23-1 The Ray Model of Light

Transcription

1 23-1 The Ray Mdel f Light We will start ur investigatin f gemetrical ptics (ptics based n the gemetry f similar triangles) by learning the basics f the ray mdel f light. We will then apply this mdel t understand reflectin and mirrrs, in this chapter, and refractin and lenses, in chapter 24. Using the triangles that result frm applying the ray mdel, we will derive equatins we can apply t prect where the image created by a mirrr r lens will be frmed. A ray is as a narrw beam f light that tends t travel in a straight line. An example f a ray is the beam f light frm a laser r laser pinter. In the ray mdel f light, a ray travels in a straight line until it hits smething, like a mirrr, r an interface between tw fferent materials. The interactin between the light ray and the mirrr r interface generally causes the ray t change rectin, at which pint the ray again travels in a straight line until it encunters smething else that causes a change in rectin. An example in which the ray mdel f light applies is shwn in Figure 23.1, in which the laser beam travels in straight lines between the mirrrs it is interacting with. Even after striking the bright bject at the center, the many beams fllw straight paths. Figure 23.1: The phtgraph shws a situatin in which the ray mdel f light applies. The laser beam fllws straight lines as it travels between each mirrr that it interacts with. (Pht cret: Digital Visin) A laser emits a single ray f light, but we can als apply the ray mdel in situatins in which a light surce sends ut many rays, in many rectins. Examples f such surces include the filaments f light bulbs, and the Sun. If we are far away frm such a surce, in relatin t the size f the surce itself, we ften treat the surce as a pint surce, and assume that the surce emits light, usually in all rectins, frm a single pint. Light bulbs, and the Sun, are ften treated as pint surces. In ther situatins, such as when we are clse t a light bulb that has a lng filament, we treat the surce as a stributed surce. Each pint n the surce can be treated as a pint surce, s a stributed surce is like a cllectin f pint surces, as shwn in Figure Figure 23.2: (a) A pint surce f light emits light unifrmly in all rectins. (b) A stributed surce f light, such as a light-bulb filament in the shape f a line, can generally be treated as a cllectin f pint surces. Wave frnts In adtin t rays f light, we will als mentin wave frnts. A wave frnt is a surface cnnecting light that was emitted by the light surce at the same time. As shwn in Figure 23.3, the wave frnts fr a pint surce are spherical shells centered n the surce, which prpagate away frm the surce at the speed f light. Fr a beam f light, like that frm a flashlight, in which the rays are parallel, the wave frnts are parallel lines that are perpencular t the beam. Chapter 23 Reflectin and Mirrrs Page 1

2 Figure 23.3: (a) Fr a pint surce, the wave frnts are spherical shells that are centered n the surce. The larger the raus f the wave frnt, the mre time has passes since the light was emitted. (b) When the rays are part f a beam f light that is traveling in a particular rectin, the wave frnts are parallel lines that are perpencular t the beam. Shadws The ray mdel f light can als be used t understand shadws. Figure 23.4 shws hw the shadw cast by a pint surce can be larger than the bject creating the shadw, while that frm parallel rays is the same size as the bject, as lng as the surface n which the shadw is cast is perpencular t the rectin f the rays. Distributed surces create mre cmplicated shadws, but they can be understd as the superpsitin f the shadws frm multiple pint surces. Figure 23.4: We can use the ray mdel f light, in which travels in straight lines, t explain shadws. When the light surce is a pint surce (a), the shadw is generally larger than the bject casting the shadw. When the light rays are all ging in the same rectin, hwever, the shadw is the same size f the bject when the shadw is cast n a surface that is perpencular t the light rays. Treating surces that d nt themselves emit light as light surces In sme cases, we will use bjects that actually emit light, such as a light bulb r the Sun, as the bjects that send light tward a mirrr r lens. In ther cases we will use bjects, such as yu, that d nt emit light themselves. Hw can we d this? In general, bjects that d nt emit light themselves are illuminated by ther light surces. As shwn in Figure 23.5, such surces can be treated as if they emit light, because they scatter much f the light incident n them in many fferent rectins. Figure 23.5: An illuminated bject can itself be treated as a surce f light, because much f the light shining n it is scattered ff the bject in all rectins. Hw we see bjects T see an bject, rays f light need either t be emitted by, r reflected frm, the bject, and then pass int ur eyes. Our brains assume that the rays f light travel in straight lines, s we trace the rays f light back until they meet at the lcatin f the bject, as shwn in Figure Figure 23.6: Objects send ut light in many rectins. We see the bject if enugh f this light enters ur eye. Only a small number f the rays are shwn fr this bject, clr-cded red fr rays frm the tp, blues fr rays frm the middle, and green fr rays frm the bttm f the bject. Related End-f-Chapter Exercises: Essential Questin 23.1: The Sun is a very large bject, much larger than the Earth. Give an example in which we can treat the Sun as a pint surce when applying the ray mdel. Give an example in which the Sun must be treated as a stributed surce f light. Chapter 23 Reflectin and Mirrrs Page 2

