Light: Reflection and Refraction · 838 CHAPTER 32 Light: Reflection and Refraction. When light is incident upon a rough surface, even microscopically rough such as this page, it

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Reflection from still water, as from a glass mirror, can be analyzed using the ray model of light.

Is this picture right side up? How can you tell? What are the clues? Notice the people and position of the Sun. Ray diagrams, which we will learn to draw in this Chapter, can provide the answer. See Example 32-3.

In this first Chapter on light and optics, we use the ray model of light to understand the formation of images by mirrors, both plane and curved (spherical). We also begin our study of refraction—how light rays bend when they go from one medium to another—which prepares us for our study in the next Chapter of lenses, which are the crucial part of so many optical instruments.

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Light: Reflection and Refraction

CHAPTER-OPENING QUESTION—Guess now!A 2.0-m-tall person is standing 2.0 m from a flat vertical mirror staring at her image. What minimum height must the mirror have if the person is to see her entire body, from the top of her head to her feet?

(a) 0.50 m.(b) 1.0 m.(c) 1.5 m.(d) 2.0 m.(e) 2.5 m.

The sense of sight is extremely important to us, for it provides us with a large part of our information about the world. How do we see? What is the something called light that enters our eyes and causes the sensation of sight? How does light behave so that we can see everything that we do?

We saw in Chapter 31 that light can be considered a form of electromagnetic radiation. We now examine the subject of light in detail in the next four Chapters.

We see an object in one of two ways: (1) the object may be a source of light, such as a lightbulb, a flame, or a star, in which case we see the light emitted directly from the source; or, more commonly, (2) we see an object by light reflected from it. In the latter case, the light may have originated from the Sun, artificial lights, or a campfire. An understanding of how objects emit light was not achieved until the 1920s, and will be discussed in Chapter 37. How light is reflected from objects was understood earlier, and will be discussed here, in Section 32-2.

CONTENTS32-1 The Ray Model of Light32-2 Reflection; Image Formation

by a Plane Mirror32-3 Formation of Images by

Spherical Mirrors32-4 Index of Refraction32-5 Refraction: Snell’s Law32-6 Visible Spectrum and

Dispersion32-7 Total Internal Reflection; Fiber

Optics*32-8 Refraction at a Spherical

Surface

837

This bundle

FIGURE 32-1 Light rays come from each single point on an object. A small bundle of rays leaving one point is shown entering a person’s eye.

32—1 The Ray Model of LightA great deal of evidence suggests that light travels in straight lines under a wide variety of circumstances. For example, a source of light like the Sun casts distinct shadows, and the light from a laser pointer appears to be a straight line. In fact, we infer the positions of objects in our environment by assuming that light moves from the object to our eyes in straight-line paths. Our orientation to the physical world is based on this assumption.

This reasonable assumption is the basis of the ray model of light. This model assumes that light travels in straight-line paths called light rays. Actually, a ray is an idealization; it is meant to represent an extremely narrow beam of light. When we see an object, according to the ray model, light reaches our eyes from each point on the object. Although light rays leave each point in many different directions, normally only a small bundle of these rays can enter an observer’s eye, as shown in Fig. 32-1. If the person’s head moves to one side, a different bundle of rays will enter the eye from each point.

We saw in Chapter 31 that light can be considered as an electromagnetic wave. Although the ray model of light does not deal with this aspect of light (we discuss the wave nature of light in Chapters 34 and 35), the ray model has been very successful in describing many aspects of light such as reflection, refraction, and the formation of images by mirrors and lenses.1 Because these explanations involve straight-line rays at various angles, this subject is referred to as geometric optics.

32—2 Reflection; Image Formation by a Plane Mirror

When light strikes the surface of an object, some of the light is reflected. The rest can be absorbed by the object (and transformed to thermal energy) or, if the object is transparent like glass or water, part can be transmitted through. For a very smooth shiny object such as a silvered mirror, over 95% of the light may be reflected.

When a narrow beam of light strikes a flat surface (Fig. 32-2), we define the angle of incidence, 6h to be the angle an incident ray makes with the normal (perpendicular) to the surface, and the angle of reflection, 0r , to be the angle the reflected ray makes with the normal. It is found that the incident and reflected rays lie in the same plane with the normal to the surface, and that

the angle of reflection equals the angle of incidence, 0r = 0\.

This is the law of reflection, and it is depicted in Fig. 32-2. It was known to the ancient Greeks, and you can confirm it yourself by shining a narrow flashlight beam or a laser pointer at a mirror in a darkened room.

FIGURE 32-2 Law of reflection: (a) Shows a 3-D view of an incident ray being reflected at the top of a flat surface; (b) shows a side or “end-on” view, which we will usually use because of its clarity.

Incident light ray

Normal to surface

III

Analp n f I A n rrl*=» n f

Normal to surface

Reflected light ray

fIn ignoring the wave properties of light we must be careful that when the light rays pass by objects or through apertures, these must be large compared to the wavelength of the light (so the wave phenomena of interference and diffraction, as discussed in Chapter 15, can be ignored), and we ignore what happens to the light at the edges of objects until we get to Chapters 34 and 35.

838 CHAPTER 32 Light: Reflection and Refraction

When light is incident upon a rough surface, even microscopically rough such as this page, it is reflected in many directions, as shown in Fig. 32-3. This is called diffuse reflection. The law of reflection still holds, however, at each small section of the surface. Because of diffuse reflection in all directions, an ordinary object can be seen at many different angles by the light reflected from it. When you move your head to the side, different reflected rays reach your eye from each point on the object (such as this page), Fig. 32-4a. Let us compare diffuse reflection to reflection from a mirror, which is known as specular reflection. (“Speculum” is Latin for mirror.) When a narrow beam of light shines on a mirror, the light will not reach your eye unless your eye is positioned at just the right place where the law of reflection is satisfied, as shown in Fig. 32-4b. This is what gives rise to the special image-forming properties of mirrors.

FIGURE 32-3 Diffuse reflection from a rough surface.

Eye at bothThe eye here does not see _ , reflected light The eye here

does see

FIGURE 32-4 A narrow beam of light shines on (a) white paper, and (b) a mirror. In part(a), you can see with your eye the white light reflected at various positions because of diffuse reflection. But in part (b), you see the reflected light only when your eye is placed correctly (0r = 6[); mirror reflection is also known as specular reflection. (Galileo, using similar arguments, showed that the Moon must have a rough surface rather than a highly polished surface like a mirror, as some people thought.)

EXAMPLE 32-1 Reflection from flat mirrors. Two flat mirrors are perpendicular to each other. An incoming beam of light makes an angle of 15° with the first mirror as shown in Fig. 32-5a. What angle will the outgoing beam make with the second mirror?APPROACH We sketch the path of the beam as it reflects off the two mirrors, and draw the two normals to the mirrors for the two reflections. We use geometry and the law of reflection to find the various angles.

FIGURE 32-5 Example 32-1.

SOLUTION In Fig. 32-5b, + 15° = 90°, so Qx = 75°; by the law of reflection02 = 0i = 75° too. The two normals to the two mirrors are perpendicular to each other, so 02 + 03 + 90° = 180° as for any triangle. Thus03 = 180° - 90° - 75° = 15°. By the law of reflection, 04 = 03 = 15°, so05 = 75° is the angle the reflected ray makes with the second mirror surface. NOTE The outgoing ray is parallel to the incoming ray. Red reflectors on bicycles and cars use this principle.

When you look straight into a mirror, you see what appears to be yourself as well as various objects around and behind you, Fig. 32-6. Your face and the other objects look as if they are in front of you, beyond the mirror. But what you see in the mirror is an image of the objects, including yourself, that are in front of the mirror.

FIGURE 32-6 When you look in a mirror, you see an image of yourself and objects around you. You don’t see yourself as others see you, because left and right appear reversed in the image.

SECTION 32-2 Reflection; Image Formation by a Plane Mirror 839

A “plane” mirror is one with a smooth flat reflecting surface. Figure 32-7 shows how an image is formed by a plane mirror according to the ray model. We are viewing the mirror, on edge, in the diagram of Fig. 32-7, and the rays are shown reflecting from the front surface. (Good mirrors are generally made by putting a highly reflective metallic coating on one surface of a very flat piece of glass.) Rays from two different points on an object (the bottle on the left in Fig. 32-7) are shown: two rays are shown leaving from a point on the top of the bottle, and two more from a point on the bottom. Rays leave each point on the object going in many directions, but only those that enclose the bundle of rays that enter the eye from each of the two points are shown. Each set of diverging rays that reflect from the mirror and enter the eye appear to come from a single point (called the image point) behind the mirror, as shown by the dashed lines. That is, our eyes and brain interpret any rays that enter an eye as having traveled straight- line paths. The point from which each bundle of rays seems to come is one point on the image. For each point on the object, there is a corresponding image point.

Plane mirror

FIGURE 3 2 -7 Formation of a virtual image by a plane mirror.

© - P H Y S I C S A P P L I E DHow tall a mirror do you need to

see a reflection o f your entire self?

840 CHAPTER 32

Let us concentrate on the two rays that leave point A on the object in Fig. 32-7, and strike the mirror at points B and B'. We use geometry for the rays at B. The angles ADB and CDB are right angles; and because of the law of reflection, dx = 0r at point B. Therefore, angles ABD and CBD are also equal. The two triangles ABD and CBD are thus congruent, and the length AD = CD. That is, the image appears as far behind the mirror as the object is in front. The image distance, dx (perpendicular distance from mirror to image, Fig. 32-7), equals the object distance, d0 (perpendicular distance from object to mirror). From the geometry, we can also see that the height of the image is the same as that of the object.

The light rays do not actually pass through the image location itself in Fig. 32-7. (Note where the red lines are dashed to show they are our projections, not rays.) The image would not appear on paper or film placed at the location of the image. Therefore, it is called a virtual image. This is to distinguish it from a real image in which the light does pass through the image and which therefore could appear on film or in an electronic sensor, and even on a white sheet of paper or screen placed at the position of the image. Our eyes can see both real and virtual images, as long as the diverging rays enter our pupils. We will see that curved mirrors and lenses can form real images, as well as virtual. A movie projector lens, for example, produces a real image that is visible on the screen.

EXAMPLE 32-2 How tall must a full-length mirror be? A woman 1.60 m tall stands in front of a vertical plane mirror. What is the minimum height of the mirror, and how close must its lower edge be to the floor, if she is to be able to see her whole body? Assume her eyes are 10 cm below the top of her head.APPROACH For her to see her whole body, light rays from the top of her head and from the bottom of her foot must reflect from the mirror and enter her eye: see Fig. 32-8. We don’t show two rays diverging from each point as we did in Fig. 32-7, where we wanted to find where the image is. Now that we know the image is the same distance behind a plane mirror as the object is in front, we only need to show one ray leaving point G (top of head) and one ray leaving point A (her toe), and then use simple geometry.

