Geometric Optics Part 1 PDF

? This surgeon performing microsurgery needs a sharp, magnified view o...

? This surgeon performing microsurgery needs a sharp, magnified view of the surgical site. To obtain this, she’s wearing glasses with magnify- ing lenses that must be (i) at a particular distance from her eye; (ii) at a particular distance from the object being magnified; (iii) both (i) and (ii); (iv) neither (i) nor (ii). 34 GEOMETRIC OPTICS Y LEARNING GOALS our reflection in the bathroom mirror, the view of the moon through a telescope, an insect seen through a magnifying lens—all of these are Looking forward at … examples of images. In each case the object that you’re looking at appears 34.1 How a plane mirror forms an image. to be in a different place than its actual position: Your reflection is on the other 34.2 Why concave and convex mirrors form side of the mirror, the moon appears to be much closer when seen through a images of different kinds. telescope, and an insect seen through a magnifying lens appears more distant (so 34.3 How images can be formed by a curved your eye can focus on it easily). In each case, light rays that come from a point on interface between two transparent materials. an object are deflected by reflection or refraction (or a combination of the two), 34.4 What aspects of a lens determine the type so they converge toward or appear to diverge from a point called an image point. of image that it produces. Our goal in this chapter is to see how this is done and to explore the different 34.5 What determines the field of view of a kinds of images that can be made with simple optical devices. camera lens. To understand images and image formation, all we need are the ray model of 34.6 What causes various defects in human light, the laws of reflection and refraction (Section 33.2), and some simple geom- vision, and how they can be corrected. etry and trigonometry. The key role played by geometry in our analysis explains 34.7 The principle of the simple magnifier. why we give the name geometric optics to the study of how light rays form 34.8 How microscopes and telescopes work. images. We’ll begin our analysis with one of the simplest image-forming optical devices, a plane mirror. We’ll go on to study how images are formed by curved Looking back at … mirrors, by refracting surfaces, and by thin lenses. Our results will lay the foun- 33.2 Reflection and refraction. dation for understanding many familiar optical instruments, including camera lenses, magnifiers, the human eye, microscopes, and telescopes. 34.1 REFLECTION AND REFRACTION AT A PLANE SURFACE Before discussing what is meant by an image, we first need the concept of object as it is used in optics. By an object we mean anything from which light rays radiate. This light could be emitted by the object itself if it is self-luminous, like the glowing filament of a light bulb. Alternatively, the light could be emitted by another source (such as a lamp or the sun) and then reflected from the object; 1111 1112 CHAPTER 34 Geometric Optics 34.1 Light rays radiate from a point an example is the light you see coming from the pages of this book. Figure 34.1 object P in all directions. For an observer shows light rays radiating in all directions from an object at a point P. Note that to see this object directly, there must be light rays from the object reach the observer’s left and right eyes at different angles; no obstruction between the object and the observer’s eyes. these differences are processed by the observer’s brain to infer the distance from the observer to the object. The object in Fig. 34.1 is a point object that has no physical extent. Real objects with length, width, and height are called extended objects. To start with, we’ll consider only an idealized point object, since we can always think of an P extended object as being made up of a very large number of point objects. Suppose some of the rays from the object strike a smooth, plane reflecting surface (Fig. 34.2). This could be the surface of a material with a different index of refraction, which reflects part of the incident light, or a polished metal surface that reflects almost 100% of the light that strikes it. We will always draw the 34.2 Light rays from the object at point P reflecting surface as a black line with a shaded area behind it, as in Fig. 34.2. are reflected from a plane mirror. The Bathroom mirrors have a thin sheet of glass that lies in front of and protects the reflected rays entering the eye look as reflecting surface; we’ll ignore the effects of this thin sheet. though they had come from image point P′. According to the law of reflection, all rays striking the surface are reflected at an angle from the normal equal to the angle of incidence. Since the surface is plane, the normal is in the same direction at all points on the surface, and we have specular reflection. After the rays are reflected, their directions are the same as though they had come from point P′. We call point P an object point and point P′ the corresponding image point, and we say that the reflecting sur- P P′ face forms an image of point P. An observer who can see only the rays reflected from the surface, and who doesn’t know that he’s seeing a reflection, thinks that the rays originate from the image point P′. The image point is therefore Image point: Object point: apparent a convenient way to describe the directions of the various reflected rays, just source of rays source of as the object point P describes the directions of the rays arriving at the surface reflected rays before reflection. Plane mirror If the surface in Fig. 34.2 were not smooth, the reflection would be diffuse. Rays reflected from different parts of the surface would go in uncorrelated direc- tions (see Fig. 33.6b), and there would be no definite image point P′ from which all reflected rays seem to emanate. You can’t see your reflection in a tarnished piece of metal because its surface is rough; polishing the metal smoothes the surface so that specular reflection occurs and a reflected image becomes visible. 