3 Answer t Essential Questin 23.1: T explain the frmatin f yur wn shadw n a sunny day, we can treat the Sun as a pint surce lcated 150 millin km away. On the ther hand, the shadw that the Earth casts n the Mn during a lunar eclipse has a very dark regin (the umbra) and a semi-dark regin (the penumbra). This mre cmplex shadw pattern can be partly explained by treating the Sun as a stributed surce The Law f Reflectin; Plane Mirrrs A ray f light that reflects frm a surface beys a very simple rule, knwn as the law f reflectin. See, als, the illustratins in Figure The Law f Reflectin: fr a ray f light reflecting frm a surface, the angle f incidence is equal t the angle f reflectin. These angles are generally measured frm the nrmal (perpencular) t the surface. Figure 23.7: In each case, the ray beys the Law f Reflectin, in that the angle f incidence, measured frm the nrmal, is equal t the angle f reflectin. In the specific examples shwn, bth the incident ray and the reflected ray are at an angle, measured frm the nrmal, f (a) 60, (b) 45, and (c) 30. A surface acts as a mirrr when the Law f Reflectin is fllwed n a large scale, as shwn in Figure 23.8 (a). In that case, the whle beam f light, with many parallel rays, reflects as expected accrng t the law. This is knwn as specular reflectin: mirrr-like reflectin that preserves the wave-frnt structure. In Figure 23.8 (b), hwever, the surface des nt seem t bey the law f reflectin. If we lk at the magnified view, in (c), hwever, we see that the surface is irregular. The law f reflectin is beyed fr each ray invidually, but the irregularities in the surface cause the rays t mve ff in many fferent rectins after being reflected. This is knwn as ffuse reflectin: reflectin in which the wave frnts are nt preserved. Diffuse reflectin explains why sme surfaces that appear t be flat, such as a table r a rad, d nt act as mirrrs. As far as light is cncerned, these surfaces are far frm flat. Figure 23.8: (a) Specular reflectin frm a flat mirrr, in which all rays reflect at the same angle. (b) Many flat surfaces exhibit ffuse reflectin, in which rays reflect at fferent angles. (c) A magnified view f the situatin in (b). Even thugh the surface may appear flat t us, the surface is actually quite irregular as far as light is cncerned. The surface in Figure 23.8(a) is knwn as a plane mirrr. Cmmn examples are the mirrrs in every bathrm. When we lk at urselves in such a mirrr, where d we see ur image? Hw large is the image? T answer such questins, we can use a ray agram t determine where an image is frmed and what its characteristics are. EXPLORATION 23.2 Using a ray agram t find the lcatin f an image Figure 23.9: An arrw lcated sme stance in frnt f a plane mirrr. Step 1 An arrw is placed in frnt f a vertical plane mirrr, as shwn in Figure Sketch tw rays f light, which travel in fferent rectins, that leave the tip f the arrw and reflect ff the mirrr. Shw the rectin f these rays after they reflect frm the mirrr. Chapter 23 Reflectin and Mirrrs Page 3

4 Figure 23.10: A selectin f reflected rays. In each case, the reflected rays are drawn in the rectin that is cnsistent with the law f reflectin. Figure shws a number f rays leaving the arrw and reflecting frm the mirrr, beying the law f reflectin. One f these, the hrizntal ray, in red, that strikes the mirrr at a 0 angle f incidence (measured frm the nrmal t the mirrr) is special, in that it reflects back alng the path the ray came in n, and is thus easy t draw. Hwever, yu d nt need t use this particular ray in the ray agram any tw rays f light that reflect frm the mirrr can be used. Step 2 The pint where the reflected rays meet is where the tip f the image is lcated. Sketch the image f the arrw. The reflected rays verge as they travel away frm the mirrr t the left, but if we extend the reflected rays back thrugh the mirrr t the right, we find a pint where they intersect. This is where the image f the tip f the arrw is lcated. Nte that all the reflected rays, when they are extended back, pass thrugh this pint, which is why we can use any tw reflected rays t create the ray agram. Because the base f the arrw, which we call the bject, is lcated n the principal axis (the hrizntal line bisecting the mirrr), we knw that the base f the image will als be lcated there, s we draw an image f the arrw between the pint where the image f the tip is and the principal axis. Figure 23.11: Fr a plane mirrr, the reflected rays must be extended back thrugh the mirrr t find the lcatin where they meet. Because the rays left the tip f the arrw, the pint where the reflected rays meet is the lcatin f the tip f the image f the arrw. Step 3 Prve that the image is lcated as shwn in Figure by drawing tw mre ray agrams, ne shwing the lcatin f the image f the midpint f the arrw, and ne shwing the lcatin f the image f the base (bttm) f the arrw. Figure cmbines the ray agram fr the tip with thse f the arrw s base and midpint, shwing that the image really is at the lcatin shwn in Figure Figure 23.12: We can sketch a ray agram fr any pint n an bject t find the lcatin f its image. In this case, the red rays are fr the tip f the arrw, the blue fr the midpint, and the green fr the base. Step 4 Cmpare the reflected rays in Figure t the rays in Figure Our brains cannt tell the fference between the tw situatins, which is why we see an image f the arrw frmed at the lcatin shwn in Figure Figure 23.13: Our brains trace the reflected rays back alng straight lines until they meet, and we see an image at that lcatin. Key idea fr ray agrams: The lcatin f the image f any pint n an bject, when the image is created by a mirrr, can be fund by drawing rays f light that leave that pint n the bject and reflect frm the mirrr. The rectin f the reflected rays must be cnsistent with the law f reflectin. The pint where the reflected rays meet is where the image f that pint is. Related End-f-Chapter Exercises: 1, 2, 4, 5 Essential Questin 23.2: First, make a prectin. When the arrw in Explratin 20.2 is mved clser t the mirrr, will its image be larger, smaller, r the same size as the image we fund in Step 2 abve? Sketch a new ray agram t check yur prectin. Chapter 23 Reflectin and Mirrrs Page 4

