24-4 Image Formation by Thin Lenses

Transcription

1 24-4 Image Frmatin by Thin Lenses Lenses, which are imprtant fr crrecting visin, fr micrscpes, and fr many telescpes, rely n the refractin f light t frm images. As with mirrrs, we draw ray agrams t help us t understand hw such images are frmed. Let s first begin by lking at what a lens des t a set f parallel rays f light, such as the five rays in Figure Figure 24.16: Five parallel rays f light. Hw can we change the rectin f the five rays f light in Figure s that they all pass thrugh the pint labeled F? Ray 3 already passes thrugh pint F, s we dn t need t change its rectin at all. We can deflect ray 2 with a triangular piece f glass, as shwn in Figure Passing frm air int the glass prism, the ray deflects tward the nrmal at that surface, while when it emerges back int the air the deflectin is away frm the nrmal at the secnd surface. We can fllw a similar prcess fr ray 1, except that we need t prduce a larger change in rectin fr ray 1 cmpared t ray 2. The glass prism we use fr ray 1 thus has its sides at a greater angle frm the vertical. Fr ray 4, we use an identical prism t that used fr ray 2, except that we invert it, and fr ray 5 the prism is identical t the prism fr ray 1, but inverted. Figure 24.17: We can deflect fur f the five rays, s that all five rays meet at pint F, by using prisms f the apprpriate shapes t prduce the deflectin required fr each beam. Nw, nt nly d we want all five rays t cnverge n pint F, but we want them t take equal times t travel frm the vertical dashed line in Figure t pint F. Remember that the light travels mre slwly in glass than in air. Because rays 1 and 5 travel the greatest stance, they need t pass thrugh the least amunt f glass. Ray 3 travels the shrtest stance, s we need t delay ray 3 by having it pass thrugh the thickest piece f glass. We can d this withut deflecting ray 3 by using a piece f glass with vertical sides, s that ray 3 is incident alng the nrmal. Rather than using varius invidual glass rectangles and prisms t d the jb, we can use a single piece f glass that is thickest in the middle. This piece f glass gets thinner, and its surfaces curve farther away frm the vertical, as yu mve away frm ray 3 (that is, as yu mve away frm the principal axis). This is shwn in Figure we use a lens. Pint F is a fcal pint f the lens. Lenses allw light t pass thrugh frm left-t-right r frm right-t-left, s a lens has tw fcal pints, ne n each side f the lens. Figure 24.18: A cnvex lens with surfaces that are spherical arcs brings all parallel rays t ne f the fcal pints, and ensures that the parallel rays take the same time t travel frm the vertical line t this fcal pint. Chapter 24 Refractin and Lenses Page 1

2 Hw des a cncave lens, which is thinner in the center than at the edges, affect parallel rays? As shwn in Figure 24.19, such lenses generally verge parallel rays away frm the fcal pint that is n the same side f the lens that the light cmes frm. Figure 24.19: The influence f a verging lens n a set f parallel rays. EXPLORATION 24.4 Using a ray agram t find the lcatin f an image When drawing ray agrams fr lenses, we fllw a prcess similar t that fr mirrrs. Figure 24.20: An bject lcated sme stance in frnt f a cnverging lens. Step 1 Figure shws an bject in frnt f a cnverging lens. Sketch tw rays f light, which travel in fferent rectins, that leave the tip (the tp) f the bject and pass thrugh the lens. Shw the rectin f these rays after they are refracted by the lens. Figure shws three rays that leave the tip f the bject and which are incident n the lens. Because the ray in red is parallel t the principal axis, it is refracted by the lens t pass thrugh the fcal pint that is t the right f the lens. The ray in green travels alng the line cnnecting the tip f the bject t the fcal pint n the left f the lens. This ray is refracted s that it is parallel t the principal axis. We knw this because f the reversibility f light rays if we reversed the rectin f the ray, it wuld cme frm the right and be refracted by the lens t pass thrugh the fcal pint n the right. Finally, the ray in blue passes thrugh the center f the lens withut changing rectin. This is an apprximatin, which is valid as lng as the lens is thin s the ray enters and exits the lens very clse t the principal axis. Figure 24.21: Three f the rays that are refracted by the lens. In general, we draw a ray changing rectin nce, inside the lens. Each ray really changes rectin twice, nce at each air-glass interface. We are drawing the ray s path incrrectly within the lens, but it is crrect utside f the lens, which is where it really matters. Step 2 The pint where the refracted rays meet is where the tip f the image is lcated. Use this infrmatin t sketch the image f the bject. In this situatin, the refracted rays meet at a pint t the right f the lens, belw the principal axis. Thus, we draw an inverted image between the pint where the image f the tip is (where the rays meet) and the principal axis. This is a real image, because the rays pass thrugh the image. All the rays take the same time t travel frm the tip f the bject t the tip f the image. Figure 24.22: Because the rays riginate at the tip f the bject, the pint where the refracted rays meet is the lcatin f the tip f the image f the bject. The base f the bject is n the principal axis, s the base f the image is n the principal axis, t. Key idea fr ray agrams: The lcatin f the image f any pint n an bject, when the image is created by a lens, can be fund by drawing rays f light that leave that pint n the bject and are refracted by the lens. The pint where the refracted rays meet (r where they appear t verge frm) is where the image f that pint is. Related End-f-Chapter Exercises: 13, 49 Essential Questin 24.4: Are the three rays in Figure the nly rays that pass thrugh the tip f the image? Explain. Chapter 24 Refractin and Lenses Page 2