FIGURE 32-8 Seeing oneself in a mirror. Example 32-2.

A D C

SOLUTION First consider the ray that leaves her foot at A, reflects at B, and enters the eye at E. The mirror needs to extend no lower than B. The angle of reflection equals the angle of incidence, so the height BD is half of the height AE. Because AE = 1.60 m - 0.10 m = 1.50 m, then BD = 0.75 m. Similarly, if the woman is to see the top of her head, the top edge of the mirror only needs to reach point F, which is 5 cm below the top of her head (half of GE = 10 cm). Thus, DF = 1.55 m, and the mirror needs to have a vertical height of only (1.55 m - 0.75 m) = 0.80 m. The mirror’s bottom edge must be 0.75 m above the floor.NOTE We see that a mirror, if positioned well, need be only half as tall as a person for that person to see all of himself or herself.

I EXERCISE A Does the result of Example 32-2 depend on your distance from the mirror? (Try it.) EXERCISE B Return to the Chapter-Opening Question, page 837, and answer it again now. Try to explain why you may have answered differently the first time.EXERCISE C Suppose you are standing about 3 m in front of a mirror in a hair salon. You can see yourself from your head to your waist, where the end of the mirror cuts off the rest of your image. If you walk closer to the mirror (a) you will not be able to see any more of your image; (b) you will be able to see more of your image, below your waist;(c) you will see less of your image, with the cutoff rising to be above your waist.

CONCEPTUAL EXAMPLE 32-3~l Is the photo upside down? Close examination of the photograph on the first page of this Chapter reveals that in the top portion, the image of the Sun is seen clearly, whereas in the lower portion, the image of the Sun is partially blocked by the tree branches. Show why the reflection is not the same as the real scene by drawing a sketch of this situation, showing the Sun, the camera, the branch, and two rays going from the Sun to the camera (one direct and one reflected). Is the photograph right side up?RESPONSE We need to draw two diagrams, one assuming the photo on p. 837 is right side up, and another assuming it is upside down. Figure 32-9 is drawn assuming the photo is upside down. In this case, the Sun blocked by the tree would be the direct view, and the full view of the Sun the reflection: the ray which reflects off the water and into the camera travels at an angle below the branch, whereas the ray that travels directly to the camera passes through the branches. This works. Try to draw a diagram assuming the photo is right side up (thus assuming that the image of the Sun in the reflection is higher above the horizon than it is as viewed directly). It won’t work. The photo on p. 837 is upside down.

Also, what about the people in the photo? Try to draw a diagram showing why they don’t appear in the reflection. [Hint: Assume they are not sitting on the edge of poolside, but back from the edge a bit.] Then try to draw a diagram of the reverse (i.e., assume the photo is right side up so the people are visible only in the reflection). Reflected images are not perfect replicas when different planes (distances) are involved.


SECTION 32-2 841

FIGURE 32-10 Mirrors with convex and concave spherical surfaces. Note that 0r = 6[ for each ray.

3 2 —3 Formation of Images by Spherical Mirrors

Reflecting surfaces do not have to be flat. The m ost com m on curved m irrors are spherical, which m eans they form a section of a sphere. A spherical m irror is called convex if the reflection takes place on the ou ter surface of the spherical shape so tha t the center of the m irror surface bulges ou t tow ard the viewer (Fig. 3 2 -10a). A m irror is called concave if the reflecting surface is on the inner surface of the sphere so that the m irror surface sinks away from the viewer (like a “cave”), Fig. 32-10b. Concave m irrors are used as shaving o r cosm etic m irrors (Fig. 32-11 a) because they magnify, and convex m irrors are som etim es used on cars and trucks (rearview m irrors) and in shops (to watch for thieves), because they take in a wide field of view (Fig. 3 2 - l lb ) .

FIGURE 32-11 (a) A concave cosmetic mirror gives a magnified image, (b) A convex mirror in a store reduces image size and so includes a wide field of view.

(a) (b)

Focal Point and Focal LengthTo see how spherical m irrors form images, we first consider an object that is very far from a concave mirror. For a distant object, as shown in Fig. 32-12, the rays from each point on the object that strike the m irror will be nearly parallel. For an object infinitely fa r away (the Sun and stars approach this), the rays w ould be precisely parallel.

essentially parallel.FIGURE 32-12 If the object’s distance is large compared to the size of the mirror (or lens), the rays are nearly parallel.They are parallel for an object at infinity (oo).

(b)


Now consider such parallel rays falling on a concave mirror as in Fig. 32-13. The law of reflection holds for each of these rays at the point each strikes the mirror. As can be seen, they are not all brought to a single point. In order to form a sharp image, the rays must come to a point. Thus a spherical mirror will not make as sharp an image as a plane mirror will. However, as we show below, if the mirror is small compared to its radius of curvature, so that a reflected ray makes only a small angle with the incident ray (20 in Fig. 32-14), then the rays will cross each other at very nearly a single point, or focus. In the case shown in Fig. 32-14, the incoming rays are parallel to the principal axis, which is defined as the straight line perpendicular to the curved surface at its center (line CA in Fig. 32-14). The point F, where incident parallel rays come to a focus after reflection, is called the focal point of the mirror. The distance between F and the center of the mirror, length FA, is called the focal length, / , of the mirror. The focal point is also the image point for an object infinitely far away along the principal axis. The image of the Sun, for example, would be at F.

FIGURE 32-13 Parallel rays striking a concave spherical mirror do not intersect (or focus) at precisely a single point. (This “defect” is referred to as “spherical aberration.”)

FIGURE 32-14 Rays parallel to the principal axis of a concave spherical mirror come to a focus at F, the focal point, as long as the mirror is small in width as compared to its radius of curvature, r, so that the rays are “paraxial”— that is, make only small angles with the horizontal axis.

Now we will show, for a mirror whose reflecting surface is small compared to its radius of curvature, that the rays very nearly meet at a common point, F, and we will also calculate the focal length / . In this approximation, we consider only rays that make a small angle with the principal axis; such rays are called paraxial rays, and their angles are exaggerated in Fig. 32-14 to make the labels clear. First we consider a ray that strikes the mirror at B in Fig. 32-14. The point C is the center of curvature of the mirror (the center of the sphere of which the mirror is a part). So the dashed line CB is equal to r, the radius of curvature, and CB is normal to the mirror’s surface at B. The incoming ray that hits the mirror at B makes an angle 6 with this normal, and hence the reflected ray, BF, also makes an angle 0 with the normal (law of reflection). Note that angle BCF is also 0 as shown. The triangle CBF is isosceles because two of its angles are equal. Thus we have length CF = BF. We assume the mirror surface is small compared to the mirror’s radius of curvature, so the angles are small, and the length FB is nearly equal to length FA. In this approximation, FA = FC. But FA = / , the focal length, and CA = 2 X FA = r. Thus the focal length is half the radius of curvature:

[spherical mirror] (32-1)

We assumed only that the angle 0 was small, so this result applies for all other incident paraxial rays. Thus all paraxial rays pass through the same point F, the focal point.

Since it is only approximately true that the rays come to a perfect focus at F, the more curved the mirror, the worse the approximation (Fig. 32-13) and the more blurred the image. This “defect” of spherical mirrors is called spherical aberration; we will discuss it more with regard to lenses in Chapter 33. A parabolic mirror, on the other hand, will reflect the rays to a perfect focus. However, because parabolic shapes are much harder to make and thus much more expensive, spherical mirrors are used for most purposes. (Many astronomical telescopes use parabolic mirrors.) We consider here only spherical mirrors and we will assume that they are small compared to their radius of curvature so that the image is sharp and Eq. 32-1 holds.

SECTION 32-3 Formation of Images by Spherical Mirrors 843

FIGURE 32-15 Rays leave point O' on the object (an arrow). Shown are the three most useful rays for determining where the image I' is formed. [Note that our mirror is not small compared to / , so our diagram will not give the precise position of the image.]

RAY D I A G R A MF i n d i n g t h e i m a g e p o s i t i o n

f o r a c u r v e d m i r r o r

P R O B L E M S O L V I N GI m a g e p o i n t i s w h e r e

r e f l e c t e d r a y s i n t e r s e c t

Image Formation—Ray DiagramsWe saw that for an object at infinity, the image is located at the focal point of a concave spherical mirror, where / = r/2. But where does the image lie for an object not at infinity? First consider the object shown as an arrow in Fig. 32-15a, which is placed between F and C at point O (O for object). Let us determine where the image will be for a given point O' at the top of the object.

(a) Ray 1 goes out from O' parallel to the axis and reflects through F.

(b) Ray 2 goes through F and then reflects back parallel to the axis.

(c) Ray 3 is chosen perpendicular to mirror, and so must reflect back on itself and go through C (center of curvature).

Diverging rays heading toward eye

844 CHAPTER 32

To do this we can draw several rays and make sure these reflect from the mirror such that the angle of reflection equals the angle of incidence. Many rays could be drawn leaving any point on an object, but determining the image position is simplified if we deal with three particularly simple rays. These are the rays labeled 1, 2, and 3 in Fig. 32-15 and we draw them leaving object point O' as follows:

Ray 1 is drawn parallel to the axis; therefore after reflection it must pass along a line through F (Fig. 32-15a).Ray 2 leaves O' and is made to pass through F (Fig. 32-15b); therefore it must reflect so it is parallel to the axis.Ray 3 passes through C, the center of curvature (Fig. 32-15c); it is along a radius of the spherical surface and is perpendicular to the mirror, so it is reflected back on itself.

All three rays leave a single point O' on the object. After reflection from a (small) mirror, the point at which these rays cross is the image point I'. All other rays from the same object point will also pass through this image point. To find the image point for any object point, only these three types of rays need to be drawn. Only two of these rays are needed, but the third serves as a check.

We have shown the image point in Fig. 32-15 only for a single point on the object. Other points on the object are imaged nearby, so a complete image of the object is formed, as shown by the dashed arrow in Fig. 32-15c. Because the light actually passes through the image itself, this is a real image that will appear on a piece of paper or film placed there. This can be compared to the virtual image formed by a plane mirror (the light does not actually pass through that image, Fig. 32-7).

The image in Fig. 32-15 can be seen by the eye when the eye is placed to the left of the image, so that some of the rays diverging from each point on the image (as point I') can enter the eye as shown in Fig. 32-15c. (See also Figs. 32-1 and 32-7.)

Mirror Equation and MagnificationImage points can be determined, roughly, by drawing the three rays as just described, Fig. 32-15. But it is difficult to draw small angles for the “paraxial” rays as we assumed. For more accurate results, we now derive an equation that gives the image distance if the object distance and radius of curvature of the mirror are known. To do this, we refer to Fig. 32-16. The object distance, dQ, is the distance of the object (point O) from the center of the mirror. The image distance, d\, is the distance of the image (point I) from the center of the mirror. The height of the object OO' is called hQ and the height of the image, I 'l, is hx.