34.3 Light rays from the object at point P A plane refracting surface also forms an image (Fig. 34.3). Rays coming from are refracted at the plane interface. The point P are refracted at the interface between two optical materials. When the refracted rays entering the eye look as though they had come from image point P′. angles of incidence are small, the final directions of the rays after refraction are the same as though they had come from an image point P′ as shown. In Section 33.2 When na 7 nb, P′ is closer to the surface than P; we described how this effect makes underwater objects appear closer to the surface for na 6 nb, the reverse is true. than they really are (see Fig. 33.9). na 7 nb nb In both Figs. 34.2 and 34.3 the rays do not actually pass through the image point P′. Indeed, if the mirror in Fig. 34.2 is opaque, there is no light at all on its right side. If the outgoing rays don’t actually pass through the image point, we call the image a virtual image. Later we will see cases in which the outgoing rays really do pass through an image point, and we will call the resulting image a P P′ real image. The images that are formed on a projection screen, on the electronic sensor in a camera, and on the retina of your eye are real images. Object point: Image point: apparent Image Formation by a Plane Mirror source of rays source of refracted rays Let’s concentrate for now on images produced by reflection; we’ll return to refraction later in the chapter. Figure 34.4 shows how to find the precise location of the virtual image P′ that a plane mirror forms of an object at P. The diagram shows two rays diverging from an object point P at a distance s to the left of a plane mirror. We call s the object distance. The ray PV is perpendicular to the mirror surface, and it returns along its original path.  34.1 Reflection and Refraction at a Plane Surface 1113 The ray PB makes an angle u with PV. It strikes the mirror at an angle of inci- 34.4 Construction for determining the dence u and is reflected at an equal angle with the normal. When we extend the location of the image formed by a plane mirror. The image point P′ is as far behind two reflected rays backward, they intersect at point P′, at a distance s′ behind the mirror as the object point P is in front the mirror. We call s′ the image distance. The line between P and P′ is perpen- of it. dicular to the mirror. The two triangles PVB and P′VB are congruent, so P and P′ are at equal distances from the mirror, and s and s′ have equal magnitudes. After reflection, all rays The image point P′ is located exactly opposite the object point P as far behind originating at P diverge from P′. Because the the mirror as the object point is from the front of the mirror. u B rays do not actually pass We can repeat the construction of Fig. 34.4 for each ray diverging from P. through P′, the image u The directions of all the outgoing reflected rays are the same as though they had h is virtual. u originated at point P′, confirming that P′ is the image of P. No matter where the P u P′ observer is located, she will always see the image at the point P′. V s s′ Object distance Image distance Sign Rules Triangles PVB and P′VB are congruent, so 0 s 0 = 0 s′ 0. Before we go further, let’s introduce some general sign rules. These may seem unnecessarily complicated for the simple case of an image formed by a plane mirror, but we want to state the rules in a form that will be applicable to all the situations we will encounter later. These will include image formation by a plane or spherical reflecting or refracting surface, or by a pair of refracting sur- 34.5 For both of these situations, the object distance s is positive (rule 1) and faces forming a lens. Here are the rules: the image distance s′ is negative (rule 2). 1. Sign rule for the object distance: When the object is on the same side of (a) Plane mirror the reflecting or refracting surface as the incoming light, object distance s Out is positive; otherwise, it is negative. goin g 2. Sign rule for the image distance: When the image is on the same side of g min the reflecting or refracting surface as the outgoing light, image distance s′ Inco P P′ is positive; otherwise, it is negative. s 7 0 s′ 6 0 3. Sign rule for the radius of curvature of a spherical surface: When the center of curvature C is on the same side as the outgoing light, the radius In both of these specific cases: of curvature is positive; otherwise, it is negative. Object distance s is Image distance s′ is positive because the negative because the Figure 34.5 illustrates rules 1 and 2 for two different situations. For a mirror object is on the same image is NOT on the the incoming and outgoing sides are always the same; for example, in Figs. 34.2, side as the incoming same side as the light. outgoing light. 34.4, and 34.5a they are both on the left side. For the refracting surfaces in Figs. 34.3 and 34.5b the incoming and outgoing sides are on the left and right (b) Plane refracting interface sides, respectively, of the interface between the two materials. (Note that other s 7 0 textbooks may use different rules.) s′ 6 0 P P′ In Figs. 34.4 and 34.5a the object distance s is positive because the object Incomin g point P is on the incoming side (the left side) of the reflecting surface. The Ou image distance s′ is negative because the image point P′ is not on the outgoing tgo ing side (the left side) of the surface. The object and image distances s and s′ are related by s = -s′ (plane mirror) (34.1) For a plane reflecting or refracting surface, the radius of curvature is infinite 34.6 Construction for determining the and not a particularly interesting or useful quantity; in these cases we really don’t height of an image formed by reflection at need sign rule 3. But this rule will be of great importance when we study image a plane reflecting surface. formation by curved reflecting and refracting surfaces later in the chapter. For a plane mirror, PQV and P′Q′V are con- gruent, so y = y′ and the object and image are the same size (the lateral magnification is 1). Image of an Extended Object: Plane Mirror Object Image Next we consider an extended object with finite size. For simplicity we often Q V′ Q′ consider an object that has only one dimension, like a slender arrow, oriented u y y′ parallel to the reflecting surface; an example is the arrow PQ in Fig. 34.6. The u u P u V P′ distance from the head to the tail of an arrow oriented in this way is called its height. In Fig. 34.6 the height is y. The image formed by such an extended object s s′ is an extended image; to each point on the object, there corresponds a point on 1114 CHAPTER 34 Geometric Optics the image. Two of the rays from Q are shown; all the rays from Q appear to diverge from its image point Q′ after reflection. The image of the arrow is the line P′Q′, with height y′. Other points of the object PQ have image points between P′ and Q′. The triangles PQV and P′Q′V are congruent, so the object PQ and image P′Q′ have the same size and orientation, and y = y′. 34.7 The image formed by a plane mirror The ratio of image height to object height, y′>y, in any image-forming situa- is virtual, erect, and reversed. It is the tion is called the lateral magnification m; that is, same size as the object. An image made by a plane mirror is reversed Lateral magnification y′ Image height back to front: the image thumb P′R′ and object in an image-forming situation m = (34.2) thumb PR point in opposite directions (toward y Object height each other). Q′ S′ For a plane mirror y = y′, so the lateral magnification m is unity. When you look R′ at yourself in a plane mirror, your image is the same size as the real you. P′ In Fig. 34.6 the image arrow points in the same direction as the object arrow; Q we say that the image is erect. In this case, y and y′ have the same sign, and S Image the lateral magnification m is positive. The image formed by a plane mirror R is always erect, so y and y′ have both the same magnitude and the same sign; P from Eq. (34.2) the lateral magnification of a plane mirror is always m = +1. Later we will encounter situations in which the image is inverted; that is, the Object image arrow points in the direction