5 Answer t Essential Questin 23.2: Images frm plane mirrrs are always the same size as the riginal bject. We can see this in the ray agram in Figure Figure 23.14: Cmpare this figure t Figure Mving the bject clser t the mirrr mves the image clser t the mirrr, but the height f the image is still equal t the height f the bject Spherical Mirrrs: Ray Diagrams Let us mve nw frm plane mirrrs t spherical mirrrs, which curve like the surface f a sphere. Spherical mirrrs can be cnvex, such as the mirrrs n the passenger side f cars, r cncave, such as shaving r makeup mirrrs. Unlike plane mirrrs, which always prduce an image that is the same size as the bject, the image in a cnvex mirrr is always smaller than the bject, while the image in a cncave mirrr can be larger, smaller, r the same size as the bject. The fcal pint f a spherical mirrr is defined by what the mirrr des t a set f rays f light that are parallel t ne anther and t the principal axis f the mirrr. As shwn in Figure 23.15, a cncave mirrr reflects the rays s they cnverge t pass thrugh the fcal pint, F. A cnvex mirrr, in cntrast, reflects parallel rays s that they verge away frm the fcal pint. Nte that each ray beys the law f reflectin when it reflects frm the mirrr. The lcatin f the fcal pint depends n the curvature f the mirrr. The smaller the raus f curvature f the mirrr, the clser the fcal pint is t the mirrr s surface. Figure 23.15: Fcal pints are shwn fr fur fferent mirrrs. In (a) and (b), cncave mirrrs reflect parallel rays s that they cnverge t a single pint called the fcal pint. In (c) and (d), cnvex mirrrs reflect parallel rays s that they verge away frm the mirrr s fcal pint. The lcatin f the fcal pint depends n the mirrr s raus f curvature. In each case, C is the mirrr s center f curvature, R is the raus f curvature, and F is the fcal pint. What we shw here fr spherical mirrrs is an apprximatin, valid fr rays that are nt t far frm the principal axis, in relatin t the magnitude f the mirrr s fcal length. A mirrr actually needs t have a parablic shape t reflect all parallel rays thrugh ne pint (r away frm ne pint, fr a verging mirrr). The fact that spherical mirrrs d nt really bring all such rays t a pint (r verge them away frm a pint) is a defect called spherical aberratin. Fcal length f a spherical mirrr: The fcal pint f a spherical mirrr is lcated halfway between the surface f the mirrr and the mirrr s center f curvature. Thus, the fcal length f a spherical mirrr has a magnitude f R/2, where R is the raus f curvature f the mirrr. By cnventin, an bject that verges parallel light rays has a negative fcal length ( f ), while an bject that cnverges parallel light rays has a psitive fcal length. Thus: R R Fr a cncave mirrr: f =+. (Eq. 23.1) Fr a cnvex mirrr: f =. (Eq. 23.2) 2 2 In the limit that the raus f curvature appraches infinity, the mirrr becmes a plane mirrr and the fcal length is either + infinity r infinity (it des nt matter which sign is used). Chapter 23 Reflectin and Mirrrs Page 5

