L ENSES. Lenses Spherical refracting surfaces. n 1 n 2

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1 Lenses 2 L ENSES 2. Sherical reracting suraces In order to start discussing lenses uantitatively, it is useul to consider a simle sherical surace, as shown in Fig. 2.. Our lens is a semi-ininte rod with one sherical surace, made o a material o reractive index n 2 greater than the surrounding material (n <n 2 ). Qualitatively, we know what will haen based on the law o reraction. Rays emanating rom a distant object laced at O will iminge on the sherical surace, bend toward the rincile axis (toward the surace normal), and converge at a oint inside the rod, orming a real image. But where? O I n n 2 Figure 2.: A sherical reracting surace. uer: Rays incident rom a distant object O are reracted toward the rincile axis, and ocused at a oint I. lower: Construction or determining the relative image and object distances in terms o the radius o curvature and reractive indices. P θ O α V d θ2 β C γ I R With a bit o geometry, we can igure out exactly where the image must orm, given the object 338

2 2. Sherical reracting suraces 339 distance and the radius o the sherical surace. Reerring to the second ortion o Fig. 2., let the object (O) and image (I) distances be and, resectively, measured rom the intersection o the rincile axis with the sherical surace (V ). The center o the shere o radius R making u the surace is at C. Trivially, a ray drawn rom O through the rincile axis ass through V, C, and I. Now, draw a ray leaving the object and intersecting the surace at oint P, OP. At the oint P, we draw surace normal and tangent lines to deine the angle o incidence θ and the angle o reraction θ 2. The reracted ray will be bent toward the rincile axis, intersecting it at oint I. This ray P I makes an angle α with the rincile axis. Recall that any line erendicular to the surace o a circle must ass through the center o the circle. Thus, i we extend the normal drawn at oint P, it must intersect oint C, orming ray P C, which makes an angle β with the rincile axis. Now we have everything labeled that we need, all that is let is to ind a relationshi between,, and R. First, we can use right triangle OP C. The angles OP C, α, and β making u this triangle must add u to 80. We also know that that the angles θ and OP C by themselves deine a straight line, and must thereore add u to 80 as well. Thus: α + β + OP C = 80 (2.) θ + OP C = 80 (2.2) = θ = α + β (2.3) Slowly, we are reducing the number o unknown uantities. Now examine the triangle P CI. We know that the angles θ 2, γ, and P CI must add u to 80. Further, we know that β and P CI must together make 80, since they deine the line OI. Putting these acts together: θ 2 + γ + P CI = 80 (2.4) β + P CI = 70 (2.5) = θ 2 = β γ (2.6) Euations 2.3 and 2.6 give us the angles o incidence (θ ) and reraction (θ 2 ) in terms o the interior angles α, β, and γ which can be more easily related to the distances o interest, viz.,,, and R. Beore we can do that, we have one trick u our sleeve: we haven t yet used Snell s law: n sin θ = n 2 sin θ 2 (2.7)

3 Sherical reracting suraces I we substitute euations 2.3 and 2.6 into this exression, we have: n sin (α + β) = n 2 sin (β γ) (2.8) We can aly the sum and dierence identities or sin (a±b) to this, which yields the ollowing: n sin α cos β + n cos α sin β = n 2 sin β cos γ n 2 cos β sin γ (2.9) n cos α (tan α cos β + sin β) = n 2 cos γ (sin β cos β tan γ) ( ) ( tan α n cos α sin β tan β + = n 2 cos γ sin β tan γ ) tan β ( ) ( tan α n cos α sin β tan β + = n 2 cos γ sin β tan γ ) (β 0) tan β ( ) ( tan α n cos α tan β + = n 2 cos γ tan γ ) (β 0) (2.0) tan β For the last line, we must take care that β 0, otherwise canceling the sin β terms would be division by zero - strictly not allowed. This is not a roblem - β is only zero or the trivial case o the ray traveling on the rincile axis, which we already know how to deal with. In order to roceed urther, we need to make a crucial aroximation. Namely, we assume that the object is very distant relative to the radius o the sherical surace, R, and we only consider rays incident near the rincile axis, d R. I this is true, then the tangents o α, β, and γ can be nicely aroximated: tan α d OV = d tan β d V C = d R tan γ d Basically, we have just decided to ignore the tiny distance between oint V and the intersection o P V with the rincile axis. Qualitatively, these aroximations seem reasonable. It would be euivalent to say that we only consider large and small α - the same aroximations result i α is small, so too are β and γ. Using these aroximations, E. 2.0 reduces to:

4 2. Sherical reracting suraces 34 ( n cos α + d/ ) ( = n 2 cos γ d/ ) d/r d/r ( n cos α + R ) ( = n 2 cos γ R ) (2.) Now, given that the angles α and β are suosed to be tiny and the object distance large, we know that d and d. Thus, the ratios d/ and d/ will be very small comared to. We can use this act to simliy things even urther. Using the same logic behind the tangent aroximations, we ind cos α, and cos γ cos α cos γ d = d = + d 2 / = 2 + d 2 / 2 + d 2 / = 2 + d 2 / 2 Thus, so long as d/ and d/ are very small (and their suares are even smaller), we can simly ignore the cosine terms, which leaves us: ( n + R ) ( = n 2 R ) (2.2) R n + n R 2 = n 2 n (2.3) = n + n 2 = n 2 n R (2.4) This is the result we desire: the image and object distances are simly related by the radius o curvature o the sherical surace, and the indices o reraction o the lens material and its surrounding. Sherical reracting suraces: n + n 2 = n 2 n R (2.5) Here is the image distance inside the dense material n 2, and is the object distance in the less dense material n (n <n 2 ). The results holds or rays not ar rom the rincile axis.

5 Sherical Lenses 2.. Flat Reracting Suraces I we let R tend toward ininity, R, our sherical surace becomes a lat one. i I R tends toward ininity, then /R tends toward zero, and our sherical lens euation reduces to: Flat reracting suraces: = n 2 n (2.6) Here is the image distance inside the dense material n 2, and is the object distance in the less dense material n (n <n 2 ). This derives a result with imortant everyday conseuences: since n 2 n, then. This is why, when looking into a ool o water, objects are actually much arther below the surace than we think they are. 2.2 Sherical Lenses Armed with a knowledge o sherical reracting suraces, we can move on to sherical lenses. All o the lenses we will consider can be deined only by the suraces o sheres, hence the name. Figure 2.2 shows how one can construct either biconvex (uer) or biconcave (lower) sherical lenses, deined by the intersection and region between two sheres, resectively. (a) rincial axis R R 2 (b) R R 2 Figure 2.2: Sherical lenses can also be either concave or convex, and their suraces are deined by the suraces o two sheres. (a) Biconvex lenses are ormed by the intersection o two sheres, and (b) biconcave lenses are ormed by the region between two sheres. When R =R 2, the lens is sherically symmetric. How can we analyze a lens like this? A lens can be considered the combination o two sherical interaces, so all we need to do is use our solution to the case o the sherical reracting surace and i One can say that the radius o curvature o a lat lane is ininite, or euivalently, that a lane is just the surace o a shere with ininite radius.