opposite to that of the object arrow. For an inverted image, y and y′ have opposite signs, and the lateral magnification m is negative. The object in Fig. 34.6 has only one dimension. Figure 34.7 shows a three- 34.8 The image formed by a plane mirror dimensional object and its three-dimensional virtual image formed by a plane is reversed; the image of a right hand is a left hand, and so on. (The hand is resting mirror. The object and image are related in the same way as a left hand and a on a horizontal mirror.) Are images of the right hand. letters I, H, and T reversed? CAUTION Reflections in a plane mirror At this point, you may be asking, “Why does a plane mirror reverse images left and right but not top and bottom?” This question is quite misleading! As Fig. 34.7 shows, the up-down image P′Q′ and the left-right image P′S′ are parallel to their objects and are not reversed at all. Only the front-back image P′R′ is reversed relative to PR. Hence it’s most correct to say that a plane mirror reverses back to front. When an object and its image are related in this way, the image is said to be reversed; this means that only the front-back dimension is reversed. ❙ The reversed image of a three-dimensional object formed by a plane mirror is the same size as the object in all its dimensions. When the transverse dimensions of object and image are in the same direction, the image is erect. Thus a plane mirror always forms an erect but reversed image (Fig. 34.8). 34.9 Images P 1= and P 2= are formed by a An important property of all images formed by reflecting or refracting sur- single reflection of each ray from the faces is that an image formed by one surface or optical device can serve as the object at P. Image P 3= , located by treating object for a second surface or device. Figure 34.9 shows a simple example. either of the other images as an object, is formed by a double reflection of each ray. Mirror 1 forms an image P 1= of the object point P, and mirror 2 forms another image P 2= , each in the way we have just discussed. But in addition, the image P 1= Image of object P Image of image P1′ formed by mirror 1 formed by mirror 2 formed by mirror 1 serves as an object for mirror 2, which then forms an image of this object at point P 3= as shown. Similarly, mirror 1 uses the image P 2= formed P1′ P3′ by mirror 2 as an object and forms an image of it. We leave it to you to show that Mirror 1 this image point is also at P 3=. The idea that an image formed by one device can act as the object for a second device is of great importance. We’ll use it later in P P2′ this chapter to locate the image formed by two successive curved-surface refrac- tions in a lens. We’ll also use it to understand image formation by combinations Image of object P formed by mirror 2 of lenses, as in a microscope or a refracting telescope. Mirror 2 TEST YOUR UNDERSTANDING OF SECTION 34.1 If you walk directly toward a plane mirror at a speed v, at what speed does your image approach you? (i) Slower than v; (ii) v; (iii) faster than v but slower than 2v; (iv) 2v; (v) faster than 2v. ❙  34.2 Reflection at a Spherical Surface 1115 34.2 REFLECTION AT A SPHERICAL SURFACE A plane mirror produces an image that is the same size as the object. But there are many applications for mirrors in which the image and object must be of dif- ferent sizes. A magnifying mirror used when applying makeup gives an image that is larger than the object, and surveillance mirrors (used in stores to help spot shoplifters) give an image that is smaller than the object. There are also applications of mirrors in which a real image is desired, so light rays do indeed pass through the image point P′. A plane mirror by itself cannot perform any of 34.10 (a) A concave spherical mirror these tasks. Instead, curved mirrors are used. forms a real image of a point object P on the mirror’s optic axis. (b) The eye sees some of the outgoing rays and perceives Image of a Point Object: Spherical Mirror them as having come from P′. We’ll consider the special (and easily analyzed) case of image formation by a (a) Construction for finding the position P′ of an image formed by a concave spherical mirror spherical mirror. Figure 34.10a shows a spherical mirror with radius of curva- ture R, with its concave side facing the incident light. The center of curvature For a spherical mirror, B of the surface (the center of the sphere of which the surface is a part) is at C, and a + b = 2f. the vertex of the mirror (the center of the mirror surface) is at V. The line CV is Point u object u h R called the optic axis. Point P is an object point that lies on the optic axis; for the a f b moment, we assume that the distance from P to V is greater than R. C P′ V P Ray PV, passing through C, strikes the mirror normally and is reflected back Center of Optic axis d on itself. Ray PB, at an angle a with the axis, strikes the mirror at B, where the curvature Vertex angles of incidence and reflection are u. The reflected ray intersects the axis at point P′. We will show shortly that all rays from P intersect the axis at the s and s′ are both positive. s′ same point P′, as in Fig. 34.10b, provided that the angle a is small. Point P′ is s therefore the image of object point P. Unlike the reflected rays in Fig. 34.1, the reflected rays in Fig. 34.10b actually do intersect at point P′, then diverge from (b) The paraxial approximation, which holds P′ as if they had originated at this point. Thus P′ is a real image. for rays with small a To see the usefulness of having a real image, suppose that the mirror is in a darkened room in which the only source of light is a self-luminous object at P. If you place a small piece of photographic film at P′, all the rays of light coming from point P that reflect off the mirror will strike the same point P′ on the film; V P P′ when developed, the film will show a single bright spot, representing a sharply focused image of the object at point P. This principle is at the heart of most astronomical telescopes, which use large concave mirrors to make photographs of celestial objects. With a plane mirror like that in Fig. 34.2, the light rays never actually pass through the image point, and the image can’t be recorded on film. All rays from P that have a small angle a pass Real images are essential for photography. through P′, forming a real image. Let’s now locate the real image point P′ in Fig. 34.10a and prove that all rays from P intersect at P′ (provided that their angle with the optic axis is small). The object distance, measured from the vertex V, is s; the image distance, also 34.11 The sign rule for the radius of a measured from V, is s′. The signs of s, s′, and the radius of curvature R are deter- spherical mirror. mined by the sign rules given in Section 34.1. The object point P is on the same side as the incident light, so according to sign rule 1, s is positive. The image point P′ is on the same side as the reflected light, so according to sign rule 2, the image distance s′ is also positive. The center of curvature C is on the same side Center of curvature on R 7 0 same side as outgoing as the reflected light, so