6 Figure 23.16: (a) Fr light rays that are parallel t the principal axis, a cnverging mirrr re-unites the wave frnt (in purple) at the fcal pint. Frm there, the wave frnt verges almst as if the fcal pint is a pint surce. (b) A verging mirrr deflects the parallel rays s the wave frnts appear t verge frm the mirrr s fcal pint. Fllwing the wave frnts Figure shws what spherical mirrrs d t wave frnts. Fr the cnverging mirrr, the waves take the same time t get frm the left t the fcal pint. Fr the verging mirrr, nce the waves reflect frm the mirrr, it is as if they left the fcal pint at the same time. EXPLORATION 23.3 Ray agram fr a cnvex mirrr We will fllw a prcess similar t that f plane mirrrs t draw a ray agram fr a cnvex (verging) mirrr. Figure 23.17: An bject, represented by an arrw, is placed in frnt f a cnvex (verging) mirrr. The mirrr s center f curvature is shwn. Step 1 First, lcate the mirrr s fcal pint. Then, draw a light ray that leaves the tip f the bject (its tp) and ges parallel t the principal axis. Shw hw this parallel ray reflects frm the mirrr. The fcal pint is halfway between the pint where the principal axis intersects the mirrr, and the center f curvature. Fr a cnvex mirrr, all parallel rays appear t verge frm the fcal pint, s we draw the reflected ray reflecting alng a line that takes it rectly away frm the fcal pint. Figure 23.18: The parallel ray reflects frm the mirrr in such a way that it travels rectly away frm the fcal pint. Step 2 Sketch a secnd ray that leaves the tip f the bject and reflects frm the mirrr. Using yur reflected rays, draw the image. One useful ray, in blue in Figure 23.19, reflects frm the pint n the mirrr that the principal axis passes thrugh. At that lcatin, the reflectin is like that frm a vertical plane mirrr. Anther useful ray, in green, ges straight tward the mirrr s center f curvature. This ray has a 0 angle f incidence, and thus reflects back alng the same line. The reflected rays verge n the left f the mirrr, but we can extend them back t meet n the right side f the mirrr, shwing us where the tip f the image is. The image, smaller than the bject, is drawn frm the tip dwn t the principal axis. Figure 23.19: In adtin t the parallel ray, tw ther rays are easy t draw the reflectin fr. The ray (in blue) that strikes the mirrr at the principal axis reflects as if the mirrr was a vertical plane mirrr. The ray in green travels rectly tward the center f curvature, striking the mirrr at 90 t the surface, and thus reflecting straight back. Key idea: As with a plane mirrr, when a number f rays leave the same pint n an bject and reflect frm a spherical mirrr, the crrespnng pint n the image is lcated at the intersectin f the reflected rays. Related End-f-Chapter Exercises: 6, 9, 18, 60, 61. Essential Questin 23.3: (a) Mfy the ray agram in Figure t shw what happens t the image when the bject is mved clser t the mirrr. (b) Add several mre rays (leaving the tip f the bject) t yur mfied ray agram, shwing what the rays d when they reflect frm the mirrr. Hw d yu knw hw t draw the reflected rays? Chapter 23 Reflectin and Mirrrs Page 6

7 Answer t Essential Questin 23.3: (a) The parallel ray is nice t wrk with, because its reflectin des nt change rectin when the bject is mved left r right. The ray that strikes the mirrr at the principal axis, hwever, cmes in at a larger angle f incidence, and thus reflects at a larger angle. As shwn in Figure 23.20, when the bject is clser t the mirrr, the image increases in size and mves clser t the mirrr. Figure 23.20: The new ray agram shws that mving the bject clser t the mirrr results in the image increasing in size (but remaining smaller than the bject) and mving clser t the surface f the mirrr. (b) We have tw ways f knwing hw t draw the reflected rays prperly. First, the law f reflectin must be beyed when the rays reflect frm the mirrr. Secnd, we knw that when we extend the reflected rays back, they will pass thrugh the tip f the image, which we lcated in Figure Three adtinal rays are shwn in Figure Figure 23.21: Fr all rays f light that leave the tip f the bject and reflect frm the mirrr, the reflected rays can be extended back t pass thrugh the tip f the image A Qualitative Apprach: Image Characteristics S far, we have lked at ray agrams in tw cases, the case f a plane mirrr (sectin 23-2) and that f a cnvex mirrr (sectin 23-3). In bth cases, we had t extend the reflected rays back thrugh the mirrr t get the rays t intersect, giving us an image behind the mirrr. We see this all the time with plane mirrrs. If yu stand 1.0 m in frnt f a plane mirrr, yu see an image f yurself 1.0 m behind the mirrr. Is it always true that the image created by a mirrr is lcated behind the mirrr? We will first investigate the cncave mirrr, the last case we will deal with, and we will then summarize varius image characteristics. EXAMPLE 23.4 A ray agram fr a cncave mirrr An bject, represented by an arrw, is lcated 15.0 cm in frnt f a cncave mirrr that has a fcal length f cm, as shwn in Figure Sketch a ray agram t find the lcatin f the tip f the image f the arrw, and sketch the image n the agram. Figure 23.22: An bject in frnt f a cncave mirrr. The squares in the grid measure 1.0 cm 1.0 cm. SOLUTION Let s use the same prcedure we have used previusly, starting by drawing rays f light that leave the tip f the bject. Again, ne useful ray, shwn in red, is the ray that travels parallel t the principal axis. The cncave mirrr reflects this ray s that it passes thrugh the mirrr s fcal pint. A secnd ray that we knw hw t draw is the ne, in blue, that reflects frm the mirrr at the principal axis, reflecting at that pint as if the mirrr were a vertical plane mirrr. Nte that these tw rays meet t the left f the mirrr, giving us the lcatin f the tip f the image. As usual, we draw the image f the arrw frm that pint t the principal axis, because the base f the bject is als n the principal axis. Chapter 23 Reflectin and Mirrrs Page 7