FIGURE 32-16 Diagram for deriving the mirror equation. For the derivation, we assume the mirror size is small compared to its radius of curvature.

Two rays leaving O ' are shown: O 'FBI' (same as ray 2 in Fig. 32-15) and O 'A I', which is a fourth type of ray that reflects at the center of the mirror and can also be used to find an image point. The ray O 'A I' obeys the law of reflection, so the two right triangles O 'AO and I'A I are similar. Therefore, we have

ho = do h[ d[

For the other ray shown, O 'FBI', the triangles O 'FO and AFB are also similar because the angles are equal and we use the approximation AB = hx (mirror small compared to its radius). Furthermore FA = / , the focal length of the mirror, so

ho _ OF _ dQ- f

K FA /

The left sides of the two preceding expressions are the same, so we can equate the right sides:

do _ dQ f

~di ~ f

We now divide both sides by dQ and rearrange to obtain

— H— ~ = (32-2) Mirror equationdQ di f

This is the equation we were seeking. It is called the mirror equation and relates the object and image distances to the focal length / (where / = r / 2).

The lateral magnification, ra, of a mirror is defined as the height of the image divided by the height of the object. From our first set of similar triangles above, or the first equation on this page, we can write:

h\ dim = 7 T = (32" 3)h0 dQ

The minus sign in Eq. 32-4 is inserted as a convention. Indeed, we must be careful about the signs of all quantities in Eqs. 32-2 and 32-3. Sign conventions are chosen so as to give the correct locations and orientations of images, as predicted by ray diagrams.


FIGURE 32-16 (Repeated from previous page.)

j P R O B L E M S O L V I N GSign conventions

A C A U T I O NRemember to take the reciprocal

The sign conventions we use are: the image height h{ is positive if the image is upright, and negative if inverted, relative to the object (assuming hQ is taken as positive); d± or dQ is positive if image or object is in front of the mirror (as in Fig. 32-16); if either image or object is behind the mirror, the corresponding distance is negative (an example can be seen in Fig. 32-18, Example 32-6). Thus the magnification (Eq. 32-3) is positive for an upright image and negative for an inverted image (upside down). We summarize sign conventions more fully in the Problem Solving Strategy following our discussion of convex mirrors later in this Section.

Concave Mirror Examplesb =»:fJ iMJI = l»g.■ Image in a concave mirror. A 1.50-cm-high diamond ring

is placed 20.0 cm from a concave mirror with radius of curvature 30.0 cm. Determine (a) the position of the image, and (b) its size.

APPROACH We determine the focal length from the radius of curvature (Eq. 32-1), / = r /2 = 15.0 cm. The ray diagram is basically like that shown in Fig. 32-16 (repeated here on this page), since the object is between F and C. The position and size of the image are found from Eqs. 32-2 and 32-3.SOLUTION Referring to Fig. 32-16, we have CA = r = 30.0 cm, FA = / =15.0 cm, and OA = dQ = 20.0 cm.(a) From Eq. 32-2,

1_ _ 1 _ 1

di ~ f d

1 1

15.0 cm 20.0 cm= 0.0167 cm

So di = 1/(0.0167 cm x) = 60.0 cm. Because is positive, the image is 60.0 cm in front of the mirror, on the same side as the object.(b) From Eq. 32-3, the magnification is

m = —

60.0 cm= -3.00.

20.0 cm

The image is 3.0 times larger than the object, and its height is

hi = mhQ = (-3.00)(1.5 cm) = -4.5 cm.

The minus sign reminds us that the image is inverted, as in Fig. 32-16.NOTE When an object is further from a concave mirror than the focal point, we can see from Fig. 32-15 or 32-16 that the image is always inverted and real.


CONCEPTUAL EXAMPLE 32-5 I Reversible rays. If the object in Example 32-4 is placed instead where the image is (see Fig. 32-16), where will the new image be?

RESPONSE The mirror equation is symmetric in d0 and d{. Thus the new image will be where the old object was. Indeed, in Fig. 32-16 we need only reverse the direction of the rays to get our new situation.

■ Object closer to concave mirror. A 1.00-cm-high objectis placed 10.0 cm from a concave mirror whose radius of curvature is30.0 cm. (a) Draw a ray diagram to locate (approximately) the position of the image. (b) Determine the position of the image and the magnification analytically.APPROACH We draw the ray diagram using the rays as in Fig. 32-15, page 844. An analytic solution uses Eqs. 32-1,32-2, and 32-3.SOLUTION (a) Since / = r/2 = 15.0 cm, the object is between the mirror and the focal point. We draw the three rays as described earlier (Fig. 32-15); they are shown leaving the tip of the object in Fig. 32-17. Ray 1 leaves the tip of our object heading toward the mirror parallel to the axis, and reflects through F. Ray 2 cannot head toward F because it would not strike the mirror; so ray 2 must point as if it started at F (dashed line) and heads to the mirror, and then is reflected parallel to the principal axis. Ray 3 is perpendicular to the mirror, as before. The rays reflected from the mirror diverge and so never meet at a point. They appear, however, to be coming from a point behind the mirror. This point locates the image of the tip of the arrow. The image is thus behind the mirror and is virtual. (Why?)(b) We use Eq. 32-2 to find di when dQ = 10.0 cm:

J_ _ 1 _ J_ _ __ 1________ 1__ _ 2 - 3 _ 1d{ f d0 15.0 cm 10.0 cm 30.0 cm 30.0 cm

Therefore, d{ = -30.0 cm. The minus sign means the image is behind the mirror, which our diagram also told us. The magnification is m = -d J d Q = -(-30.0cm )/(10.0cm ) = +3.00. So the image is 3.00 times larger than the object. The plus sign indicates that the image is upright (same as object), which is consistent with the ray diagram, Fig. 32-17.NOTE The image distance cannot be obtained accurately by measuring on Fig. 32-17, because our diagram violates the paraxial ray assumption (we draw rays at steeper angles to make them clearly visible).NOTE When the object is located inside the focal point of a concave mirror (d0 < f )> the image is always upright and vertical. And if the object O in Fig. 32-17 is you, you see yourself clearly, because the reflected rays at point O are diverging. Your image is upright and enlarged.

--- ^/[ i 1 i

O> ---- /

1

A i

FIGURE 32-17 Object placed within the focal point F. The image is behind the mirror and is virtual, Example 32-6. [Note that the vertical scale (height of object = 1.0 cm) is different from the horizontal (OA = 10.0 cm) for ease of drawing, and reduces the precision of the drawing.]

0 P H Y S I C S A P P L I E DSeeing yourself upright and

magnified in a concave mirror


@ P H Y S I C S A P P L I E DM a g n i f y i n g m i r r o r

( s h a v i n g / c o s m e t i c )

It is useful to compare Figs. 32-15 and 32-17. We can see that if the object is within the focal point (dQ < /) , as in Fig. 32-17, the image is virtual, upright, and magnified. This is how a shaving or cosmetic mirror is used—you must place your head closer to the mirror than the focal point if you are to see yourself right-side up (see the photograph of Fig. 32-11 a). If the object is beyond the focal point, as in Fig. 32-15, the image is real and inverted (upside down—and hard to use!). Whether the magnification has magnitude greater or less than 1.0 in the latter case depends on the position of the object relative to the center of curvature, point C. Practice making ray diagrams with various object distances.

The mirror equation also holds for a plane mirror: the focal length is / = r/2 = oo, and Eq. 32-2 gives d{ = ~dQ.

Seeing the ImageFor a person’s eye to see a sharp image, the eye must be at a place where it intercepts diverging rays from points on the image, as is the case for the eye’s position in Figs. 32-15 and 32-16. Our eyes are made to see normal objects, which always means the rays are diverging toward the eye as shown in Fig. 32-1. (Or, for very distant objects like stars, the rays become essentially parallel, as in Fig. 32-12.) If you placed your eye between points O and I in Fig. 32-16, for example, converging rays from the object OO' would enter your eye and the lens of your eye could not bring them to a focus; you would see a blurry image. [We will discuss the eye more in Chapter 33.]

FIGURE 32-18 You can see a clear inverted image of your face when you are beyond C (dQ > 2 /), because the rays that arrive at your eye are diverging. Standard rays 2 and 3 are shown leaving point O on your nose. Ray 2 (and other nearby rays) enters your eye. Notice that rays are diverging as they move to the left of image point I.

FIGURE 32-19 Convex mirror: (a) the focal point is at F, behind the mirror; (b) the image I of the object at O is virtual, upright, and smaller than the object. [Not to scale for Example 32-7.]

If you are the object OO' in Fig. 32-16, situated between F and C, and are trying to see yourself in the mirror, you would see a blur; but the person whose eye is shown in Fig. 32-16 can see you clearly. You can see yourself clearly, but upside down, if you are to the left of C in Fig. 32-16, where dQ > 2/. Why? Because then the rays reflected from the image will be diverging at your position as demonstrated in Fig. 32-18, and your eye can then focus them. You can also see yourself clearly, and right-side up, if you are closer to the mirror than its focal point (dQ < /) , as we saw in Example 32-6, Fig. 32-17.

Convex MirrorsThe analysis used for concave mirrors can be applied to convex mirrors. Even the mirror equation (Eq. 32-2) holds for a convex mirror, although the quantities involved must be carefully defined. Figure 32-19a shows parallel rays falling on a convex mirror. Again spherical aberration is significant (Fig. 32-13), unless we assume the mirror’s size is very small compared to its radius of curvature. The reflected rays diverge, but seem to come from point F behind the mirror. This is the focal point, and its distance from the center of the mirror (point A) is the focal length, / . It is easy to show that again / = r/2. We see that an object at infinity produces a virtual image in a convex mirror. Indeed, no matter where the object is


9.0

Bl

&

placed on the reflecting side of a convex mirror, the image will be virtual and upright, as indicated in Fig. 32-19b. To find the image we draw rays 1 and 3 according to the rules used before on the concave mirror, as shown in Fig. 32-19b. Note that although rays 1 and 3 don’t actually pass through points F and C, the line along which each is drawn does (shown dashed).

The mirror equation, Eq. 32-2, holds for convex mirrors but the focal length / must be considered negative, as must the radius of curvature. The proof is left as a Problem. It is also left as a Problem to show that Eq. 32-3 for the magnification is also valid.

S O L V /

^ c > ______________________________________________

Spherical Mirrors1. Always draw a ray diagram even though you are

going to make an analytic calculation—the diagram serves as a check, even if not precise. From one point on the object, draw at least two, preferably three, of the easy-to-draw rays using the rules described in Fig. 32-15. The image point is where the reflected rays intersect or appear to intersect.