according to sign rule 3, R, too, is positive; R is always P C light: R is positive. positive when reflection occurs at the concave side of a surface (Fig. 34.11). Outgoing We now use the following theorem from plane geometry: An exterior angle of a triangle equals the sum of the two opposite interior angles. Applying this theorem to triangles PBC and P′BC in Fig. 34.10a, we have f = a + u b = f + u R 6 0 C P Center of curvature not on Eliminating u between these equations gives same side as outgoing Outgoing light: R is negative. a + b = 2f (34.3) 1116 CHAPTER 34 Geometric Optics 34.12 (a), (b) Soon after the Hubble We may now compute the image distance s′. Let h represent the height of Space Telescope (HST) was placed in orbit point B above the optic axis, and let d represent the short distance from V to the in 1990, it was discovered that the concave primary mirror (also called the objective foot of this vertical line. We now write expressions for the tangents of a, b, and mirror) was too shallow by about 50 1 the f, remembering that s, s′, and R are all positive quantities: width of a human hair, leading to spherical aberration of the star’s image. (c) After h h h corrective optics were installed in 1993, tan a = tan b = tan f = s - d s′ - d R - d the effects of spherical aberration were almost completely eliminated. These trigonometric equations cannot be solved as simply as the corresponding (a) The 2.4-m-diameter primary mirror of the algebraic equations for a plane mirror. However, if the angle a is small, the Hubble Space Telescope angles b and f are also small. The tangent of an angle that is much less than one radian is nearly equal to the angle itself (measured in radians), so we can replace tan a by a, and so on, in the equations above. Also, if a is small, we can ignore the distance d compared with s′, s, and R. So for small angles we have the following approximate relationships: h h h a = b = f = s s′ R Substituting these into Eq. (34.3) and dividing by h, we obtain a general relation- ship among s, s′, and R: 1 1 2 + = (object–image relationship, spherical mirror) (34.4) s s′ R (b) A star seen with the original mirror This equation does not contain the angle a. Hence all rays from P that make sufficiently small angles with the axis intersect at P′ after they are reflected; this verifies our earlier assertion. Such rays, nearly parallel to the axis and close to it, are called paraxial rays. (The term paraxial approximation is often used for the approximations we have just described.) Since all such reflected light rays converge on the image point, a concave mirror is also called a converging mirror. Be sure you understand that Eq. (34.4), as well as many similar relationships that we will derive later in this chapter and the next, is only approximately cor- rect. It results from a calculation containing approximations, and it is valid only for paraxial rays. If we increase the angle a that a ray makes with the optic axis, the point P′ where the ray intersects the optic axis moves some- what closer to the vertex than for a paraxial ray. As a result, a spherical mirror, (c) The same star with corrective optics unlike a plane mirror, does not form a precise point image of a point object; the image is “smeared out.” This property of a spherical mirror is called spherical aberration. When the primary mirror of the Hubble Space Telescope (Fig. 34.12a) was manufactured, tiny errors were made in its shape that led to an unacceptable amount of spherical aberration (Fig. 34.12b). The performance of the telescope improved dramatically after the installation of corrective optics (Fig. 34.12c). If the radius of curvature becomes infinite 1R = q2, the mirror becomes plane, and Eq. (34.4) reduces to Eq. (34.1) for a plane reflecting surface. Focal Point and Focal Length When the object point P is very far from the spherical mirror 1s = q2, the incoming rays are parallel. (The star shown in Fig. 34.12c is an example of such a distant object.) From Eq. (34.4) the image distance s′ in this case is given by 1 1 2 R + = s′ = q s′ R 2  34.2 Reflection at a Spherical Surface 1117 The situation is shown in Fig. 34.13a. The beam of incident parallel rays con- 34.13 The focal point and focal length of verges, after reflection from the mirror, to a point F at a distance R>2 from the a concave mirror. vertex of the mirror. The point F at which the incident parallel rays converge is (a) All parallel rays incident on a spherical called the focal point; we say that these rays are brought to a focus. The distance mirror reflect through the focal point. from the vertex to the focal point, denoted by f, is called the focal length. We R (positive) see that f is related to the radius of curvature R by Focal R point f = (focal length of a spherical mirror) (34.5) 2 C F Figure 34.13b shows the opposite situation. Now the object is placed at the focal point F, so the object distance is s = f = R>2. The image distance s′ is again given by Eq. (34.4): 2 1 2 1 s = q + = = 0 s′ = q R s′ R s′ R Focal length s′ = = f 2 With the object at the focal point, the reflected rays in Fig. 34.13b are parallel to the optic axis; they meet only at a point infinitely far from the mirror, so the (b) Rays diverging from the focal point reflect image is at infinity. to form parallel outgoing rays. Thus the focal point F of a spherical mirror has the properties that (1) any incoming ray parallel to the optic axis is reflected through the focal point and R (positive) (2) any incoming ray that passes through the focal point is reflected parallel to the optic axis. For spherical mirrors these statements are true only for paraxial rays. For parabolic mirrors these statements are exactly true. Spherical or para- bolic mirrors are used in flashlights and headlights to form the light from the C F bulb into a parallel beam. Some solar-power plants use an array of plane mirrors Focal point to simulate an approximately spherical concave mirror; sunlight is collected by the mirrors and directed to the focal point, where a steam boiler is placed. (The concepts of focal point and focal length also apply to lenses, as we’ll see in Section 34.4.) s′ = q We will usually express the relationship between object and image distances R s = = f for a mirror, Eq. (34.4), in terms of the focal length f : 2 1 1 1 + = (object–image relationship, spherical mirror) (34.6) s s′ f Image of an Extended Object: Spherical Mirror Now suppose we have an object with finite size, represented by the arrow PQ in Fig. 34.14, perpendicular to the optic axis CV. The image of P formed by par- axial rays is at P′. The object distance for point Q is very nearly equal to that for point P, so the image P′Q′ is nearly straight and perpendicular to the axis. The beige and blue triangles are similar, so the 34.14 Construction for determining the position, lateral magnification is m = y′>y = - s′>s. The orientation, and height of an image formed by a Q negative value of m means that the image is inverted. concave spherical mirror. y Object P′ u V P C y′ Image u Q′ s′ R s 1118 CHAPTER 34 Geometric Optics Application Satellite Television Note that the object