8 Figure 23.23: A ray agram shwing tw f the several pssible rays we can draw t lcate the tip f the image. Wave frnts Figure shws hw the cnverging mirrr affects the wave frnts (in purple). The light leaving the tip f the bject, reflecting frm the mirrr, and arriving at the tip f the image, takes the same time n matter which path it takes. Figure 23.24: The wave frnts that leave the tip f the bject cnverge n the tip f the image the rays take the same time t reach the image. The wave frnts then verge away frm the image. T yur brain, the image lks like an bject. Image characteristics The image in Figure is quite fferent frm images we have seen earlier in the chapter. First, the image is inverted (upside dwn) cmpared t the bject. Secnd, the image is larger than the bject. Third, the image is frmed frm light rays that actually pass thrugh the image. Nte that cncave mirrrs d nt always frm images with these characteristics, as we will investigate in mre detail in sectin Fr nw, hwever, let s scuss sme general issues related t image characteristics. Upright r inverted? As we have seen, plane mirrrs and cnvex mirrrs prduce an upright image. This is an image that is in the same rientatin as the bject. An inverted image, like that in Example 23.4, is ne in which the image is upside dwn in relatin t the bject. Real r virtual? Mst f the images we see n a daily basis in mirrrs are virtual. A virtual image is ne that the light des nt actually pass thrugh. Instead, ur brains see an image there because, when we lk in the mirrr at the bject, ur brains are s used t light traveling in straight lines that we trace all the reflected rays back t their apparent surce, the pint behind the mirrr where the light appears t cme frm. Fr a single mirrr, when the image is virtual it is als upright. In Example 23.4, we saw a situatin in which the light rays passed thrugh the mirrr, creating a real image. Real images, frm cncave mirrrs, have a three-mensinal quality that virtual images d nt have, and it is wrth ging ut f yur way t see ne. Fr a single mirrr, when the image is real it is als inverted. Larger r smaller? A plane mirrr, as we investigated earlier, prduces an image that is always the same size as the bject. Cnvex (verging) mirrrs, n the ther hand, always prduce an image that is smaller than the bject. Cncave (cnverging) mirrrs, as we will investigate further in sectin 23-6, can prduce an image that is larger, smaller, r the same size as the bject. We will scuss these ideas in a mre quantitative way when we define the magnificatin f a mirrr, in sectin Related End-f-Chapter Exercises: 7, 37, 38. Essential Questin 23.4: Cnsider the ray agram in Figure If an bject f the same size f the image was placed at the image s psitin, where wuld its image be lcated? Chapter 23 Reflectin and Mirrrs Page 8

9 Answer t Essential Questin 23.4: The image wuld be lcated where the bject in Figure is lcated. This demnstrates an imprtant fact abut light rays they are reversible. As Figure shws, we can simply reverse the rectin f the rays frm Figure t btain the apprpriate ray agram. Nte that we can d this nly when the image is a real image. Figure 23.25: When the image is a real image, the ray agram is reversible A Quantitative Apprach: The Mirrr Equatin The branch f ptics that invlves mirrrs and lenses is generally called gemetrical ptics, because it is based n the gemetry f similar triangles. Let s investigate this gemetry, and use it t derive an imprtant relatinship between the image stance, bject stance, and the fcal length. Let s lk again at the ray agram we drew in Figure f sectin 23-4, shwn again here in Figure Figure 23.26: The ray agram we cnstructed in sectin 23-4, fr an bject in frnt f a cncave mirrr. Remve the red rays, and examine the tw triangles in Figure 23.26, ne shaded green and ne shaded yellw, bunded by the blue rays, the principal axis, and the bject and image. The tw triangles are similar, because the three angles in ne triangle are the same as the three angles in the ther triangle. We can nw define the fllwing variables: d is the bject stance, the stance f the bject frm the center f the mirrr; d i is the image stance, the stance f the image frm the center f the mirrr; h is the height f the bject; h i is the height f the image. Figure 23.27: Similar triangles, bunded by the principal axis, the bject and image, and the blue ray that reflects frm the mirrr. Using the fact that the ratis f the lengths f crrespnng sides in similar triangles are equal, we find that: hi h = d. (Equatin 23.3) The image height is negative because the image is inverted, which is why we need the minus sign in the equatin. Let s nw return t Figure 23.26, and remve the blue rays. This gives us the shaded similar triangles shwn in Figure Figure 23.28: Similar triangles, with tw sides bunded by the principal axis and the red ray, and a third side that is equal t the bject height (pink triangle) r the image height (blue triangle). We use an apprximatin, which is valid as lng as the bject height is relatively small, that the length f the pink triangle is f, the fcal length. Again, using the fact that the ratis f the lengths f crrespnng sides in similar triangles are equal, we find that: Chapter 23 Reflectin and Mirrrs Page 9