3 Answer t Essential Questin 24.4: The three rays in Figure are easy t draw, because we knw what they d after passing thrugh the lens. Hwever, as shwn in Figure 24.23, all rays that leave the tip f the bject and which are refracted by the lens will cnverge at the tip f the image. Figure 24.23: All rays that leave the tip f the bject and pass thrugh the lens will cnverge at the tip f the image Lens Cncepts In general, a lens has a larger index f refractin than the meum that surrunds it. In that case, a lens that is thicker in the center than the ends is a cnverging lens (it cnverges parallel rays tward a fcal pint), while a lens that is thinner in the middle than at the ends is a verging lens (it verges parallel rays away frm a fcal pint). Like a cncave mirrr, cnverging lenses can prduce a real image r a virtual image, and the image can be larger, smaller, r the same size as the bject. Like a cnvex mirrr, verging lenses can nly prduce a virtual image that is smaller than the bject. As with mirrrs, the fcal pint f a lens is defined by what the lens des t a set f rays f light that are parallel t ne anther and t the principal axis f the mirrr. As we scussed in Sectin 24-4, a cnverging lens generally refracts the rays s they cnverge t pass thrugh a fcal pint, F. A verging lens, in cntrast, refracts parallel rays s that they verge away frm a fcal pint. Because f spersin (the fact that the index f refractin f the lens material depends n wavelength), a lens generally has slightly fferent fcal pints fr fferent clrs f light. This range f fcal pints is a defect called chrmatic aberratin. Fcal length f a lens with spherical surfaces: The fcal length f a lens depends n the curvature f the tw surfaces f the lens, the index f refractin f the lens material, and n the index f refractin f the surrunng meum. The fcal length is given by: = ( nlens nmeum ) +, (Equatin 24.5: The lensmaker s equatin) f R1 R2 where the tw R s represent the rai f curvature f the tw lens surfaces. A raus is psitive if the surface is cnvex, and negative if the surface is cncave. Cmpanies that make eyeglasses explit the lensmaker s equatin in creating lenses f the desired fcal length. By chsing a lens material that has a high index f refractin, a smaller raus f curvature (and thus a thinner and lighter lens) can be used, cmpared t a lens made frm glass with a smaller index f refractin. The factr f (n lens n meum ) in Equatin 24.5 has an interesting implicatin. First, cnsider the agram in Figure 24.24, which shws a familiar situatin f a lens, made frm material with an index f refractin larger than that f air, surrunded by air. This lens causes the parallel rays t change rectin s that they pass thrugh the fcal pint n the right, and (n lens n meum ) is psitive, s the fcal length f the lens is psitive. Figure 24.24: A ray agram fr a set f parallel rays encuntering a cnvex lens, made f plastic, surrunded by air. Chapter 24 Refractin and Lenses Page 3