and image arrows have different sizes, y and y′, respectively, Dishes A dish antenna used to receive and that they have opposite orientation. In Eq. (34.2) we defined the lateral satellite TV broadcasts is actually a concave magnification m as the ratio of image size y′ to object size y: parabolic mirror. The waves are of much lower frequency than visible light (1.2 to y′ 1.8 * 1010 Hz compared with 4.0 to m = 7.9 * 1014 Hz), but the laws of reflection are y the same. The transmitter in orbit is so far away that the arriving waves have essentially Because triangles PVQ and P′VQ′ in Fig. 34.14 are similar, we also have the parallel rays, as in Fig. 34.13a. The dish reflects relationship y>s = -y′>s′. The negative sign is needed because object and image the waves and brings them to a focus at a feed are on opposite sides of the optic axis; if y is positive, y′ is negative. Therefore horn, from which they are “piped” to a decoder that extracts the signal. y′ s′ m = = - (lateral magnification, spherical mirror) (34.7) y s Dish If m is positive, the image is erect in comparison to the object; if m is negative, the image is inverted relative to the object, as in Fig. 34.14. For a plane mirror, s = -s′, so y′ = y and m = +1; since m is positive, the image is erect, and since 0 m 0 = 1, the image is the same size as the object. CAUTION Lateral magnification can be less than 1 Although the ratio of image size to Feed horn object size is called the lateral magnification, the image formed by a mirror or lens may Dish = segment of a curved mirror. Only a be larger than, smaller than, or the same size as the object. If it is smaller, then the lateral segment away from the optic axis is used so that magnification is less than unity in absolute value: 0 m 0 6 1. The image formed by an the feed horn does not block incoming waves. astronomical telescope mirror or a camera lens is usually much smaller than the object. For example, the image of the bright star shown in Fig. 34.12c is just a few millimeters across, while the star itself is hundreds of thousands of kilometers in diameter. ❙ Rays from satellite In our discussion of concave mirrors we have so far considered only objects that lie outside or at the focal point, so that the object distance s is greater than or equal to the (positive) focal length f. In this case the image point is on the same side of the mirror as the outgoing rays, and the image is real and inverted. If an object is inside the focal point of a concave mirror, so that s 6 f, the resulting Optic axis of image is virtual (that is, the image point is on the opposite side of the mirror dish from the object), erect, and larger than the object. Mirrors used when you apply Feed horn at makeup (referred to at the beginning of this section) are concave mirrors; in use, focal point the distance from the face to the mirror is less than the focal length, and you see an enlarged, erect image. You can prove these statements about concave mirrors by applying Eqs. (34.6) and (34.7). We’ll also be able to verify these results later in this section, after we’ve learned some graphical methods for relating the posi- tions and sizes of the object and the image. SOLUTION EXAMPLE 34.1 IMAGE FORMATION BY A CONCAVE MIRROR I A concave mirror forms an image, on a wall 3.00 m in front of the mirror, of a headlamp filament 10.0 cm in front of the mirror. (a) What are the radius of curvature and focal length of the mirror? (b) What is the lateral magnification? What is the image height if the object height is 5.00 mm? 34.15 Our sketch for this problem. SOLUTION IDENTIFY and SET UP: Figure 34.15 shows our sketch. Our target variables are the radius of curvature R, focal length f, lateral magnification m, and image height y′. We are given the distances from the mirror to the object (s) and from the mirror to the image (s′). We solve Eq. (34.4) for R, and then use Eq. (34.5) to find f. Equation (34.7) yields both m and y′.  34.2 Reflection at a Spherical Surface 1119 EXECUTE: (a) Both the object and the image are on the concave Because m is negative, the image is inverted. The height of the (reflective) side of the mirror, so both s and s′ are positive; we image is 30.0 times the height of the object, or 130.0215.00 mm2 = have s = 10.0 cm and s′ = 300 cm. We solve Eq. (34.4) for R: 150 mm. 1 1 2 EVALUATE: Our sketch indicates that the image is inverted; our + = 10.0 cm 300 cm R calculations agree. Note that the object (at s = 10.0 cm) is just -1 outside the focal point 1f = 9.7 cm2. This is very similar to 1 1 R = 2a + b = 19.4 cm what is done in automobile headlights. With the filament close 10.0 cm 300 cm to the focal point, the concave mirror produces a beam of nearly The focal length of the mirror is f = R>2 = 9.7 cm. parallel rays. (b) From Eq. (34.7) the lateral magnification is s′ 300 cm m = - = - = - 30.0 s 10.0 cm SOLUTION CONCEPTUAL EXAMPLE 34.2 IMAGE FORMATION BY A CONCAVE MIRROR II In Example 34.1, suppose that the lower half of the mirror’s the mirror surface is made nonreflective (or is removed altogether), reflecting surface is covered with nonreflective soot. What effect rays from the remaining reflective surface still form an image of will this have on the image of the filament? every part of the object. Reducing the reflecting area reduces the light energy reach- SOLUTION ing the image point, however: The image becomes dimmer. If It would be natural to guess that the image would now show only the area is reduced by one-half, the image will be one-half as half of the filament. But in fact the image will still show the entire bright. Conversely, increasing the reflective area makes the image filament. You can see why by examining Fig. 34.10b. Light rays brighter. To make reasonably bright images of faint stars, astro- coming from any object point P are reflected from all parts of the nomical telescopes use mirrors that are up to several meters in mirror and converge on the corresponding image point P′. If part of diameter (see Fig. 34.12a). Convex Mirrors In Fig. 34.16a the convex side of a spherical mirror faces the incident light. The center of curvature is on the side opposite to the outgoing rays; according to sign rule 3 in Section 34.1, R is negative (see Fig. 34.11). Ray PB is reflected, with the angles of incidence and reflection both equal to u. The reflected ray, projected back- ward, intersects the axis at P′. As with a concave mirror, all rays from P that are reflected by the mirror diverge from the same point P′, provided that the angle a is small. Therefore P′ is the image of P. The object distance s is positive, the image distance s′ is negative, and the radius of curvature R is negative for a convex mirror. Figure 34.16b shows two rays diverging from the head of the arrow PQ and the virtual image P′Q′ of this arrow. The same procedure that we used for a con- cave mirror can be used to show that for a convex mirror, the expressions for the object–image relationship and the lateral magnification are 1 1 2 y′ s′ + = and m = = - s s′ R y s 34.16 Image formation by a convex mirror. (a) Construction for finding the position of an image formed (b) Construction