10 f hi =. f h Simplifying the left side, and bringing in equatin 23.3, we get: 1 =. f d Diving bth sides by d i gives: 1 1 = 1, which is generally written as: f d d = +. (Equatin 23.4: The mirrr equatin) f d The mnemnic If I d I can help yu t remember the mirrr equatin. i Often, we knw the fcal length f and the bject stance d, s equatin 23.4 can be slved fr d i, the image stance: d f = (Equatin 23.5: The mirrr equatin, slved fr the image stance) d f Sign cnventins We derived the mirrr equatin abve by using a specific case invlving a cncave mirrr. The equatin can als be applied t a plane mirrr, a cnvex mirrr, and all situatins invlving a cncave mirrr if we use the fllwing sign cnventins. The fcal length is psitive fr a cncave mirrr, and negative fr a cnvex mirrr. The image stance is psitive if the image is n the reflective side f the mirrr (a real image), and negative if the image is behind the mirrr (a virtual image). The image height is psitive when the image is abve the principal axis, and negative when the image is belw the principal axis. A similar rule applies t the bject height. Magnificatin The magnificatin, m, is defined as the rati f the height f the image ( h i ) t the height f the bject ( h ). Making use f Equatin 23.3, we can write the magnificatin as: hi m = =. (Equatin 23.6: Magnificatin) h d The relative sizes f the image and bject are as fllws: The image is larger than the bject if m > 1. The image and bject have the same size if m = 1. The image is smaller than the bject if m < 1. The sign f the magnificatin tells us whether the image is upright (+) r inverted ( ) cmpared t the bject. Related End-f-Chapter Exercises: Essential Questin 23.4: As yu are analyzing a spherical mirrr situatin, yu write an equatin that states: 1 = What is the value f 1/f in this situatin? What is f? f + 12 cm + 24 cm Chapter 23 Reflectin and Mirrrs Page 10

11 Answer t Essential Questin 23.4: T add fractins yu need t find a cmmn denminatr cm = + = + =. This gives f = = 8.0 cm. f + 12 cm + 24 cm + 24 cm + 24 cm + 24 cm Analyzing the Cncave Mirrr In sectin 23-4, we drew ne ray agram fr a cncave mirrr. Let s investigate the range f ray agrams we can draw fr such a mirrr. EXPLORATION 23.6 Ray agrams fr a cncave mirrr Step 1 Draw a ray agram fr an bject lcated 40 cm frm a cncave mirrr that has a raus f curvature f 20 cm. Verify the image lcatin n yur agram with the mirrr equatin. In drawing a ray agram, it is helpful t knw where the mirrr s fcal pint is. Fr a spherical mirrr, the fcal pint is halfway between the mirrr s center f curvature and the pint at which the principal axis intersects the mirrr. Thus, the fcal length in this case is +10 cm. In Figure 23.29, tw rays are shwn. One is the parallel ray, which leaves the tip f the bject, travels parallel t the principal axis, and reflects frm the mirrr s that it passes thrugh the fcal pint. The secnd ray reflects ff the mirrr at the pint at which the principal axis meets the mirrr, reflecting as if the mirrr was a vertical plane mirrr. Figure 23.29: A ray agram fr the situatin in which the bject is far frm the mirrr. The squares in the grid measure 1.0 cm 1.0 cm. Applying the mirrr equatin, in the frm f equatin 23.5, t find the image stance: d ( 40 cm ) ( +10 cm) 400 cm 2 f + = = = = cm. d f ( 40 cm ) ( +10 cm) 30 cm This image stance is cnsistent with the ray agram in Figure Step 2 Repeat step 1, with the bject nw mved t the center f curvature. The parallel ray fllws the same path as it d Figure As shwn in Figure 23.30, the ray (in blue) that reflects frm the center f the mirrr fllws a fferent path, because shifting the bject changes the angle f incidence fr that ray. This situatin is a special case. When the bject is lcated at the center f curvature, the image is inverted, als at the center f curvature, and the same size as the bject because the bject and image are the same stance frm the mirrr. Figure 23.30: When the bject is at the mirrr s center f curvature, s is the image. Applying the mirrr equatin t find the image stance, we get: ( ) ( ) ( ) ( ) d 20 cm +10 cm 200 cm 2 f + = = = =+ 20 cm, matching the ray agram. d f 20 cm +10 cm 10 cm Chapter 23 Reflectin and Mirrrs Page 11

12 Step 3 Repeat step 1, with the bject 15 cm frm the mirrr. N matter where the bject is, the parallel ray fllws the same path. The path f the secnd ray, in blue, depends n the bject s psitin. The ray agram (Figure 23.31) shws that the image is real, inverted, larger than the bject, and abut twice as far frm the mirrr as the bject. Figure 23.31: A ray agram fr a situatin in which the bject is between the mirrr s center f curvature and its fcal pint. Applying the mirrr equatin gives: d ( 15 cm ) ( +10 cm) 150 cm 2 f + = = = =+ 30 cm, matching the ray agram. d f 15 cm +10 cm 5.0 cm ( ) ( ) Step 4 Repeat step 1, with the bject at the mirrr s fcal pint. As shwn in Figure 23.32, the tw reflected rays are parallel t ne anther, and never meet. In such a case the image is frmed at infinity. Figure 23.32: A ray agram fr a situatin in which the bject is at the fcal pint. Applying the mirrr equatin gives: d ( 10 cm ) ( +10 cm) 100 cm 2 f + = = = =+, which agrees with the ray agram. d f 10 cm +10 cm 0 cm ( ) ( ) Step 5 Repeat step 1, with the bject 5.0 cm frm the mirrr. When the bject is clser t the mirrr than the fcal pint, the tw reflected rays verge t the left f the mirrr, and they must be extended back t meet n the right f the mirrr. The result is a virtual, upright image that is larger than the bject, as shwn in Figure Figure 23.33: A ray agram fr a situatin in when the bject is between the mirrr s surface and its fcal pint. gives: Applying the mirrr equatin t find the image stance ( ) ( ) ( ) ( ) d 5.0 cm +10 cm 50 cm 2 f + = = = = 10 cm. d f 5.0 cm +10 cm 5.0 cm Recalling the sign cnventin that a negative image stance is cnsistent with a virtual image, the result frm the mirrr equatin is cnsistent with the ray agram. Key idea fr cncave mirrrs: Depenng n where the bject is placed relative t a cncave mirrr s fcal pint, the mirrr can frm an image f the bject that is real r virtual. If the image is real, it can be larger than, smaller than, r the same size as the bject. If the image is virtual, the image is larger than the bject. Related End-f-Chapter Exercises: 23 and Essential Questin 23.6: When an bject is placed 20 cm frm a spherical mirrr, the image frmed by the mirrr is larger than the bject. What kind f mirrr is it? What, if anything, can yu say abut the mirrr s fcal length? Chapter 23 Reflectin and Mirrrs Page 12