4 What happens t the rays if the meum surrunng the lens has the same index f refractin as the lens material (perhaps we immerse the lens in sme kind f il)? In this case, the factr f (n lens n meum ) is zer, s the fcal length is infinite. An infinite fcal length means that the lens des nt change the rectin f the parallel rays at all! Ging further, if the lens is immersed in a meum that has a larger index f refractin than the lens material, (n lens n meum ) is negative and s is the fcal length: parallel rays wuld be verged in this situatin. Thus, depenng n the situatin, a lens with a cnvex shape can be a cnverging lens, a verging lens, r neither. T minimize any ambiguity, fr the rest f this chapter we will refer t lenses by their functin (cnverging r verging) rather than their shape (cnvex r cncave). EXPLORATION 24.5 Ray agram fr a verging lens We will fllw a prcess similar t that f mirrrs t draw a ray agram fr a verging lens, starting with the situatin in Figure The ray agram will shw us where the image f an bject is. Figure 24.25: An bject in frnt f a verging lens. Step 1 Draw a ray f light that leaves the tip f the bject (the tp f the arrw) and ges parallel t the principal axis (this is knwn as the parallel ray). Shw hw this ray is refracted by the lens. Fr a verging lens, all parallel rays appear t verge frm the fcal pint n the side f the lens that the light cmes frm, s we draw the refracted ray (see Figure 24.26) refracting alng a line that takes it rectly away frm that fcal pint. Figure 24.26: The parallel ray refracts t travel rectly away frm the fcal pint n the side f the lens the light came frm. Step 2 Sketch a secnd ray that leaves the tip f the bject and is refracted by the lens. Using the refracted rays, draw the image. One useful ray, shwn in blue in Figure 24.27, passes thrugh the lens withut changing rectin. This is smething f an apprximatin, but the thinner the lens, the mre accurate this is. Anther useful ray, in green, ges straight tward the fcal pint n the right f the lens. This ray refracts s as t emerge frm the lens ging parallel t the principal axis. The refracted rays verge t the right f the lens, but we can extend them back t meet n the left side f the lens, shwing us where the tip f the image is. Figure 24.27: In adtin t the parallel ray, tw ther rays are easy t draw the refracted rays fr. The ray (in blue) that passes thrugh the center f the lens is undeflected, apprximately. The ray in green travels rectly tward the fcal pint n the far side f the lens, and is refracted s it emerges frm the lens traveling parallel t the principal axis. If yu lk at the bject thrugh the lens, yur brain interprets the light as cming frm the image. Key idea: When a number f rays leave the same pint n an bject and are refracted by a lens, the crrespnng pint n the image is lcated at the intersectin f the refracted rays. Related End-f-Chapter Exercises: 11, 12, 52. Essential Questin 24.5: Starting with Figure 24.27, shw a few mre rays f light leaving the tip f the bject and being refracted by the lens. Hw d yu knw hw t draw the refracted rays? Chapter 24 Refractin and Lenses Page 4

5 Answer t Essential Questin 24.5: T draw the refracted rays prperly, we knw that when we extend the refracted rays back, they will pass thrugh the tip f the image, which we lcated in Figure Three adtinal rays are shwn in Figure Figure 24.28: Fr all rays f light that leave the tip f the bject and reflect frm the mirrr, the refracted rays can be extended back t pass thrugh the tip f the image A Quantitative Apprach: The Thin-Lens Equatin Even thugh mirrrs and lenses frm images using cmpletely fferent principles (the law f reflectin versus Snell s law), we use the same equatin t relate fcal length, bject stance, and image stance, fr bth mirrrs and lenses. This surprising result cmes frm the fact that the frmatin f images with bth mirrrs and lenses can be understd using the gemetry f similar triangles. Let s lk at hw that wrks fr lenses. Let s lk at the ray agram we drew in Figure f sectin 24-4, shwn again here in Figure Figure 24.29: The ray agram we cnstructed in sectin 24-4, fr an bject in frnt f a cnverging lens. Remve the red rays, and examine the tw triangles in Figure 24.30, 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 24.30: Similar triangles, bunded by the principal axis, the bject and image, and the blue ray that passes thrugh the center f the lens. Using the fact that the ratis f the lengths f crrespnng sides in similar triangles are equal, we find that: hi h = d. (Equatin 24.6) 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 24.29, and remve the blue rays. This gives us the shaded similar triangles shwn in Figure Figure 24.31: 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). Chapter 24 Refractin and Lenses Page 5