for finding the magnification of an image by a convex mirror formed by a convex mirror Norma Q As with a concave spherical mirror, l u B V′ y′ s′ u R is negative. Q′ m = y = - s y h u R y′ a V b f u u P P' C P u V P′ C s is positive; s s′ s′ is negative. s s′ 1120 CHAPTER 34 Geometric Optics 34.17 The focal point and focal length of (a) Paraxial rays incident on a convex spherical (b) Rays aimed at the virtual focal point a convex mirror. mirror diverge from a virtual focal point. are parallel to the axis after reflection. R (negative) R (negative) F C F C Virtual focal point R R s′ = = f s = = f DATA SPEAKS s = q 2 s′ = q 2 Image Formation by Mirrors When students were given a problem involving image formation by mirrors, These expressions are exactly the same as Eqs. (34.4) and (34.7) for a concave more than 59% gave an incorrect mirror. Thus when we use our sign rules consistently, Eqs. (34.4) and (34.7) are response. Common errors: valid for both concave and convex mirrors. Not using the law of reflection properly. When R is negative (convex mirror), incoming rays that are parallel to the For a mirror (plane or curved), the inci- optic axis are not reflected through the focal point F. Instead, they diverge as dent and reflected rays make the same angle with the normal to the mirror. though they had come from the point F at a distance f behind the mirror, as The reflected ray originates at the point shown in Fig. 34.17a. In this case, f is the focal length, and F is called a virtual where the incident ray hits the mirror. focal point. The corresponding image distance s′ is negative, so both f and R are Confusion about lateral magnification. negative, and Eq. (34.5), f = R>2, holds for convex as well as concave mirrors. The lateral magnification m depends In Fig. 34.17b the incoming rays are converging as though they would meet at the on only the ratio of image distance s′ virtual focal point F, and they are reflected parallel to the optic axis. to object distance s. If s′ and s have In summary, Eqs. (34.4) through (34.7), the basic relationships for image for- different values but have the same ratio in two situations, then the value of m is mation by a spherical mirror, are valid for both concave and convex mirrors, the same. provided that we use the sign rules consistently. SOLUTION EXAMPLE 34.3 SANTA’S IMAGE PROBLEM Santa checks himself for soot, using his reflection in a silvered SOLUTION Christmas tree ornament 0.750 m away (Fig. 34.18a). The diame- ter of the ornament is 7.20 cm. Standard reference texts state that IDENTIFY and SET UP: Figure 34.18b shows the situation. Santa he is a “right jolly old elf,” so we estimate his height to be 1.6 m. is the object, and the surface of the ornament closest to him acts Where and how tall is the image of Santa formed by the ornament? as a convex mirror. The relationships among object distance, Is it erect or inverted? image distance, focal length, and magnification are the same as 34.18 (a) The ornament forms a (a) (b) virtual, reduced, erect image of Santa. (b) Our sketch of two of the rays forming the image.  34.2 Reflection at a Spherical Surface 1121 for concave mirrors, provided we use the sign rules consistently. virtual. The image is about halfway between the front surface of The radius of curvature and the focal length of a convex mirror the ornament and its center. are negative. The object distance is s = 0.750 m = 75.0 cm, and From Eq. (34.7), the lateral magnification and the image height Santa’s height is y = 1.6 m. We solve Eq. (34.6) to find the image are distance s′, and then use Eq. (34.7) to find the lateral magnifica- tion m and the image height y′. The sign of m tells us whether the y′ s′ -1.76 cm m = = - = - = 0.0234 image is erect or inverted. y s 75.0 cm EXECUTE: The radius of the mirror (half the diameter) is y′ = my = 10.0234211.6 m2 = 3.8 * 10-2 m = 3.8 cm R = - 17.20 cm2>2 = - 3.60 cm, and the focal length is f = R>2 = - 1.80 cm. From Eq. (34.6), EVALUATE: Our sketch indicates that the image is erect so both m 1 1 1 1 1 and y′ are positive; our calculations agree. When the object dis- = - = - s′ f s - 1.80 cm 75.0 cm tance s is positive, a convex mirror always forms an erect, virtual, reduced, reversed image. For this reason, convex mirrors are used s′ = - 1.76 cm at blind intersections, for surveillance in stores, and as wide-angle Because s′ is negative, the image is behind the mirror—that is, rear-view mirrors for cars and trucks. (Many such mirrors read on the side opposite to the outgoing light (Fig. 34.18b)—and it is “Objects in mirror are closer than they appear.”) Graphical Methods for Mirrors In Examples 34.1 and 34.3, we used Eqs. (34.6) and (34.7) to find the position and size of the image formed by a mirror. We can also determine the properties of the image by a simple graphical method. This method consists of finding the point of intersection of a few particular rays that diverge from a point of the object (such as point Q in Fig. 34.19) and are reflected by the mirror. Then (ignoring aberrations) all rays from this object point that strike the mirror will intersect at the same point. For this construction we always choose an object point that is not on the optic axis. Four rays that we can usually draw easily are shown in Fig. 34.19. These are called principal rays. 1. A ray parallel to the axis, after reflection, passes through the focal point F of a concave mirror or appears to come from the (virtual) focal point of a convex mirror. 2. A ray through (or proceeding toward) the focal point F is reflected parallel to the axis. 3. A ray along the radius through or away from the center of curvature C intersects the surface normally and is reflected back along its original path. 4. A ray to the vertex V is reflected forming equal angles with the optic axis. 34.19 The graphical method of locating an image formed by a spherical mirror. The colors of the rays are for identification only; they do not refer to specific colors of light. (a) Principal rays for concave mirror (b) Principal rays for convex mirror 1 Q Q 1 1 3 4 4 2 Q′ 2 2 3 C P′ F 4 V P P V P′ F C 2 4 Q′ 4 3 1 1 Ray parallel to axis reflects through focal point. 1 Reflected parallel ray appears to come from focal point. 2 Ray through focal point reflects parallel to axis. 2 Ray toward focal point reflects parallel to axis. 3 Ray through center of curvature intersects the surface normally 3 As with concave mirror: Ray radial to center of curvature intersects and reflects along its original path. the surface normally and reflects along its original path. 4 Ray to vertex reflects symmetrically around optic axis. 4 As with concave mirror: Ray to vertex reflects symmetrically around optic axis. 1122 CHAPTER 34 Geometric Optics Once we have found the position of the image point by means of the intersection of any two of these principal rays 11, 2, 3, 42, we can draw the path of any other ray from the object point to the same image point. CAUTION Principal rays are not the only rays Although we’ve emphasized the principal rays, in fact any ray from the object that strikes the mirror will pass through the image point (for a real image) or appear to originate from the image point (for a virtual image). Usually, you need to draw only the principal rays in order to locate the image. ❙ PROBLEM-SOLVING STRATEGY 34.1 IMAGE FORMATION BY MIRRORS IDENTIFY the relevant concepts: Problems involving image for- image point, as in Fig. 34.19b. We recommend drawing the mation by mirrors can be solved in two ways: using principal-ray extensions with broken lines. diagrams and using equations. A successful problem solution uses 4. Measure the resulting diagram to obtain the magnitudes of the both approaches. target variables. 