13 Answer t Essential Questin 23.6: The mirrr must be cncave, because a cnvex mirrr cannt prduce an image that is larger than the bject. A cncave mirrr prduces an image larger than the bject nly when the bject is between the mirrr and twice the fcal length. S, twice the fcal length must be at least +20 cm, and the fcal length must be at least +10 cm. All we can say is that the fcal length is greater than r equal t +10 cm An Example Prblem Let s begin by scussing a general apprach we can use t slve prblems invlving mirrrs. We will then apply the methd t a particular situatin. A general methd fr slving prblems invlving mirrrs 1. Sketch a ray agram, shwing rays leaving the tip f the bject and reflecting frm the mirrr. Where the reflected rays meet is where the tip f the image is lcated. The ray agram gives us qualitative infrmatin abut the lcatin and size f the image and abut the characteristics f the image. 2. Apply the mirrr equatin and/r the magnificatin equatin. Make sure that the signs yu use match thse listed in the sign cnventin in sectin The equatins prvide quantitative infrmatin abut the lcatin and size f the image and abut the image characteristics. 3. Check the results f applying the equatins with yur ray agram, t see if the equatins and the ray agram give cnsistent results. Rays that are easy t draw the reflectins fr T lcate an image n a ray agram, yu need a minimum f tw rays. If yu draw mre than tw rays, hwever, yu can check the image lcatin yu find with the first tw rays. Remember, t, that yu can draw any number f rays reflecting frm the mirrr, and that all the rays shuld bey the law f reflectin. There are at least fur rays that are easy t draw the reflectins fr. These rays are shwn n Figure 23.34, and include: 1. The ray, in red, that ges parallel t the principal axis, and reflects s that it passes thrugh the fcal pint (cncave mirrr), r away frm the fcal pint (cnvex mirrr). 2. The ray, in blue, that reflects frm the pint n the principal axis that intersects the surface f the mirrr. The principal axis is perpencular t the surface f the mirrr, s the angle between the incident ray and the principal axis is the same as the angle between the reflected ray and the principal axis. 3. The ray, in green, that travels alng the straight line cnnecting the tip f the bject and the mirrr s center f curvature. This ray is incident n the mirrr alng the nrmal t the mirrr s surface, and thus reflects straight back alng the same line. 4. The ray, in purple, that travels alng the straight line cnnecting the tip f the bject and the fcal pint. This ray reflects t g parallel t the principal axis. Figure 23.34: An example f the fur rays that are easy t draw the reflectins fr. All the reflected rays meet at the tip f the image. Chapter 23 Reflectin and Mirrrs Page 13

14 EXAMPLE 23.7 Applying the general methd When yu stand in frnt f a mirrr that has a raus f curvature f 40 cm, yu see an image that is half yur size. What kind f mirrr is it? Hw far frm the mirrr are yu? Sketch a ray agram t check yur calculatins. SOLUTION In this case, let s first apply the equatins and then draw the ray agram. The mirrr culd be cnvex, because cnvex mirrrs always prduce images that are smaller than the bject. A cnvex mirrr prduces a virtual, upright image, s the sign f the magnificatin is psitive. Applying the magnificatin equatin, we get: 1 d m =+ = i 2 d, which tells us that 1 2 = d. Fr a cnvex mirrr, the fcal length is R/2, which in this case is 20 cm. Applying the mirrr equatin: = + = = f d d d d. Thus, we find that d = f =+ 20 cm, and we can shw that d = 10 cm. The ray agram fr this situatin is shwn in Figure 23.35, cnfirming the calculatins. Figure 23.35: A ray agram fr the slutin invlving a cnvex mirrr. Each bx n the grid measures 2 cm 2 cm. The slutin abve is nly ne f the pssible answers. The mirrr culd als be cncave, because a cncave mirrr can prduce a real, inverted image, s the sign f the magnificatin is negative. Applying the magnificatin equatin, we get: m = =, which tells us that =. 2 d d Fr a cncave mirrr the fcal length is +R/2, which in this case is +20 cm. Applying the mirrr equatin: = + = + =. f d d d d Thus, we find that d = 3f =+ 60 cm, and we can shw that d = + 30 cm. The ray agram fr this situatin is shwn in Figure 23.36, again cnfirming the calculatins abve. Figure 23.36: A ray agram fr the situatin invlving a cncave mirrr. Each bx n the grid measures 2 cm 2 cm. Related End-f-Chapter Exercises: 16, 17, 19, 20, 43. Essential Questin 23.7: Return t the situatin described in Example Wuld there still be tw slutins if the image was larger than the bject? Explain. i i Chapter 23 Reflectin and Mirrrs Page 14