6 Again, using the fact that the ratis f the lengths f crrespnng sides in similar f hi triangles are equal, we find that: =. f h Simplifying the left side, and bringing in equatin 24.3, we get: 1 =. f d Diving bth sides by d i gives: 1 1 = 1, which is generally written as: f d d i = +. (Equatin 24.7: The thin-lens equatin) f d The mnemnic If I d I can help yu t remember the thin-lens equatin. Often, we knw the fcal length f and the bject stance d, s equatin 24.4 can be slved fr d i, the image stance: d f = (Equatin 24.8: The thin-lens equatin, slved fr the image stance) d f Sign cnventins We derived the lens equatin abve by using a specific case invlving a cnvex lens. The equatin can be applied t all situatins invlving a cnvex lens r a cncave lens if we use the fllwing sign cnventins. The fcal length is psitive fr a cnverging lens, and negative fr a verging lens. The image stance is psitive, and the image is real, if the image is n the side f the lens the light passes thrugh t, and negative, and the image is virtual, if the image is n the side the light cmes frm. 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 24.6, we can write the magnificatin as: hi m = =. (Equatin 24.9: 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 24.6: As yu are analyzing a thin-lens 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 24 Refractin and Lenses Page 6

7 Answer t Essential Questin 24.6: 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 a Cnverging Lens In sectin 24-4, we drew ne ray agram fr a cnverging lens. Let s investigate the range f ray agrams we can draw fr such a lens. Nte the similarities between a cnverging lens and a cncave mirrr. This sectin is very much a parallel f sectin 23-6, in which we analyzed the range f images frmed by a cncave mirrr. EXPLORATION 24.7 Ray agrams fr a cnverging lens Step 1 Draw a ray agram fr an bject lcated 40 cm frm a cnverging lens that has a fcal length f +10 cm. Verify the image lcatin n yur agram with the thin-lens equatin. Tw rays are shwn in Figure One is the parallel ray, which leaves the tip f the bject, travels parallel t the principal axis, and is refracted by the lens t pass thrugh the fcal pint n the far side f the lens. The secnd ray passes straight thrugh the center f the lens, undeflected. Figure 24.32: 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 thin-lens equatin, in the frm f equatin 24.5, 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 twice the fcal length frm the lens. We draw the same tw rays again, with the parallel ray (in red) being refracted s that it passes thrugh the fcal pint n the far side f the lens, and the secnd ray (in blue) passing undeflected (apprximately) thrugh the center f the lens. As shwn in Figure 24.33, this situatin is a special case. When the bject is lcated at twice the fcal length frm the lens, the image is inverted, als at twice the fcal length frm the lens (n the ther side f the lens), and the same size as the bject because the bject and image are the same stance frm the lens. Figure 24.33: When the bject is at twice the fcal length frm the lens, s is the image. Applying the thin-lens 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 24 Refractin and Lenses Page 7

8 Step 3 Repeat step 1, with the bject 15 cm frm the mirrr. N matter what the bject stance is, the parallel ray always des the same thing, being refracted by the lens t pass thrugh the fcal pint n the far side. The path f the secnd ray, in blue, depends n the bject s psitin. The ray agram (Figure 24.34) shws that the image is real, inverted, larger than the bject, and abut twice as far frm the lens as the bject. Figure 24.34: A ray agram fr a situatin in which the bject is between twice the fcal length frm the lens and the fcal pint. Applying the thin-lens 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 a fcal pint. As shwn in Figure 24.35, the tw refracted rays are parallel t ne anther, and never meet. In such a case the image is frmed at infinity. Figure 24.35: A ray agram fr a situatin in which the bject is at the fcal pint. Applying the thin-lens 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 lens. When the bject is clser t the lens than the fcal pint, the refracted rays verge t the right f the lens, and they must be extended back t meet n the left f the lens. The result is a virtual, upright image that is larger than the bject, as shwn in Figure If yu lk at the bject thrugh the lens, yur brain interprets the light as cming frm the image. Figure 24.36: A ray agram fr a situatin in when the bject is between the lens and its fcal pint. Applying the thin-lens equatin gives: 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 thin-lens equatin is cnsistent with the ray agram. Key idea fr cnverging lenses: Depenng n where the bject is relative t the fcal pint f a cnverging lens, the lens 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: 44, 50, 53. Essential Questin 24.7: When an bject is placed 20 cm frm a lens, the image frmed by the lens is real. What kind f lens is it? What, if anything, can yu say abut the lens fcal length? Chapter 24 Refractin and Lenses Page 8