5. Solve for the target variables by using Eq. (34.6), 1>s + 1>s′ = SET UP the problem: Identify the target variables. One of them 1>f , and the lateral magnification equation, Eq. (34.7), as is likely to be the focal length, the object distance, or the image appropriate. Apply the sign rules given in Section 34.1 to object distance, with the other two quantities given. and image distances, radii of curvature, and object and image EXECUTE the solution as follows: heights. 1. Draw a large, clear principal-ray diagram if you have enough 6. Use the sign rules to interpret the results that you deduced information. from your ray diagram and calculations. Note that the same 2. Orient your diagram so that incoming rays go from left to right. sign rules (given in Section 34.1) work for all four cases in this Draw only the principal rays; color-code them as in Fig. 34.19. chapter: reflection and refraction from plane and spherical If possible, use graph paper or quadrille-ruled paper. Use a surfaces. ruler and measure distances carefully! A freehand sketch will not give good results. EVALUATE your answer: Check that the results of your calcula- 3. If your principal rays don’t converge at a real image point, you tions agree with your ray-diagram results for image position, may have to extend them straight backward to locate a virtual image size, and whether the image is real or virtual. SOLUTION EXAMPLE 34.4 CONCAVE MIRROR WITH VARIOUS OBJECT DISTANCES A concave mirror has a radius of curvature with absolute value F does not strike the mirror. In this case the outgoing rays are 20 cm. Find graphically the image of an object in the form of an parallel, corresponding to an infinite image distance. In case (d), arrow perpendicular to the axis of the mirror at object distances of the outgoing rays diverge; they have been extended backward to (a) 30 cm, (b) 20 cm, (c) 10 cm, and (d) 5 cm. Check the construc- the virtual image point Q′, from which they appear to diverge. tion by computing the size and lateral magnification of each image. Case (d) illustrates the general observation that an object placed inside the focal point of a concave mirror produces a virtual image. Measurements of the figures, with appropriate scaling, give SOLUTION the following approximate image distances: (a) 15 cm; (b) 20 cm; IDENTIFY and SET UP: We must use graphical methods and cal- (c) q or - q (because the outgoing rays are parallel and do not culations to analyze the image made by a mirror. The mirror is converge at any finite distance); (d) -10 cm. To compute these concave, so its radius of curvature is R = +20 cm and its focal distances, we solve Eq. (34.6) for s′ and insert f = 10 cm : length is f = R>2 = + 10 cm. Our target variables are the image 1 1 1 distances s′ and lateral magnifications m corresponding to four (a) + = s′ = 15 cm 30 cm s′ 10 cm cases with successively smaller object distances s. In each case we solve Eq. (34.6) for s′ and use m = - s′>s to find m. 1 1 1 (b) + = s′ = 20 cm 20 cm s′ 10 cm EXECUTE: Figure 34.20 shows the principal-ray diagrams for the four cases. Study each of these diagrams carefully and confirm 1 1 1 (c) + = s′ = q 1or - q2 that each numbered ray is drawn in accordance with the rules given 10 cm s′ 10 cm earlier (under “Graphical Methods for Mirrors”). Several points 1 1 1 are worth noting. First, in case (b) the object and image distances (d) + = s′ = -10 cm 5 cm s′ 10 cm are equal. Ray 3 cannot be drawn in this case because a ray from Q through the center of curvature C does not strike the mirror. The signs of s′ tell us that the image is real in cases (a) and (b) and In case (c), ray 2 cannot be drawn because a ray from Q through virtual in case (d).  34.3 Refraction at a Spherical Surface 1123 The lateral magnifications measured from the figures are The signs of m tell us that the image is inverted in cases (a) and approximately (a) - 12 ; (b) - 1; (c) q or - q; (d) +2. From (b) and erect in case (d). Eq. (34.7), 15 cm 1 EVALUATE: Notice the trend of the results in the four cases. When (a) m = - = - 30 cm 2 the object is far from the mirror, as in Fig. 34.20a, the image is 20 cm smaller than the object, inverted, and real. As the object distance s (b) m = - = -1 decreases, the image moves farther from the mirror and gets larger 20 cm (Fig. 34.20b). When the object is at the focal point, the image is at q cm (c) m = - = - q 1or + q2 infinity (Fig. 34.20c). When the object is inside the focal point, the 10 cm image becomes larger than the object, erect, and virtual (Fig. 34.20d). -10 cm You can confirm these conclusions by looking at objects reflected (d) m = - = +2 5 cm in the concave bowl of a shiny metal spoon. 34.20 Using principal-ray diagrams to locate the image P′Q′ made by a concave mirror. (a) Construction for s = 30 cm (b) Construction for s = 20 cm All principal rays can be drawn. Ray 3 (from Q through C ) cannot The image is inverted. be drawn because it does not strike Q the mirror. Q 1 1 4 4 32 The image 2 C P′ F is inverted. P P′ V V P 3 C F 2 4 Q′ 2 Q′ 1 4 s and s′ are equal. 1 s′ s = s′ = 20 cm s = 30 cm (c) Construction for s = 10 cm (d) Construction for s = 5 cm Ray 2 (from Q through F ) cannot be drawn because it does not Q′ strike the mirror. 2 Q 3 Q 2 4 1 3 1 C C 4 F F V P 3 P V P′ 3 Parallel The image is 1 outgoing rays 1 virtual and correspond to an 4 erect. 4 infinite image distance. s = 10 cm s′ s′ = q s = 5 cm TEST YOUR UNDERSTANDING OF SECTION 34.2 A cosmetics mirror is designed so that your reflection appears right-side up and enlarged. (a) Is the mirror concave or convex? (b) To see an enlarged image, what should be the distance from the mirror (of focal length f ) to your face? (i) 0 f 0 ; (ii) less than 0 f 0 ; (iii) greater than 0 f 0. ❙ 34.3 REFRACTION AT A SPHERICAL SURFACE As we mentioned in Section 34.1, images can be formed by refraction as well as by reflection. To begin with, let’s consider refraction at a spherical surface—that is, at a spherical interface between two optical materials with different indexes of refraction. This analysis is directly applicable to some real optical systems, such as the human eye. It also provides a stepping-stone for the analysis of lenses, which usually have two spherical (or nearly spherical) surfaces. 