15 Answer t Essential Questin 23.7: Yes, there wuld still be tw slutins, but we wuld nt have a slutin assciated with a cnvex mirrr, because the cnvex mirrr cannt prduce an image that is larger than the bject. Instead, bth slutins wuld be assciated with a cncave mirrr. One slutin wuld be fr a real image, and the ther slutin wuld be fr a virtual image. Chapter Summary Essential Idea: Reflectin and Mirrrs. T understand the image that is created by a mirrr, we make use f a simple mdel f light called the ray mdel. In the ray mdel, a ray f light travels in a straight line until it encunters an bject. When a ray f light reflects frm an bject, the light beys the Law f Reflectin. The Law f Reflectin The angle f incidence is equal t the angle f reflectin. These angles are generally measured frm the nrmal t the surface. Image Frmatin Fr a mirrr t frm an image f an bject, light rays must leave the bject and be reflected by the mirrr. If the rays leaving a single pint n the bject are reflected s that they pass thrugh a single pint, a real image is frmed. If, instead, such reflected rays appear t verge frm a single pint behind the mirrr (as is the case fr a typical bathrm mirrr), a virtual image is frmed. Ray Diagrams When drawing a ray agram, we generally shw rays leaving the tip f the bject and reflecting frm the mirrr. Where the reflected rays meet is where the tip f the image is lcated. The ray agram gives us qualitative infrmatin abut the lcatin and size f the image and abut the image characteristics. All rays bey the Law f Reflectin when they reflect frm the mirrr, but sme reflected rays are particularly easy t draw. A summary f fur such rays is given in sectin Plane and Spherical Mirrrs Type f mirrr Fcal length Image characteristics Plane The image is virtual, upright, the same size as the bject, and the same stance behind the mirrr that the bject is in frnt f the mirrr. Cnvex (verging) R/2, R being the mirrr s raus f curvature The image is virtual, upright, smaller than the bject, and lcated between the mirrr and the mirrr s fcal pint. Cncave (cnverging) +R/2 The image can be real r virtual, and larger than, smaller than, r the same size as the bject. See the table belw fr details. Table 23.1: A summary f the mirrrs we investigated in this chapter. Chapter 23 Reflectin and Mirrrs Page 15

16 Images frmed by a Cncave (Cnverging) Mirrr Object psitin Image psitin Image characteristics At the fcal pint. Real image with height f zer. Mving frm tward the center f curvature. Mving frm the fcal pint tward the center f curvature. The image is real, inverted, and smaller than the bject. The image mves clser t the center f curvature, and increases in height, as the bject is mved clser t the center f curvature. At the center f curvature. At the center f curvature. The image is real, inverted, and the same size as the bject. Mving frm the center f curvature tward the fcal pint. Mving frm the center f curvature tward infinity. The image is real, inverted, and larger than the bject. The image mves farther frm the mirrr, and increases in height, as the bject is mved clser t the fcal pint. At the fcal pint. At infinity. The image is at infinity, and is infinitely tall. Clser t the mirrr than the fcal pint. Behind the mirrr The image is virtual, upright, and larger than the bject. The image mves clser t the mirrr, and decreases in height, as the bject is mved clser t the mirrr. Table 23.1: A summary f the image psitins and characteristics fr varius bject psitins with a cncave mirrr. The mirrr equatin The mirrr equatin relates the bject stance, d, the image stance, d i, and the mirrr s fcal length, f. The mnemnic If I d I can help yu t remember the mirrr equatin = +. (Equatin 23.4: The mirrr equatin) f d d i d = d f f (Equatin 23.5: The mirrr equatin, slved fr the image stance) Sign cnventins The fcal length is psitive fr a cncave mirrr, and negative fr a cnvex mirrr. The image stance is psitive if the image is n the reflective side f the mirrr (a real image), and negative if the image is behind the mirrr (a virtual image). The image height is psitive when the image is abve the principal axis, and negative when the image is belw the principal axis. A similar rule applies t the bject height. The image height is psitive when the image is upright, and negative when the image is inverted. A similar rule applies t the bject height. Magnificatin The magnificatin, m, is the rati f the image height ( h i ) t the bject height ( h ). hi m = =. (Equatin 23.6: Magnificatin) h d The image is larger than the bject if m > 1. The image and bject have the same size if m = 1. The image is smaller than the bject if m < 1. The magnificatin is psitive if the image is upright, and negative if the image is inverted. Chapter 23 Reflectin and Mirrrs Page 16