2. Apply the mirror equation, Eq. 32-2, and the magnification equation, Eq. 32-3. It is crucially important to follow the sign conventions—see the next point.

f Object distances are positive for material objects, but can be negative

3. Sign Conventions(a) When the object, image, or focal point is on the

reflecting side of the mirror (on the left in our drawings), the corresponding distance is positive. If any of these points is behind the mirror (on the right) the corresponding distance is negatived

(b) The image height hi is positive if the image is upright, and negative if inverted, relative to the object (hQ is always taken as positive).

4. Check that the analytical solution is consistent with the ray diagram.

in systems with more than one mirror or lens—see Section 33-3.

EXAMPLE 32-7 Convex rearview mirror. An external rearview car mirror is @ P H Y S I C S a p p l i e d

convex with a radius of curvature of 16.0 m (Fig. 32-20). Determine the location Convex rearview mirrorof the image and its magnification for an object 10.0 m from the mirror.APPROACH We follow the steps of the Problem Solving Strategy explicitly.SOLUTION1. Draw a ray diagram. The ray diagram will be like Fig. 32-19b, but the large

object distance (dQ = 10.0 m) makes a precise drawing difficult. We have a convex mirror, so r is negative by convention.

2. Mirror and magnification equations. The center of curvature of a convex mirror is behind the mirror, as is its focal point, so we set r = -16.0 m so that the focal length is / = r/ 2 = -8.0 m. The object is in front of the mirror, dQ = 10.0 m. Solving the mirror equation, Eq. 32-2, for 1 /d { gives

1 1 1 1 1 - 10.0 - 8.0 18di f dQ - 8.0 m 10.0 m 80.0 m 80.0 m

Thus di = — 80.0m/18 = -4.4 m. Equation 32-3 gives the magnification

d{ (-4 .4 m)m = ~ T 0 = ~ = + 0 ’4 4 '

3. Sign conventions. The image distance is negative, -4.4 m, so the image is behind the mirror. The magnification is m = +0.44, so the image is upright (same orientation as object) and less than half as tall as the object.

4. Check. Our results are consistent with Fig. 32-19b.

Convex rearview mirrors on vehicles sometimes come with a warning that objects are closer than they appear in the mirror. The fact that d{ may be smaller than dQ (as in Example 32-7) seems to contradict this observation. The real reason the object seems farther away is that its image in the convex mirror is smaller than it would be in a plane mirror, and we judge distance of ordinary objects such as other cars mostly by their size.



TABLE 32-1 Indices of Refraction1 32—4 Index of RefractionMaterial

Vacuum Air (at STP)WaterEthyl alcoholGlass

Fused quartz Crown glass Light flint

Lucite or PlexiglasSodium chlorideDiamond

1.00001.0003

1.33

1.36

1.461.52 1.58

1.51

1.53

2.42

fA = 589i

A C A U T I O NA n g l e s d i a n d d 2 a r e m e a s u r e d f r o m

t h e p e r p e n d i c u l a r , n o t f r o m s u r f a c e

We saw in Chapter 31 that the speed of light in vacuum is

c = 2.99792458 X 108 m/s,

which is usually rounded off to

3.00 X 108 m/s

when extremely precise results are not required.In air, the speed is only slightly less. In other transparent materials such as glass

and water, the speed is always less than that in vacuum. For example, in water light travels at about fc. The ratio of the speed of light in vacuum to the speed v in a given material is called the index of refraction, n, of that material:

cn = —• v (32-4)

The index of refraction is never less than 1, and values for various materials are given in Table 32-1. For example, since n = 2.42 for diamond, the speed of light in diamond is

V n

(3.00 X 108m/s)2.42

= 1.24 X 108m/s.

As we shall see later, n varies somewhat with the wavelength of the light—except in vacuum—so a particular wavelength is specified in Table 32-1, that of yellow light with wavelength A = 589 nm.

That light travels more slowly in matter than in vacuum can be explained at the atomic level as being due to the absorption and reemission of light by atoms and molecules of the material.

3 2 -5 Refraction: Snell's LawWhen light passes from one transparent medium into another with a different index of refraction, part of the incident light is reflected at the boundary. The remainder passes into the new medium. If a ray of light is incident at an angle to the surface (other than perpendicular), the ray changes direction as it enters the new medium. This change in direction, or bending, is called refraction.

Figure 32-21 a shows a ray passing from air into water. Angle 61 is the angle the incident ray makes with the normal (perpendicular) to the surface and is called the angle of incidence. Angle 02 is the angle of refraction, the angle the refracted ray makes with the normal to the surface. Notice that the ray bends toward the normal when entering the water. This is always the case when the ray enters a medium where the speed of light is less (and the index of refraction greater, Eq. 32-4). If light travels from one medium into a second where its speed is greater, the ray bends away from the normal; this is shown in Fig. 32-21b for a ray traveling from water to air.

Normal

Retimedmy

FIGURE 32-21 Refraction.(a) Light refracted when passing from air ( n i ) into water (n 2 ): n 2 > n \ .

(b) Light refracted when passing from water («i) into air ( n 2 ) : > n 2 .

NormalJ

Refracted I

Refractedray

H2 >H|(a) Ray hends toward 1

; ,1 > n l (b) Ray bonds away from


FIGURE 32-22 Ray diagram showing why a person’s legs look shorter when standing in waist-deep water: the path of light traveling from the bather’s foot to the observer’s eye bends at the water’s surface, and our brain interprets the light as having traveled in a straight line, from higher up (dashed line).

Refraction is responsible for a number of common optical illusions. For example, a person standing in waist-deep water appears to have shortened legs. As shown in Fig. 32-22, the rays leaving the person’s foot are bent at the surface. The observer’s brain assumes the rays to have traveled a straight-line path (dashed red line), and so the feet appear to be higher than they really are. Similarly, when you put a straw in water, it appears to be bent (Fig. 32-23).

Snell's LawThe angle of refraction depends on the speed of light in the two media and on the incident angle. An analytical relation between Qx and 02 in Fig- 32-21 was arrived at experimentally about 1621 by Willebrord Snell (1591-1626). It is known as Snell’s law and is written:

riismOx = n2s in02. (32-5)6i is the angle of incidence and d2 is the angle of refraction; nx and n2 are the respective indices of refraction in the materials. See Fig. 32-21. The incident and refracted rays lie in the same plane, which also includes the perpendicular to the surface. Snell’s law is the law of refraction. (Snell’s law was derived in Section 15-10 where Eq. 15-19 is just a combination of Eqs. 32-5 and 32-4. We also derive it in Chapter 34 using the wave theory of light.)

It is clear from Snell’s law that if n2 > n1, then 02 < #i • That is, if light enters a medium where n is greater (and its speed is less), then the ray is bent toward the normal. And if n2 < nx, then 02 > 0i, so the ray bends away from the normal. This is what we saw in Fig. 32-21.

EXERCISE D Light passes from a medium with n = 1.3 into a medium with n = 1.5. Is the light bent toward or away from the perpendicular to the interface?

Refraction through flat glass. Light traveling in air strikes a flat piece of uniformly thick glass at an incident angle of 60°, as shown in Fig. 32-24. If the index of refraction of the glass is 1.50, (a) what is the angle of refraction 0A in the glass; (b) what is the angle 0B at which the ray emerges from the glass?APPROACH We apply Snell’s law at the first surface, where the light enters the glass, and again at the second surface where it leaves the glass and enters the air. SOLUTION (a) The incident ray is in air, so nx = 1.00 and n2 = 1.50. Applying Snell’s law where the light enters the glass (0! = 60°) gives

sin0A = ^ sin 60° = 0.5774,

so 0A = 35.3°.(b) Since the faces of the glass are parallel, the incident angle at the second surface is just 0A (simple geometry), so sin 0A = 0.5774. At this second interface, rii = 1.50 and n2 = 1.00. Thus the ray re-enters the air at an angle 0B(=d2) given by

1.50sin0B = T o o sin0A = ° '866’

and 0B = 60°. The direction of a light ray is thus unchanged by passing through a flat piece of glass of uniform thickness.NOTE This result is valid for any angle of incidence. The ray is displaced slightly to one side, however. You can observe this by looking through a piece of glass (near its edge) at some object and then moving your head to the side slightly so that you see the object directly. It “jumps.”

FIGURE 32-23 A straw in water looks bent even when it isn’t.

SNELL’S LAW(LAW OF REFRACTION)

FIGURE 32-24 Light passing through a piece of glass (Example 32-8).

f /“Image” (where object object appears to be) when

viewed through the

SECTION 32-5 851


FIGURE 32-26 The spectrum of visible light, showing the range of wavelengths for the various colors as seen in air. Many colors, such as brown, do not appear in the spectrum; they are made from a mixture of wavelengths.

EXAMPLE 32-9 Apparent depth of a pool. A swimmer has dropped her goggles to the bottom of a pool at the shallow end, marked as 1.0 m deep. But the goggles don’t look that deep. Why? How deep do the goggles appear to be when you look straight down into the water?APPROACH We draw a ray diagram showing two rays going upward from a point on the goggles at a small angle, and being refracted at the water’s (flat) surface, Fig. 32-25. The two rays traveling upward from the goggles are refracted away from the normal as they exit the water, and so appear to be diverging from a point above the goggles (dashed lines), which is why the water seems less deep than it actually is.SOLUTION To calculate the apparent depth d' (Fig. 32-25), given a real depth d = 1.0 m, we use Snell’s law with ^ = 1.33 for water and n2 = 1.0 for air:

sin 02 = Wi sin •We are considering only small angles, so sin 0 « tan 0 « 0, with 0 in radians. So Snell’s law becomes

02 ~ Wi 0 i .

From Fig. 32-25, we see that

X X02 ~ tan 02 = — and 6X « tan d1 = — ■

d d

Putting these into Snell’s law, 02 ~ n161, we get

or.. d 1.0 m

d ~ — = ~ r ^ = 0.75 m. nx 1.33

The pool seems only three-fourths as deep as it actually is.

3 2 -6 Visible Spectrum and DispersionAn obvious property of visible light is its color. Color is related to the wavelengths or frequencies of the light. (How this was discovered will be discussed in Chapter 34.) Visible light—that to which our eyes are sensitive—has wavelengths in air in the range of about 400 nm to 750 nm.f This is known as the visible spectrum, and within it lie the different colors from violet to red, as shown in Fig. 32-26. Light with wavelength shorter than 400 nm (= violet) is called ultraviolet (UV), and light with wavelength greater than 750 nm (= red) is called infrared (IR).* Although human eyes are not sensitive to UV or IR, some types of photographic film and digital cameras do respond to them.