1124 CHAPTER 34 Geometric Optics 34.21 Construction for finding the position na 6 nb nb of the image point P′ of a point object P B formed by refraction at a spherical surface. ua s, s′, and R are positive. The materials to the left and right of the interface have indexes of refraction na and nb , h R ub respectively. In the case shown here, P a V f b P′ na 6 nb. d C s s′ Image of a Point Object: Spherical Refracting Surface In Fig. 34.21 a spherical surface with radius R forms an interface between two materials with different indexes of refraction na and nb. The surface forms an image P′ of an object point P; we want to find how the object and image dis- tances (s and s′) are related. We will use the same sign rules that we used for spherical mirrors. The center of curvature C is on the outgoing side of the sur- face, so R is positive. Ray PV strikes the vertex V and is perpendicular to the sur- face (that is, to the plane that is tangent to the surface at the point of incidence V). It passes into the second material without deviation. Ray PB, making an angle a with the axis, is incident at an angle ua with the normal and is refracted at an angle ub. These rays intersect at P′, a distance s′ to the right of the vertex. The fig- ure is drawn for the case na 6 nb. Both the object and image distances are positive. We are going to prove that if the angle a is small, all rays from P intersect at the same point P′, so P′ is the real image of P. We use much the same approach as we did for spherical mirrors in Section 34.2. We again use the theorem that an exterior angle of a triangle equals the sum of the two opposite interior angles; applying this to the triangles PBC and P′BC gives ua = a + f f = b + ub (34.8) From the law of refraction, na sin ua = nb sin ub Also, the tangents of a, b, and f are h h h tan a = tan b = tan f = (34.9) s + d s′ - d R - d For paraxial rays, ua and ub are both small in comparison to a radian, and we may approximate both the sine and tangent of either of these angles by the angle itself (measured in radians). The law of refraction then gives na ua = nb ub Combining this with the first of Eqs. (34.8), we obtain na ub = 1a + f2 nb When we substitute this into the second of Eqs. (34.8), we get na a + nb b = 1nb - na2f (34.10) Now we use the approximations tan a = a, and so on, in Eqs. (34.9) and also ignore the small distance d; those equations then become h h h a = b = f = s s′ R  34.3 Refraction at a Spherical Surface 1125 na 6 nb nb 34.22 Construction for determining the Q height of an image formed by refraction at s and s′ are positive. a spherical surface. In the case shown here, na 6 nb. y ua C P′ P ub y′ V Q′ s s′ Finally, we substitute these into Eq. (34.10) and divide out the common factor h. We obtain na nb nb - na (object–image relationship, + = (34.11) s s′ R spherical refracting surface) This equation does not contain the angle a, so the image distance is the same for all paraxial rays emanating from P; this proves that P′ is the image of P. To obtain the lateral magnification m for this situation, we use the construction in Fig. 34.22. We draw two rays from point Q, one through the center of curva- ture C and the other incident at the vertex V. From the triangles PQV and P′Q′V, y -y′ tan ua = tan ub = s s′ and from the law of refraction, na sin ua = nb sin ub For small angles, tan ua = sin ua tan ub = sin ub so finally na y nb y′ = - or s s′ y′ na s′ (lateral magnification, m = = - (34.12) y nb s spherical refracting surface) Equations (34.11) and (34.12) apply to both convex and concave refracting sur- faces, provided that you use the sign rules consistently. It doesn’t matter whether nb is greater or less than na. To verify these statements, construct diagrams like Figs. 34.21 and 34.22 for the following three cases: (i) R 7 0 and na 7 nb , 34.23 Light rays refract as they pass (ii) R 6 0 and na 6 nb , and (iii) R 6 0 and na 7 nb. Then in each case, use through the curved surfaces of these water your diagram to again derive Eqs. (34.11) and (34.12). droplets. Here’s a final note on the sign rule for the radius of curvature R of a surface. For the convex reflecting surface in Fig. 34.16, we considered R negative, but the convex refracting surface in Fig. 34.21 has a positive value of R. This may seem inconsistent, but it isn’t. The rule is that R is positive if the center of curvature C is on the outgoing side of the surface and negative if C is on the other side. For the convex reflecting surface in Fig. 34.16, R is negative because point C is to the right of the surface but outgoing rays are to the left. For the convex refracting surface in Fig. 34.21, R is positive because both C and the outgoing rays are to the right of the surface. Refraction at a curved surface is one reason gardeners avoid watering plants at midday. As sunlight enters a water drop resting on a leaf (Fig. 34.23), the light rays are refracted toward each other as in Figs. 34.21 and 34.22. The sunlight that strikes the leaf is therefore more concentrated and able to cause damage. 1126 CHAPTER 34 Geometric Optics An important special case of a spherical refracting surface is a plane surface between two optical materials. This corresponds to setting R = q in Eq. (34.11). In this case, na nb + = 0 (plane refracting surface) (34.13) s s′ To find the lateral magnification m for this case, we combine this equation with the general relationship, Eq. (34.12), obtaining the simple result m = 1 That is, the image formed by a plane refracting surface always has the same lateral size as the object, and it is always erect. An example of image formation by a plane refracting surface is the appear- ance of a partly submerged drinking straw or canoe paddle. When viewed from some angles, the submerged part appears to be only about three-quarters of its actual distance below the surface. (We commented on the appearance of a sub- merged object in Section 33.2; see Fig. 33.9.) SOLUTION EXAMPLE 34.5 IMAGE FORMATION BY REFRACTION I A cylindrical glass rod (Fig. 34.24) has index of refraction 1.52. It 34.24 The glass rod in air forms a real image. is surrounded by air. One end is ground to a hemispherical surface with radius R = 2.00 cm. A small object is placed on the axis of na = 1.00 (air) nb = 1.52 the rod, 8.00 cm to the left of the vertex. Find (a) the image dis- P C P′ tance and (b) the lateral magnification. R = 2.00 cm SOLUTION s = 8.00 cm s′ IDENTIFY and SET UP: This problem uses the ideas of refraction at a curved surface. Our target variables are the image distance s′ and the lateral magnification m. Here material a is air 1na = 1.002 and material b is the glass of which the rod is made 1nb = 1.522. (b) From Eq. (34.12), We are given s = 8.00 cm. The center of curvature of the spheri- 11.002111.3 cm2 na s′ cal surface is on the outgoing side of the surface, so the radius is m = - = - = - 0.929 positive: R = + 2.00 cm. We solve Eq. (34.11) for s′, and we use nb s 11.52218.00 cm2 Eq. (34.12) to find m. EVALUATE: Because the image distance s′ is positive, the image is EXECUTE: (a) From Eq. (34.11), formed 11.3 cm to the right of the vertex (on the outgoing side), as Fig. 34.24 shows. The value of m tells us that the image is some- 1.00 1.52 1.52 - 1.00 + = what smaller than the object and that it is inverted. If the object is 8.00 cm s′ +2.00 cm an arrow 1.000 mm high, pointing upward, the image is an arrow s′ = + 11.3 cm 0.929 mm high, pointing downward.