9 Answer t Essential Questin 24.7: The lens must be cnverging, because a verging lens cannt prduce an image that is larger than the bject. A cnverging lens prduces a real image nly when the bject stance is larger than the fcal length, s the fcal length in this case must be psitive but less than 20 cm An Example Prblem Invlving a Lens Let s begin by scussing a general apprach we can use t slve prblems invlving a lens. We will then apply the methd t a particular situatin. A general methd fr slving prblems invlving a lens 1. Sketch a ray agram, shwing rays leaving the tip f the bject and being refracted by the lens. Where the refracted 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 thin-lens 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. Yu can draw any number f rays being refracted by the lens, but sme are easier t draw than thers because we knw exactly where the refracted rays g fr these rays. Such rays are shwn n Figure 24.37, and include: 1. The ray, in red, that ges parallel t the principal axis, and refracts t pass thrugh the fcal pint n the far side f the lens (cnverging lens), r away frm the fcal pint n the near side f the lens (verging lens). 2. The ray, in blue, that passes straight thrugh the center f the lens withut being deflected. 3. The ray, in green, that travels alng the straight line cnnecting the tip f the bject and the fcal pint nt assciated with the first ray. This ray is refracted by the lens t g parallel t the principal axis. Figure 24.37: An example f the three rays that are easy t draw the refracted rays fr. When yu lk at the bject thrugh the lens, yur brain interprets the light as traveling in straight lines as if it emanated frm the image. Chapter 24 Refractin and Lenses Page 9

10 EXAMPLE 24.8 Applying the general methd When yu lk at a cat thrugh a lens that has its fcal pints at stances f 24 cm n either side f the lens, yu see an image f the cat that is 1.5 times as large as the cat. Hw far is the cat frm the lens? 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 lens is clearly a cnverging lens, because nly cnverging lenses prduce images that are larger than the bject. One pssibility is that the lens prduces a virtual, upright image, s the sign f the magnificatin is psitive. Applying the magnificatin equatin, we get: 1 1 m =+ 1.5 =, which tells us that =. d 1.5d Applying the thin-lens equatin: = + = = =. f d d 1.5d 3d 3d 3d Thus, we find that 3d = f =+ 24 cm, s d =+ 8.0 cm and we can shw that d = 12 cm. The ray agram fr this situatin is shwn in Figure 24.38, cnfirming the calculatins. Figure 24.38: A ray agram fr the slutin invlving a virtual image. Each bx n the grid measures 2 cm 2 cm. i The slutin abve is nly ne f the pssible answers. The image culd als be real and inverted, s the sign f the magnificatin is negative. Applying the magnificatin equatin, we get: 1 1 m = 1.5 =, which tells us that =+. d 1.5d Applying the mirrr equatin: = + = + = + =. f d d 1.5d 3d 3d 3d Thus, we find that d 5 = f =+ 40 cm, and we can shw that d i = + 60 cm. 3 The ray agram fr this situatin is shwn in Figure 24.39, again cnfirming the calculatins abve. These tw rays, and all rays that travel frm the tip f the bject t the tip f the image, take the same time t get there. Figure 24.39: A ray agram fr the situatin invlving a real image. Each bx n the grid measures 2 cm 2 cm. Related End-f-Chapter Exercises: Essential Questin 24.8: Return t the situatin described in Example Wuld there still be tw slutins if the image was smaller than the bject? Explain. Chapter 24 Refractin and Lenses Page 10