A prism can separate white light into a rainbow of colors, as shown in Fig. 32-27. This happens because the index of refraction of a material depends on the wavelength, as shown for several materials in Fig. 32-28. White light is a

400 nm 500 nm fi(K) nm 7(XJ nm J------------ 1------------ 1------------ 1------------ 1------------ 1------------ 1------------ 1

f Sometimes the angstrom (A) unit is used when referring to light: 1 A = 1 X 10 10 m. Then visible light falls in the wavelength range of 4000 A to 7500 A.*The complete electromagnetic spectrum is illustrated in Fig. 31-12.


Violet BlueWavelength (nm) Green Yellow Orange Red

FIGURE 32-27 White light passing through a prism is broken down into its constituent colors.

FIGURE 32-28 Index of refraction as a function of wavelength for various transparent solids.

mixture of all visible wavelengths, and when incident on a prism, as in Fig. 32-29, the different wavelengths are bent to varying degrees. Because the index of refraction is greater for the shorter wavelengths, violet light is bent the most and red the least, as indicated. This spreading of white light into the full spectrum is called dispersion.

Rainbows are a spectacular example of dispersion—by drops of water. You can see rainbows when you look at falling water droplets with the Sun behind you. Figure 32-30 shows how red and violet rays are bent by spherical water droplets and are reflected off the back surface of the droplet. Red is bent the least and so reaches the observer’s eyes from droplets higher in the sky, as shown in the diagram. Thus the top of the rainbow is red.

FIGURE 32-30 (a) Ray diagram explaining how a rainbow (b) is formed.

These two rays are seen by observer (not to scale)

FIGURE 32-29 White light dispersed by a prism into the visible spectrum.

® lP H Y S I C S A P P L I E DR a i n b o w s

(a) (b)

The visible spectrum, Fig. 32-26, does not show all the colors seen in nature. For example, there is no brown in Fig. 32-26. Many of the colors we see are a mixture of wavelengths. For practical purposes, most natural colors can be reproduced using three primary colors. They are red, green, and blue for direct source viewing such as TV and computer monitors. For inks used in printing, the primary colors are cyan (the color of the margin notes in this book), yellow, and magenta (the color we use for light rays in diagrams).

For any wave, its velocity v is related to its wavelength A and frequency / by v = /A (Eq. 15-1 or 31-14). When a wave travels from one material into another, the frequency of the wave does not change across the boundary since a point (an atom) at the boundary oscillates at that frequency. Thus if light goes from air into a material with index of refraction n, the wavelength becomes (recall Eq. 32^):

= 7 = ~ f = A (32"6>f n f n

where A is the wavelength in vacuum or air and \ n is the wavelength in the material with index of refraction n. SECTION 32-6 853

CONCEPTUAL EXAMPLE 32-10 I Observed color of light under water.We said that color depends on wavelength. For example, for an object emitting 650 nm light in air, we see red. But this is true only in air. If we observe this same object when under water, it still looks red. But the wavelength in water A„ is (Eq. 32-6) 650nm/1.33 = 489 nm. Light with wavelength 489 nm would appear blue in air. Can you explain why the light appears red rather than blue when observed under water?

RESPONSE It must be that it is not the wavelength that the eye responds to, but rather the frequency. For example, the frequency of 650 nm red light in air is / = c/A = (3.0 X 108m/s)/(650 X 10_9m) = 4.6 X 1014Hz, and does not change when the light travels from one medium to another. Only A changes.

NOTE If we classified colors by frequency, the color assignments would be valid for any material. We typically specify colors by wavelength in air (even if less general) not just because we usually see objects in air, but because wavelength is what is commonly measured (it is easier to measure than frequency).

3 2 -7 Total Internal Reflection; Fiber Optics

When light passes from one material into a second material where the index of refraction is less (say, from water into air), the light bends away from the normal, as for rays I and J in Fig. 32-31. At a particular incident angle, the angle of refraction will be 90°, and the refracted ray would skim the surface (ray K) in this

FIGURE 32-31 Since n 2 < « i , light rays are totally internally reflected if the incident angle > 6q , as for ray L. If di < 6C, as for rays I and J, only a part of the light is reflected, and the rest is refracted.

case. The incident angle at which this occurs is called the critical angle, 0C. From Snell’s law, 0C is given by

A C A U T I O NT o t a l i n t e r n a l r e f l e c t i o n

( o c c u r s o n l y i f r e f r a c t i v e

i n d e x i s s m a l l e r b e y o n d b o u n d a r y )

n2 n2sm 6C = — sin 90° = — nx ni (32-7)

For any incident angle less than 0C, there will be a refracted ray, although part of the light will also be reflected at the boundary. However, for incident angles greater than 0C, Snell’s law would tell us that sin02 is greater than 1.00. Yet the sine of an angle can never be greater than 1.00. In this case there is no refracted ray at all, and all o f the light is reflected, as for ray L in Fig. 32-31. This effect is called total internal reflection. Total internal reflection can occur only when light strikes a boundary where the medium beyond has a lower index of refraction.

EXERCISE E Fill a sink with water. Place a waterproof watch just below the surface with the watch’s flat crystal parallel to the water surface. From above you can still see the watch reading. As you move your head to one side far enough what do you see, and why?


CONCEPTUAL EXAMPLE 32-11 I View up from under water. Describe what a person would see who looked up at the world from beneath the perfectly smooth surface of a lake or swimming pool.RESPONSE For an air-water interface, the critical angle is given by

1.00sin 0C = — — = 0.750. c 1.33

Therefore, 0C = 49°. Thus the person would see the outside world compressed into a circle whose edge makes a 49° angle with the vertical. Beyond this angle, the person would see reflections from the sides and bottom of the lake or pool (Fig. 32-32).

4 . FIGURE 32-32 (a) Light rays, and (b) view looking upward from beneath the water (the surface of the water must be very smooth). Example 32-11.

Diamonds achieve their brilliance from a combination of dispersion and total internal reflection. Because diamonds have a very high index of refraction of about 2.4, the critical angle for total internal reflection is only 25°. The light dispersed into a spectrum inside the diamond therefore strikes many of the internal surfaces of the diamond before it strikes one at less than 25° and emerges.

After many such reflections, the light has traveled far enough that the colors have become sufficiently separated to be seen individually and brilliantly by the eye after leaving the diamond.

Many optical instruments, such as binoculars, use total internal reflection within a prism to reflect light. The advantage is that very nearly 100% of the light is reflected, whereas even the best mirrors reflect somewhat less than 100%. Thus the image is brighter, especially after several reflections. For glass with n = 1.50,0C = 41.8°. Therefore, 45° prisms will reflect all the light internally, if oriented as shown in the binoculars of Fig. 32-33.

EXERCISE F If 45.0° plastic lenses are used in binoculars, what minimum index of refraction must the plastic have?

Fiber OpticsTotal internal reflection is the principle behind fiber optics. Glass and plastic fibers as thin as a few micrometers in diameter are common. A bundle of such tiny fibers is called a light pipe or cable, and light1" can be transmitted along it with almost no loss because of total internal reflection. Figure 32-34 shows how light traveling down a thin fiber makes only glancing collisions with the walls so that total internal reflection occurs. Even if the light pipe is bent into a complicated shape, the critical angle still won’t be exceeded, so light is transmitted practically undiminished to the other end. Very small losses do occur, mainly by reflection at the ends and absorption within the fiber.

Important applications of fiber-optic cables are in communications and medicine. They are used in place of wire to carry telephone calls, video signals, and computer data. The signal is a modulated light beam (a light beam whose intensity can be varied) and data is transmitted at a much higher rate and with less loss and less interference than an electrical signal in a copper wire. Fibers have been developed that can support over one hundred separate wavelengths, each modulated to carry up to 10 gigabits (lO10 bits) of information per second. That amounts to a terabit (lO12 bits) per second for the full one hundred wavelengths.

FIGURE 32-33 Total internal reflection of light by prisms in binoculars.

FIGURE 32-34 Light reflected totally at the interior surface of a glass or transparent plastic fiber.

0 P H Y S I C S A P P L I E DF i b e r o p t i c s i n c o m m u n i c a t i o n s

f Fiber optic devices use not only visible light but also infrared light, ultraviolet light, and microwaves.

SECTION 32-7 Total Internal Reflection; Fiber Optics 855

P H Y S I C S A P P L I E D

FIGURE 32-35 (a) How a fiber-optic image is made, (b) Example of a fiber-optic device inserted through the mouth to view the gastrointestinal tract, with image on screen.

The sophisticated use of fiber optics to transmit a clear picture is particularly useful in medicine, Fig. 32-35. For example, a patient’s lungs can be examined by inserting a light pipe known as a bronchoscope through the mouth and down the bronchial tube. Light is sent down an outer set of fibers to illuminate the lungs. The reflected light returns up a central core set of fibers. Light directly in front of each fiber travels up that fiber. At the opposite end, a viewer sees a series of bright and dark spots, much like a TV screen—that is, a picture of what lies at the opposite end. Lenses are used at each end. The image may be viewed directly or on a monitor screen or film. The fibers must be optically insulated from one another, usually by a thin coating of material with index of refraction less than that of the fiber. The more fibers there are, and the smaller they are, the more detailed the picture. Such instruments, including bronchoscopes, colonoscopes (for viewing the colon), and endoscopes (stomach or other organs), are extremely useful for examining hard-to-reach places.

32—8 Refraction at a Spherical SurfaceWe now examine the refraction of rays at the spherical surface of a transparent material. Such a surface could be one face of a lens or the cornea of the eye. To be general, let us consider an object which is located in a medium whose index of refraction is n1, and rays from each point on the object can enter a medium whose index of refraction is n2. The radius of curvature of the spherical boundary is R, and its center of curvature is at point C, Fig. 32-36. We now show that all rays leaving a point O on the object will be focused at a single point I, the image point, if we consider only paraxial rays: rays that make a small angle with the axis.

FIGURE 32-36 Rays from a point O on an object will be focused at a single image point I by a spherical boundary between two transparent materials (n2 > ni), as long as the rays make small angles with the axis.

To do so, we consider a single ray that leaves point O as shown in Fig. 32-37. From Snell’s law, Eq. 32-5, we have

sin Qx = n2 sin 02 •We are assuming that angles d1,d2, a, /3, and 7 are small, so sin 0 ~ 0 (in radians), and Snell’s law becomes, approximately,

All 1̂ = ^2 ̂ 2 •

Also, /3 + 0 = 180° and 02 + 7 + (f> = 180°, so

/3 = 7 + 02 .

Similarly for triangle OPC,

= a + j3.

FIGURE 32-37 Diagram for showing that all paraxial rays from O focus at the same point I (n2 > ni).


