diameter of the image of the bulb. (d) Find thelight intensity at the image.

An object is placed 12.0 cm to the left of a diverging lensof focal length $ 6.00 cm. A converging lens of focallength 12.0 cm is placed a distance d to the right of the di-verging lens. Find the distance d so that the final image isat infinity. Draw a ray diagram for this case.

68. An observer to the right of the mirror–lens combinationshown in Figure P36.68 sees two real images that are thesame size and in the same location. One image is upright

67.

Object

Mirror LensImages

Figure P36.68

F2

C

F1

Figure P36.72

73. A compound microscope has an objective of focal length0.300 cm and an eyepiece of focal length 2.50 cm. Ifan object is 3.40 mm from the objective, what is themagnification? (Suggestion: Use the lens equation for theobjective.)

74. Two converging lenses having focal lengths of 10.0 cmand 20.0 cm are located 50.0 cm apart, as shown in


Figure P36.74. The final image is to be located betweenthe lenses at the position indicated. (a) How far to theleft of the first lens should the object be? (b) What is theoverall magnification? (c) Is the final image upright orinverted?

P36.76). If a strawberry is placed on the lower mirror, animage of the strawberry is formed at the small opening atthe center of the top mirror. Show that the final image isformed at that location and describe its characteristics.(Note: A very startling effect is to shine a flashlight beamon this image. Even at a glancing angle, the incoming lightbeam is seemingly reflected from the image! Do youunderstand why?)

77. An object 2.00 cm high is placed 40.0 cm to the left of aconverging lens having a focal length of 30.0 cm. Adiverging lens with a focal length of $ 20.0 cm is placed110 cm to the right of the converging lens. (a) Deter-mine the position and magnification of the final image.(b) Is the image upright or inverted? (c) What If?Repeat parts (a) and (b) for the case where the secondlens is a converging lens having a focal length of& 20.0 cm.

78. Two lenses made of kinds of glass having different refractiveindices n1 and n2 are cemented together to form what iscalled an optical doublet. Optical doublets are often used tocorrect chromatic aberrations in optical devices. The firstlens of a doublet has one flat side and one concave side ofradius of curvature R. The second lens has two convex sidesof radius of curvature R. Show that the doublet can bemodeled as a single thin lens with a focal length described by

Answers to Quick Quizzes

36.1 At C. A ray traced from the stone to the mirror and thento observer 2 looks like this:

1f

"2n2 $ n1 $ 1

R

f2 (20.0 cm)f1 (10.0 cm)

Final imageObject

p 31.0 cm50.0 cm

Figure P36.74

Strawberry

Small hole

Figure P36.76

© M

ichae

l Lev

in/O

pti-G

one

Asso

ciate

s

75. A cataract-impaired lens in an eye may be surgicallyremoved and replaced by a manufactured lens. The focallength required for the new lens is determined bythe lens-to-retina distance, which is measured by a sonar-like device, and by the requirement that the implantprovide for correct distant vision. (a) Assuming thedistance from lens to retina is 22.4 mm, calculate thepower of the implanted lens in diopters. (b) Because noaccommodation occurs and the implant allows forcorrect distant vision, a corrective lens for close workor reading must be used. Assume a reading distance of33.0 cm and calculate the power of the lens in thereading glasses.

76. A floating strawberry illusion is achieved with two para-bolic mirrors, each having a focal length 7.50 cm, facingeach other so that their centers are 7.50 cm apart (Fig.

C

2 1

36.2 False. The water spots are 2 m away from you and yourimage is 4 m away. You cannot focus your eyes on both atthe same time.

36.3 (b). A concave mirror will focus the light from a largearea of the mirror onto a small area of the paper, result-ing in a very high power input to the paper.

Answers to Quick Quizzes 1175

36.6 (a). No matter where O is, the rays refract into the airaway from the normal and form a virtual image betweenO and the surface.

36.7 (b). Because the flat surfaces of the plane have infiniteradii of curvature, Equation 36.15 indicates that thefocal length is also infinite. Parallel rays strikingthe plane focus at infinity, which means that they remainparallel after passing through the glass.

36.8 (b). If there is a curve on the front surface, the refrac-tion will differ at that surface when the mask is worn inair and water. In order for there to be no difference inrefraction (for normal incidence), the front of the maskshould be flat.

36.9 (a). Because the light reflecting from a mirror does notenter the material of the mirror, there is no opportunityfor the dispersion of the material to cause chromaticaberration.

36.10 (a). If the object is brought closer to the lens, the imagemoves farther away from the lens, behind the plane ofthe film. In order to bring the image back up to thefilm, the lens is moved toward the object and away fromthe film.

36.11 (c). The Sun’s rays must converge onto the paper. A far-sighted person wears converging lenses.

n1 < n2

I

n2

R

O

qp

n1

36.4 (b). A convex mirror always forms an image with a mag-nification less than one, so the mirror must be concave.In a concave mirror, only virtual images are upright.This particular photograph is of the Hubble SpaceTelescope primary mirror.

36.5 (d). When O is far away, the rays refract into the mater-ial of index n2 and converge to form a real image as inFigure 36.18. For certain combinations of R and n2 as Omoves very close to the refracting surface, the incidentangle of the rays increases so much that rays are nolonger refracted back toward the principal axis. This results in a virtual image as shown below:

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