6 2.2 Sherical Lenses 343 aly it twice. First, we ind the image due to (or instance) the let-hand sherical surace, and the image ormed by that surace serves as the object or the right-hand sherical surace. This is shown in Fig. 2.3, where we consider a lens o thickness d ormed by overlaing sheres o radii R and R 2, both o which are made o a material o reractive index n 2. Surrounding the model lens is a material o reractive index n. n R P R 2 O n 2 I = O 2 Figure 2.3: Our model sherical lens is built out o two searate sherical reracting suraces. d First, consider only the object on the right-hand side by itsel. Light rom oint O a distance rom the sherical surace, reaches the sherical interace at oint P. Since we are only worrying in the end about the region where the two sherical suraces overla, we resume that the light is not reracted on the way rom O to P. Ater reraction, the ray is reracted toward oint I on the rincile axis. Since this is just reraction rom a sherical surace as we solved above, we know n + n 2 = n 2 n R (2.7) This orms an image at oint I. This image now serves as an object or the second sherical surace - I =O 2. Now ignore the right-hand side and consider only the let-hand side. Light rom the image ormed at O 2 will be incident on the sherical surace deined by R 2 in this case. Now, since oint O 2 is on the right side o the lens, the object distance is negative, <0. This distance is related to the object distance o the irst lens,, by the thickness o the lens: = d (2.8) where we made sure to careully ollow our sign convention. Reraction rom the sherical surace R 2 can be calculated in the same way:

7 Sherical Lenses n 2 + n = n n 2 R 2 n 2 d + n = n n 2 R 2 (2.9) (2.20) Now, add Ens. 2.7 and 2.20: n + n 2 + n 2 d + n = n n 2 + n 2 n (2.2) R 2 R n + n 2 + n + n [ 2 d = (n 2 n ) ] (2.22) R R 2 This is the general euation or a sherical lens. General euation or a sherical lens: n + n 2 + n + n [ 2 d = (n 2 n ) ] R R 2 (2.23) Here R and R 2 are the radii o the sherical sections making u the lens, d is the thickness o the lens, n 2 the reractive index o the lens material, and n o the surrounding material. The result holds or rays not ar rom the rincile axis. Most o the time, we are interested in the so-called thin lens aroximation, in which we neglect the thickness o the lens. That is, we resume that the image and object distances are so large comared to the thickness o the lens,, d, that we can saely neglect d. I we let d 0, we have what is known as the lensmaker s ormula: n + n [ = (n 2 n ) ] R R 2 (2.24) We can ind the ocal length o the lens by considering the case o an extremely distant object, where we let tend toward ininity. In that case, arallel rays will be converged on to a single ocal oint, just as with a sherical mirror, which we deine to be the ocal length. Thus, we let tend toward ininity (which makes / tend toward zero), and ind the corresonding value o =. This yields the more common orm o the lensmaker s euation:

8 2.2 Sherical Lenses 345 Lensmaker s euation: = ( ) [ n2 n ] n R R 2 (2.25) here n is the index o reraction o the surrounding material, n 2 o the lens. The lens is deined by the suraces o sheres o radius R and R 2. Comaring this to the receding euation, we can also immediately relate the ocal length to the image and object distance, which yields the lens euation : Lens euation: = + (2.26) Surrise, surrise, the mirror euation is the same as the lens euation! A convex lens like the one we just considered will have a ositive ocal length. Even though we derived these lens euations or the case o a convex lens, they are valid or thin concave lenses as well, so long as they are sherical. We will consider some other tyes o lenses shortly, but we have one bit o ressing business: we still don t know the magniication actor o the lens! In order to determine the image magniication, it is easier at this oint to construct a ray diagram, just as we did with mirrors. The rules are only slightly dierent: How to construct ray diagrams: Ray is drawn arallel to the rincile axis, and reracts through one ocal oint. Ray 2 is drawn through the (other) ocal oint, and reracts arallel to the axis. Ray 3 is drawn through the center o the lens, and continues in a straight line. Figure?? shows a ray diagram or a simle convex lens. Using the geometry o this igure, we can readily igure out the magniication actor, and veriy our lens euation above to boot. P h O F α Q θ α F θ I h Figure 2.4: Image construction with a biconcave lens. Consider the triangle ormed by oints O, Q, and the ti o the object arrow. The tangent o the angle α is the object height over the object distance: tan α = h (2.27)