These three relations can be combined to yield

+ n2y = (n2 — «i)/3.Since we are considering only the case of small angles, we can write, approximately,

h h _ h “ “ T a ’ p ~ r ’ 7 ~ 1 -

where d0 and d\ are the object and image distances and h is the height as shown in Fig. 32-37. We substitute these into the previous equation, divide through by h, and obtain

n2 — rixdi "

rh + n i = d r

(32-8)

For a given object distance dQ, this equation tells us the image distance, does not depend on the angle of a ray. Hence all paraxial rays meet at the same point I. This is true only for rays that make small angles with the axis and with each other, and is equivalent to assuming that the width of the refracting spherical surface is small compared to its radius of curvature. If this assumption is not true, the rays will not converge to a point; there will be spherical aberration, just as for a mirror (see Fig. 32-13), and the image will be blurry. (Spherical aberration will be discussed further in Section 33-10.)

We derived Eq. 32-8 using Fig. 32-37 for which the spherical surface is convex (as viewed by the incoming ray). It is also valid for a concave surface—as can be seen using Fig. 32-38—if we use the following conventions:1. If the surface is convex (so the center of curvature C is on the side of the

surface opposite to that from which the light comes), R is positive; if the surface is concave (C on the same side from which the light comes) R is negative.

2. The image distance, d{, follows the same convention: positive if on the opposite side from where the light comes, negative if on the same side.

3. The object distance is positive if on the same side from which the light comes (this is the normal case, although when several surfaces bend the light it may not be so), otherwise it is negative.

For the case shown in Fig. 32-38 with a concave surface, both R and d-x are negative when used in Eq. 32-8. Note, in this case, that the image is virtual.

FIGURE 32-38 Rays from O refracted by a concave surface form a virtual image (n2 > ni). Per our conventions, R < 0, < 0, dQ > 0.

EXAMPLE 32-12 Apparent depth II. A person looks vertically down into a1.0-m-deep pool. How deep does the water appear to be?APPROACH Example 32-9 solved this problem using Snell’s law. Here we use Eq. 32-8.SOLUTION A ray diagram is shown in Fig. 32-39. Point O represents a point on the pool’s bottom. The rays diverge and appear to come from point I, the image. We have d0 = 1.0 m and, for a flat surface, R = oo. Then Eq. 32-8 becomes

1.33 1.00 _ (1.00 - 1.33)1.0 m di oo


0.

Hence d\ = — (1.0m)/(1.33) = -0.75 m. So the pool appears to be only three- fourths as deep as it actually is, the same result we found in Example 32-9. The minus sign tells us the image point I is on the same side of the surface as O, and the image is virtual. At angles other than vertical, this conclusion must be modified.

*SECTION 3 2-8 Refraction at a Spherical Surface 857

EXAMPLE 32-13 A spherical "lens." A point source of light is placed at a distance of 25.0 cm from the center of a glass sphere (n = 1.5) of radius 10.0 cm, Fig. 32-40. Find the image of the source.

APPROACH As shown in Fig. 32-40, there are two refractions, and we treat them successively, one at a time. The light rays from the source first refract from the convex glass surface facing the source. We analyze this first refraction, treating it as in Fig. 32-36, ignoring the back side of the sphere.SOLUTION Using Eq. 32-8 (assuming paraxial rays) with nx = 1.0, n2 = 1.5, R = 10.0 cm, and dQ = 25.0 cm - 10.0 cm = 15.0 cm, we solve for the image distance as formed at surface 1 ,d xX\

1 1 f ( n 2 - n x) Wl\_ J _ | ( 1.5 - 1.0 1.0 \ 1

d\\ n2\ R d j 1.5 '̂ 10.0 cm 15.0 c m / 90.0 cm

Thus, the image of the first refraction is located 90.0 cm to the left of the front surface. This image (1^ now serves as the object for the refraction occurring at the back surface (surface 2) of the sphere. This surface is concave so R = -1 0 .0 cm, and we consider a ray close to the axis. Then the object distance is d02 = 90.0 cm + 2(10.0 cm) = 110.0 cm, and Eq. 32-8 yields, with n x = 1.5, n2 = 1 .0,

d[2

1.5 1.5J _ / L 0________________1.0 V - 10.0 cm 110.0 cm

4.0110.0 cm

so dl2 = 28 cm. Thus, the final image is located a distance 28 cm from the back side of the sphere.

FIG U RE 3 2 -4 0 Example 32-13.Source

R = 10.0 cm

SummaryLight appears to travel along straight-line paths, called rays, at

a speed v that depends on the index of refraction, n, of the material; that is

(32-4)

where c is the speed of light in vacuum.When light reflects from a flat surface, the angle o f reflection

equals the angle o f incidence. This law of reflection explains why mirrors can form images.

In a plane mirror, the image is virtual, upright, the same size as the object, and is as far behind the mirror as the object is in front.

A spherical mirror can be concave or convex. A concave spherical mirror focuses parallel rays of light (light from a very distant object) to a point called the focal point. The distance of this point from the mirror is the focal length / of the mirror and

f = h (32-1)

where r is the radius of curvature of the mirror.Parallel rays falling on a convex mirror reflect from the

mirror as if they diverged from a common point behind the mirror. The distance of this point from the mirror is the focal length and is considered negative for a convex mirror.

For a given object, the position and size of the image formed by a mirror can be found by ray tracing. Algebraically,

the relation between image and object distances, d[ and dQ, and the focal length / , is given by the mirror equation:

JL i - I dQ + di f

(32-2)

The ratio of image height to object height, which equals the magnification m of a mirror, is

hi dim = 7T = - ~ T ' <32“ 3>hQ dQ

If the rays that converge to form an image actually pass through the image, so the image would appear on film or a screen placed there, the image is said to be a real image. If the rays do not actually pass through the image, the image is a virtual image.

When light passes from one transparent medium into another, the rays bend or refract. The law of refraction (Snell’s law) states that

ni sin0! = n2 sin02> (32-5)where nx and are the index of refraction and angle with the normal to the surface for the incident ray, and n2 and 02 are for the refracted ray.

When light of wavelength A enters a medium with index of refraction n, the wavelength is reduced to

K = ~ 0 2 - 6 )

The frequency does not change.


The frequency or wavelength of light determines its color. The visible spectrum in air extends from about 400 nm (violet) to about 750 nm (red).

Glass prisms break white light down into its constituent colors because the index of refraction varies with wavelength, a phenomenon known as dispersion.

Questions

When light rays reach the boundary of a material where the index of refraction decreases, the rays will be totally internally reflected if the incident angle, Qi, is such that Snell’s law would predict sin02 > 1. This occurs if d\ exceeds the critical angle 6C given by

sin 0C = — * (32-7)n\

1. What would be the appearance of the Moon if it had (a) a rough surface; (b) a polished mirrorlike surface?

2. Archimedes is said to have burned the whole Roman fleet in the harbor of Syracuse by focusing the rays of the Sun with a huge spherical mirror. Is this1’ reasonable?

3. What is the focal length of a plane mirror? What is the magnification of a plane mirror?

4. An object is placed along the principal axis of a spherical mirror. The magnification of the object is -3.0. Is the image real or virtual, inverted or upright? Is the mirror concave or convex? On which side of the mirror is the image located?

5. Using the rules for the three rays discussed with reference to Fig. 32-15, draw ray 2 for Fig. 32-19b.

6. Does the mirror equation, Eq. 32-2, hold for a plane mirror? Explain.

7. If a concave mirror produces a real image, is the image necessarily inverted?

8. How might you determine the speed of light in a solid, rectangular, transparent object?

9. When you look at the Moon’s reflection from a ripply sea, it appears elongated (Fig. 32-41). Explain.

FIGURE 32-41Question 9.

10. How can a spherical mirror have a negative object distance?11. What is the angle of refraction when a light ray is incident

perpendicular to the boundary between two transparent materials?

12. When you look down into a swimming pool or a lake, are you likely to overestimate or underestimate its depth? Explain. How does the apparent depth vary with the viewing angle? (Use ray diagrams.)

13. Draw a ray diagram to show why a stick looks bent when part of it is under water (Fig. 32-23).

Students at MIT did a feasibility study. See www.mit.edu.

14. When a wide beam of parallel light enters water at an angle, the beam broadens. Explain.

15. You look into an aquarium and view a fish inside. One ray of light from the fish as it emerges from the tank is shown in Fig. 32-42. The apparent position of the fish is also shown. In the drawing, indicate the approximate position of the actual fish. Briefly justify youranswer.


A

16. How can you “see” a round drop of water on a table even though the water is transparent and colorless?

17. A ray of light is refracted through three different materials (Fig. 32-43). Rank the materialsaccording to their index of refraction, least to greatest.


18. Can a light ray traveling in air be totally reflected when it strikes a smooth water surface if the incident angle is chosen correctly? Explain.

19. When you look up at an object in air from beneath the surface in a swimming pool, does the object appear to be the same size as when you see it directly in air? Explain.

20. What type of mirror is shown in Fig. 32-44?

FIGURE 32-Question 20.

21. Light rays from stars (including our Sun) always bend toward the vertical direction as they pass through the Earth’s atmosphere, (a) Why does this make sense? (b) What can you conclude about the apparent positions of stars as viewed from Earth?

Questions 859

http://www.mit.edu

Problems32-2 Reflection; Plane Mirrors1. (I) When you look at yourself in a 60-cm-tall plane mirror,

you see the same amount of your body whether you are close to the mirror or far away. (Try it and see.) Use ray diagrams to show why this should be true.

2. (I) Suppose that you want to take a photograph of yourself as you look at your image in a mirror 2.8 m away. For what distance should the camera lens be focused?

3. (II) Two plane mirrors meet at a 135° angle, Fig. 32-45. If light rays strike one mirror at 38° as shown, at what angle (f> do they leave the second mirror?

FIGURE 32-45Problem 3.

4. (II) A person whose eyes are 1.64 m above the floor stands 2.30 m in front of a vertical plane mirror whose bottom edge is 38 cm above the floor, Fig. 32-46. What is the horizontal distance x to the base of the wall supporting the mirror of the nearest point on the floor that can be seen reflected in frthe mirror? y £"

- 2 JO m-

1.64 m J

i 1 MK cmFIGURE 32-46Problem 4. ’■*-£—

5. (II) Show that if two plane mirrors meet at an angle 4>, a single ray reflected successively from both mirrors is deflected through an angle of 2<j> independent of the incident angle. Assume <f> < 90° and that only two reflections, one from each mirror, take place.

6. (II) Suppose you are 88 cm from a plane mirror. What area of the mirror is used to reflect the rays entering one eye from a point on the tip of your nose if your pupil diameter is 4.5 mm?

7. (II) Stand up two plane mirrors so they form a 90.0° angle as in Fig. 32-47. When you look into this double mirror, you see yourself as others see you, instead of reversed as in a single mirror. Make a ray diagram to show how this occurs.

FIGURE 32-47Problems 7 and 8.

8. (Ill) Suppose a third mirror is placed beneath the two shown in Fig. 32-47, so that all three are perpendicular to each other.(a) Show that for such a “corner reflector,” any incident ray will return in its original direction after three reflections.(b) What happens if it makes only two reflections?

32-3 Spherical Mirrors9. (I) A solar cooker, really a concave mirror pointed at the

Sun, focuses the Sun’s rays 18.8 cm in front of the mirror. What is the radius of the spherical surface from which the mirror was made?

10. (I) How far from a concave mirror (radius 24.0 cm) must an object be placed if its image is to be at infinity?

11. (I) When walking toward a concave mirror you notice that the image flips at a distance of 0.50 m. What is the radius of curvature of the mirror?

12. (II) A small candle is 35 cm from a concave mirror having a radius of curvature of 24 cm. (a) What is the focal length of the mirror? (b) Where will the image of the candle be located? (c) Will the image be upright or inverted?

13. (II) You look at yourself in a shiny 9.2-cm-diameter Christmas tree ball. If your face is 25.0 cm away from the ball’s front surface, where is your image? Is it real or virtual? Is it upright or inverted?

14. (II) A mirror at an amusement park shows an upright image of any person who stands 1.7 m in front of it. If the image is three times the person’s height, what is the radius of curvature of the mirror? (See Fig. 32-44.)

15. (II) A dentist wants a small mirror that, when 2.00 cm from a tooth, will produce a 4.0X upright image. What kind of mirror must be used and what must its radius of curvature be?

16. (II) Some rearview mirrors produce images of cars to your rear that are smaller than they would be if the mirror were flat. Are the mirrors concave or convex? What is a mirror’s radius of curvature if cars 18.0 m away appear 0.33 their normal size?

17. (II) You are standing 3.0 m from a convex security mirror in a store. You estimate the height of your image to be half of your actual height. Estimate the radius of curvature of the mirror.

18. (II) An object 3.0 mm high is placed 18 cm from a convex mirror of radius of curvature 18 cm. (a) Show by ray tracing that the image is virtual, and estimate the image distance.(b) Show that the (negative) image distance can be computed from Eq. 32-2 using a focal length of -9 .0 cm. (c) Compute the image size, using Eq. 32-3.

19. (II) The image of a distant tree is virtual and very small when viewed in a curved mirror. The image appears to be16.0 cm behind the mirror. What kind of mirror is it, and what is its radius of curvature?

20. (II) Use two techniques, (a) a ray diagram, and (b) the mirror equation, to show that the magnitude of the magnification of a concave mirror is less than 1 if the object is beyond the center of curvature C (d0 > r), and is greater than 1 if the object is within C (d0 < r).

21. (II) Show, using a ray diagram, that the magnification m of a convex mirror is m = ~d[/d0 , just as for a concave mirror. [Hint: Consider a ray from the top of the object that reflects at the center of the mirror.]

22. (II) Use ray diagrams to show that the mirror equation, Eq. 32-2, is valid for a convex mirror as long as / is considered negative.

23. (II) The magnification of a convex mirror is +0.55 X for objects 3.2 m from the mirror. What is the focal length of this mirror?


24. (II) (a) Where should an object be placed in front of a concave mirror so that it produces an image at the same location as the object? (b) Is the image real or virtual? (c) Is the image inverted or upright? (d) What is the magnification of the image?

25. (II) A 4.5-cm tall object is placed 26 cm in front of a spherical mirror. It is desired to produce a virtual image that is upright and 3.5 cm tall, (a) What type of mirror should be used? (b) Where is the image located? (c) What is the focal length of the mirror? (d) What is the radius of curvature of the mirror?

26. (II) A shaving or makeup mirror is designed to magnify your face by a factor of 1.35 when your face is placed20.0 cm in front of it. (a) What type of mirror is it?(b) Describe the type of image that it makes of your face.(c) Calculate the required radius of curvature for the mirror.

27. (II) A concave mirror has focal length / . When an object is placed a distance dQ > f from this mirror, a real image with magnification m is formed, (a) Show that m = / / ( / - dQ). (ib) Sketch m vs. dQ over the range f < d0 < +oo where / = 0.45 m. (c) For what value of dQ will the real image have the same (lateral) size as the object? (d) To obtain a real image that is much larger than the object, in what general region should the object be placed relative to the mirror?

28. (II) Let the focal length of a convex mirror be written as / = - | / | . Show that the magnification m of an object a distance dQ from this mirror is given by m = \f\/(d0 + |/|). Based on this relation, explain why your nose looks bigger than the rest of your face when looking into a convex mirror.

29. (II) A spherical mirror of focal length / produces an image of an object with magnification m. (a) Show that the object

is a distance d0 = f ( 1 — —) from the reflecting side of V m )

the mirror. (b) Use the relation in part (a) to show that, no matter where an object is placed in front of a convex mirror, its image will have a magnification in the range 0 < m < + 1 .

30. (Ill) An object is placed a distance r in front of a wall, where r exactly equals the radius of curvature of a certain concave mirror. At what distance from the wall should this mirror be placed so that a real image of the object is formed on the wall? What is the magnification of the image?

31. (Ill) A short thin object (like a short length of wire) of length i is placed along the axis of a spherical mirror (perpendicular to the glass surface). Show that its image has length I' = m2l so the longitudinal magnification is equal to —m2 where m is the normal “lateral” magnification, Eq. 32-3. Why the minus sign? [Hint: Find the image positions for both ends of the wire, and assume £ is very small.]

32-4 Index of Refraction

32. (I) The speed of light in ice is 2.29 X 108 m/s. What is the index of refraction of ice?

33. (I) What is the speed of light in (a) ethyl alcohol, (b) lucite,(c) crown glass?

34. (I) Our nearest star (other than the Sun) is 4.2 light years away. That is, it takes 4.2 years for the light to reach Earth. How far away is it in meters?

35. (I) How long does it take light to reach us from the Sun,1.50 X 108 km away?

36. (II) The speed of light in a certain substance is 88% of its value in water. What is the index of refraction of that substance?

37. (II) Light is emitted from an ordinary lightbulb filament in wave-train bursts of about 10 8 s in duration. What is the length in space of such wave trains?

32-5 Snell's Law38. (I) A diver shines a flashlight upward from beneath the

water at a 38.5° angle to the vertical. At what angle does the light leave the water?

39. (I) A flashlight beam strikes the surface of a pane of glass (n = 1.56) at a 63° angle to the normal. What is the angle of refraction?

40. (I) Rays of the Sun are seen to make a 33.0° angle to the vertical beneath the water. At what angle above the horizon is the Sun?

41. (I) A light beam coming from an underwater spotlight exits the water at an angle of 56.0°. At what angle of incidence did it hit the air-water interface from below the surface?

42. (II) A beam of light in air strikes a slab of glass (n = 1.56) and is partially reflected and partially refracted. Determine the angle of incidence if the angle of reflection is twice the angle of refraction.

43. (II) A light beam strikes a 2.0-cm-thick piece of plastic with a refractive index of 1.62 at a 45° angle. The plastic is on top of a 3.0-cm- thick piece of glass for which n = 1.47. What is the distance D in Fig. 32-48?

44.

45.


(II) An aquarium filled with water has flat glass sides whose index of refraction is 1.56. A beam of light from outside the aquarium strikes the glass at a 43.5° angle to the perpendicular (Fig. 32-49). What is the angle of this light ray when it enters (a) the glass, and then(b) the water? (c) What would Glassbe the refracted angle if the ray Airentered the water directly?


4 3 . 5 ^ ^

Water

(II) In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point2.5 m from his foot at the edge of the pool (Fig. 32-50). Where does the spot of light hit the bottom of the pool, measured from the bottom of the wall beneath his foot, if the pool is 2.1 m deep?


Problems 861

46. (II) The block of glass (n = 1.5) shown in cross section in Fig. 32-51 is surrounded by air. A ray of light enters the block at its left-hand face with incident angle 6\ and reemerges into the air from the right-hand face directed parallel to the block’s base. Determine 6 i.


47. (II) A laser beam of diameter d\ = 3.0 mm in air has an incident angle 6\ = 25° at a flat air-glass surface. If the index of refraction of the glass is n = 1.5, determine the diameter d2 of the beam after it enters the glass.

48. (II) Light is incident on an equilateral glass prism at a 45.0° angle to one face, Fig. 32-52. Calculate the angle at which light emerges from the opposite face. Assume that n = 1.54.


49. (II) A triangular prism made of crown glass (n = 1.52) with base angles of 30.0° is surrounded by air. If parallel rays are incident normally on its base as shown in Fig. 32-53, what is the angle <f> between the two emerging rays?


50. (II) Show in general that for a light beam incident on a uniform layer of transparent material, as in Fig. 32-24, the direction of the emerging beam is parallel to the incident beam, independent of the incident angle 0. Assume the air on the two sides of the transparent material is the same.

51. (Ill) A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 32-24. Show that if the incident angle 0 is small, the emerging ray is displaced a distance d = td(n — 1 )/n, where t is the thickness of the glass, 0 is in radians, and d is the perpendicular distance between the incident ray and the (dashed) line of the emerging ray (Fig. 32-24).

32-6 Visible Spectrum; Dispersion52. (I) By what percent is the speed of blue light (450 nm) less

than the speed of red light (680 nm), in silicate flint glass (see Fig. 32-28)?

53. (I) A light beam strikes a piece of glass at a 60.00° incident angle. The beam contains two wavelengths, 450.0 nm and700.0 nm, for which the index of refraction of the glass is 1.4831 and 1.4754, respectively. What is the angle between the two refracted beams?

54. (II) A parallel beam of light containing two wavelengths, Xl = 465 nm and A2 = 652 nm, enters the silicate flint glass of an equilateral prism as shown in Fig. 32-54. At what angle does each beam leave the prism (give angle with normal to the face)?See Fig. 32-28.


55. (I ll) A ray of light with wavelength A is incident from air at precisely 60° ( = 0) on a spherical water drop of radius r and index of refraction n (which depends on A). When the ray reemerges into the air from the far side of the drop, it has been deflected an angle cf> from its original direction as shown in Fig. 32-55. By how ̂much does the value \ of <f> for violet light _ ^no\(n = 1.341) differ from the value for red light (n = 1.330)?


56. (Ill) For visible light, the index of refraction n of glass is roughly 1.5, although this value varies by about 1% across the visible range. Consider a ray of white light incident from air at angle Qi onto a flat piece of glass, (a) Show that, upon entering the glass, the visible colors contained in this incident ray will be dispersed over a range A02 of refracted angles given approximately by

sin di A nA 02 ~ . : *

\ / n 2 - sin2 6i n

[Hint: For x in radians, (d/dx)(siri -1*) = i/V i - *2.](b) If Qi = 0°, what is A02 in degrees? (c) If di = 90°, what is A02 in degrees?