9 Sherical Lenses The triangle ormed by oints I, Q, and the ti o the image arrow give us another exression or tan α: tan α = h (2.28) Comaring these two exressions, and using the deinition o the magniication actor, we have our answer: Magniication or a sherical lens: M h h = = = (2.29) The last two orms are derived below. They ollow by using the lens euation (??) in the irst relationshi. Once again, the lens and mirror euations are the same - same sherical geometry, same euations. This ormula is also much more general than our derivation suggests - it is valid or any sherical lens, not just the symmetric concave one we considered here. We can also veriy the lens euation by using the geometry o the uermost ray. The triangle P QF gives us another relationshi, noting that the distance rom the center o the lens (Q) to the ocal oint (F ) is by deinition the ocal length (QF =) and P Q=h: tan θ = P Q = h (2.30) The triangle deined by F, I, and the ti o the object arrow gives us one more euation: tan θ = h (2.3) Comaring the last two euations, we have h = h = h h = (2.32) M (2.33) Now we have two dierent exressions or M, which we can combine:

10 2.3 Tyes o sherical lenses 347 M = = (2.34) = (2.35) + = (2.36) + = (2.37) = + = (2.38) A result that should be reassuring: we have now indeendently derived the lens euation. We can derive a third relationshi between the magniication and ocal length using the lens euation and our result above: = = = M = = = 2 ( ) = 2 ( ) = (2.39) (2.40) (2.4) (2.42) (2.43) (2.44) (2.45) This gives us three dierent relationshis or the magniication actor, each one involving only two o the three uantities,, and. We now have all the mathematical and geometric ammunition we need or sherical lenses o any kind. Though we derived our results or the secial case o convex lenses, they are more generally valid (it would take much more mathematics and geometry to demonstrate this, however), and hold or any sherical lenses we wish to consider. What we need to do next is igure out how dierent sorts o sherical lenses behave and what sorts o images the orm on a ualitative level. 2.3 Tyes o sherical lenses

11 Quick Questions (a) (b) (c) (d) (e) () Figure 2.5: There are a variety o common lens shaes, all essentially based on the intersection o two sheres or the sace between two sheres. (a) Double convex, (b) lano-convex, (c) convex meniscus, (d) double concave, (e) lano-concave, () and concave meniscus lenses. (a) F (b) Figure 2.6: (a) A biconvex lens converges distant light rays and ocuses them onto a oint hence the name ocusing lens. (a) A biconcave lens causes distant light rays to diverge. They aear to diverge outward rom a ocal oint on the incident side o the lens. F 2.4 Quick Questions. An object is laced to the let o a converging lens. Which o the ollowing statements are true and which are alse?. The image is always to the right o the lens 2. The image can be uright or inverted 3. The image is always smaller or the same size as the object and 2 are true, 3 is true 2 and 3 are alse, is true and 3 are alse, 2 is true 2 and 3 are true, is alse 2.5 Problems. A contact lens is made o a lastic with an index o reraction o.60. The lens has an inner radius o curvature o.99 cm and an inner radius o curvature o 2.56 cm. What is the ocal

12 2.5 Problems 349 length o the contact lens?

13 Solutions to Quick Questions 2.6 Solutions to Quick Questions. and 3 are alse, 2 is true. 2.7 Solutions to Problems. 4.9 cm. Qualitatively, we know that the lens must orm a real image in order or contact lenses to unction roerly. Thereore, we know that in the end the lens must have a ositive ocal length. Further, we know the order o magnitude o the lens, based on the size o an average human head: it must be centimeters, clearly not meters, kilometers, or micrometers! In order to attack the roblem uantitatively, we need the lensmaker s euation. [ = (n 2 n ) ] R R 2 (2.46) Since the outer surace o the lens is exosed to lain air, we may assume n =.00 there. Since we just want the reractive index o the contact lens itsel, not the contact lens in combination with the eye, we will assume the other surace is exosed to air as well. Given the reractive index o the lens material n 2 =.60 and the two radii, we need only solve or. Since we know the answer has to be ositive, we know that R 2 = 2.56 cm and R =.99 cm, not the other way around: [ = (.60.00).99 ] 2.56 (2.47) = (0.60) [0.2] cm (2.48) = (2.49) = = 4.9 cm (2.50)