32-7 Total Internal Reflection57. (I) What is the critical angle for the interface between water

and diamond? To be internally reflected, the light must start in which material?

58. (I) The critical angle for a certain liquid-air surface is 49.6°. What is the index of refraction of the liquid?

59. (II) A beam of light is emitted in a pool of water from a depth of 72.0 cm. Where must it strike the air-water interface, relative to the spot directly above it, in order that the light does not exit the water?

60. (II) A ray of light, after entering a light fiber, reflects at an angle of 14.5° with the long axis of the fiber, as in Fig. 32-56. Calculate the distance along the axis of the fiber that the light ray travels between successive reflections off the sides of the fiber. Assume that the fiber has an index of refraction of 1.55 and is 1.40 X 10_4m in diameter.

FIGURE 32-56 Problem 60.


61. (II) A beam of light is emitted 8.0 cm beneath the surface of a liquid and strikes the surface 7.6 cm from the point directly above the source. If total internal reflection occurs, what can you say about the index of refraction of the liquid?

62. (II) Figure 32-57 shows a liquid-detecting prism device that might be used inside a washing machine or other liquid-containing appliance. If no liquid covers the prism’s hypotenuse, total internal reflection of the beam from the light source produces a large signal in the light sensor. If liquid covers the hypotenuse, some light escapes from the prism into the liquid and the light sensor’s signal decreases. Thus a large signal from the light sensor indicates the absence of liquid in the reservoir. If this device is designed to detect the presence of water, determine the allowable range for the prism’s index of refraction n. Will the device work properly if the prism is constructed from (inexpensive) lucite? For lucite, n = 1.5.


64. (II) (a) What is the minimum index of refraction for a glass or plastic prism to be used in binoculars (Fig. 32-33) so that total internal reflection occurs at 45°? (b) Will binoculars work if their prisms (assume n = 1.58) are immersed in water?(c) What minimum n is needed if the prisms are immersed in water?

65. (Ill) Suppose a ray strikes the left face of the prism in Fig. 32-52 at 45.0° as shown, but is totally internally reflected at the opposite side. If the apex angle (at the top) is 6 = 60.0°, what can you say about the index of refraction of the prism?

66. (Ill) A beam of light enters the end of an optic fiber as shown in Fig. 32-59. (a) Show that we can guarantee total internal reflection at the side surface of the material (at point A), if the index of refraction is greater than about 1.42. In other words, regardless of the angle a, the light beam reflects back into the material at point A, assuming air outside.

I

FIGURE 32-59Problem 66. material

*32-8 Refraction at Spherical Surface*67. (II) A 13.0-cm-thick plane piece of glass (n = 1.58) lies on

the surface of a 12.0-cm-deep pool of water. How far below the top of the glass does the bottom of the pool seem, as viewed from directly above?

63. (II) Two rays A and B travel down a cylindrical optical fiber of diameter d = 75.0 /z,m, length £ = 1.0 km, and index of refraction ni = 1.465. Ray A travels a straight path down the fiber’s axis, whereas ray B propagates down the fiber by repeated reflections at the critical angle each time it impinges on the fiber’s boundary. Determine the extra time Af it takes for ray B to travel down the entire fiber in comparison with ray A (Fig. 32-58), assuming (a) the fiber is surrounded by air, (b) the fiber is surrounded by a cylindrical glass “cladding” with index of refraction n2 = 1.460.

75.0 /xm j-g

= 1.465


* 68. (II) A fish is swimming in water inside a thin spherical glass bowl of uniform thickness. Assuming the radius of curvature of the bowl is 28.0 cm, locate the image of the fish if the fish is located: (a) at the center of the bowl; (b) 20.0 cm from the side of the bowl between the observer and the center of the bowl. Assume the fish is small.

“69. (Ill) In Section 32-8, we derived Eq. 32-8 for a convex spherical surface with n2 > nt. Using the same conventions and using diagrams similar to Fig. 32-37, show that Eq. 32-8 is valid also for (a) a convex spherical surface with n2 < nu(b) a concave spherical surface with n2 > nt, and (c) a concave spherical surface with n2 < nx.

*70. (Ill) A coin lies at the bottom of a 0.75-m-deep pool. If a viewer sees it at a 45° angle, where is the image of the coin, relative to the coin? [Hint. The image is found by tracing back to the intersection of two rays.]

Problems 863

| General Problems71. Two identical concave mirrors are set facing each other 1.0 m

apart. A small lightbulb is placed halfway between the mirrors. A small piece of paper placed just to the left of the bulb prevents light from the bulb from directly shining on the left mirror, but light reflected from the right mirror still reaches the left mirror. A good image of the bulb appears on the left side of the piece of paper. What is the focal length of the mirrors?

72. A slab of thickness D, whose two faces are parallel, has index of refraction n. A ray of light incident from air onto one face of the slab at incident angle Qi splits into two rays A and B. Ray A reflects directly back into the air, while B travels a total distance ft within the slab before reemerging from the slab’s face a distance d from its point of entry (Fig. 32-60). (a) Derive expressions for I and d in terms of D, n, and 6 i . (b) For normal incidence (i.e., Qi = 0°) show that your expressionsyield the expect values for £ and


73. Two plane mirrors are facing each other 2.2 m apart as in Fig. 32-61. You stand 1.5 m away from one of these mirrors and look into it. You will see multiple images of yourself.(a) How far away from you are the first three images of yourself in the mirror in front of you? (b) Are these first three images facing toward you or away from you?


74. We wish to determine the depth of a swimming pool filled with water by measuring the width (x = 5.50 m) and then noting that the bottom edge of the pool is just visible at an angle of 13.0° above the horizontal as shown in Fig. 32-62. Calculate the depth of the pool.

75. A 1.80-m-tall person stands 3.80 m from a convex mirror and notices that he looks precisely half as tall as he does in a plane mirror placed at the same distance. What is the radius of curvature of the convex mirror? (Assume that sin 0 ~ 0.) [Hint: The viewing angle is half.]


76. The critical angle of a certain piece of plastic in air is 0C = 39.3°. What is the critical angle of the same plastic if it is immersed in water?

77. Each student in a physics lab is assigned to find the location where a bright object may be placed in order that a concave mirror, with radius of curvature r = 46 cm, will produce an image three times the size of the object. Two students complete the assignment at different times using identical equipment, but when they compare notes later, they discover that their answers for the object distance are not the same. Explain why they do not necessarily need to repeat the lab, and justify your response with a calculation.

78. A kaleidoscope makes symmetric patterns with two plane mirrors having a 60° angle between them as shown in Fig. 32-63. Draw the location of the images (some of them images of images) of the object placed between the mirrors.


79. When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle 5, Fig. 32-64. Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle (f>, and show that the minimum deviation angle, <5m, is related to the prism’s index of refraction n by

sin \ ((f> + Sm) n sin 0/2

[Hint: For 0 in radians, (d/d0)(sin~16) = 1/Vl - 02.]


80. If the apex angle of a prism is <f> = 72° (see Fig. 32-64), what is the minimum incident angle for a ray if it is to emerge from the opposite side (i.e., not be totally internally reflected), given n = 1.58?

81. Fermat’s principle states that “light travels between two points along the path that requires the least time, as compared to other nearby paths.” From Fermat’s principle derive (a) the law of reflection (6[ = 0r) and (b) the law of refraction (Snell’s law). [Hint: Choose two appropriate points so that a ray between them can undergo reflection or refraction. Draw a rough path for a ray between these points, and write down an expression of the time required for light to travel the arbitrary path chosen. Then take the derivative to find the minimum.]

* 82. Suppose Fig. 32-36 shows a cylindrical rod whose end has a radius of curvature R = 2.0 cm, and the rod is immersed in water with index of refraction of 1.33. The rod has index of refraction 1.53. Find the location and height of the image of an object 2.0 mm high located 23 cm away from the rod.

83.


H Numerical/ComputerH87. (II) The index of refraction, n, of crown flint glass at

different wavelengths (A) of light are given in the Table below.

A (nm) n

FIGURE 32-65Problem 83. h-d-H

84. An object is placed 15 cm from a certain mirror. The image is half the height of the object, inverted, and real. How far is the image from the mirror, and what is the radius of curvature of the mirror?

85. The end faces of a cylindrical glass rod (n = 1.51) are perpendicular to the sides. Show that a light ray entering an end face at any angle will be totally internally reflected inside the rod when it strikes the sides. Assume the rod is in air. What if it were in water?

86. The paint used on highway signs often contains small transparent spheres which provide nighttime illumination of the sign’s lettering by retro-reflecting vehicle headlight beams. Consider a light ray from air incident on one such sphere of radius r and index of refraction n. Let 6 be its incident angle, and let the ray follow the path shown in Fig. 32-66, so that the ray exits the sphere in the direction exactly antiparallel to its incoming direction. Consid- \ering only rays for which sin 0 can be approximated as 0, determine the required value for n.

10601.50586

546.11.51978

365.01.54251

312.51.5600

Make a graph of n versus A. The variation in index of refraction with wavelength is given by the Cauchy equation n = A + B /A2. Make another graph of n versus 1 /A 2 and determine the constants A and B for the glass by fitting the data with a straight line.

H88. (Ill) Consider a ray of sunlight incident from air on a spherical raindrop of radius r and index of refraction n. Defining 0 to be its incident angle, the ray then follows the path shown in Fig. 32-67, exiting the drop at a “scattering angle” <f> compared with its original incoming direction, (a) Show that (f) = 180° + 20 - 4 siiT1 (sin 0/n). (b) The parallel rays of sunlight illuminate a raindrop with rays of all possible incident angles from 0° to 90°. Plot 4> vs. 0 in the range 0° < 0 < 90°, in 0.5° steps, assuming n = 1.33 as is appropriate for water at visible-light wavelengths. (c) From your plot, you should find that a fairly large fraction of the incident angles have nearly the same scattering angle. Approximately what fraction of the possible incident angles is within roughly 1° of (f> = 139°? [This subset of incident rays is what creates the rainbow. Wavelength- dependent variations in n cause the rainbow to form at slightly different <f> for the various visible colors.]


Answers to Exercises

A: No. D : Toward.B : (b). E: The face becomes shiny: total internal reflection.C: (a). F: 1.414.

An optical fiber is a long transparent cylinder of diameter d and index of refraction n. If this fiber is bent sharply, some light hitting the side of the cylinder may escape rather than reflect back into the fiber (Fig. 32-65). What is the smallest radius r at a short bentsection for which total internal reflection will be assured for light initially travelling parallel to the axis of the fiber?

General Problems 865

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Light: Reflection and Refraction · 838 CHAPTER 32 Light: Reflection and Refraction. When light is incident upon a rough surface, even microscopically rough such as this page, it