Optics Image Formation from Mirrors and Lenses

Transcription

1 Optics Image Formation from Mirrors and Lenses Lana Sheridan De Anza College June 14, 2017

2 Last time refraction dispersion mirrors image terminology images formed by spherical mirrors

3 Overview images formed by spherical mirrors images formed by refraction images formed by lenses images formed by lens combinations

4 Mirror Equation and Thin Lens Equation The same equation can help us to find the location and magnification of the image that will be formed by curved mirrors and thin lenses. 1 f = 1 p + 1 q This is what we now want to prove, for the case of spherical concave mirrors. Last lecture we showed that for spherical concave mirrors of radius, R: f = R 2

5 Concave Mirrors and the Mirror Equation Simple geometry shows why the mirror equation is true for a e Formation concave mirror. The real image lies at the location at which the reflected rays cross. h O a C a I h' u u V Principal axis q R p Figure 36.9 The image formed by a spherical concave mirror when the object O lies outside the center of curvature C. This geometric construction is used to derive Equation We will use this ray diagram.

6 Ray Diagrams for Spherical Mirrors For a ray diagram: draw at least two rays that you know the path of accurately. For Spherical mirrors: 1 Draw a ray from the top of the object parallel to the principle axis reflected through the focal point F. 2 Draw a ray from the top of the object through the focal point and reflected parallel to the principal axis. 3 Draw a ray from the top of the object through the center of curvature C and reflected back on itself. Where the lines meet, an image is formed.

7 When the object is located between the focal point and a concave mirror surface, the image is virtual, upright, and enlarged. Examples of Ray Diagrams 36.2 Images Formed by Sphe When the object is located so that the center of curvature lies between the object and a concave mirror surface, the image is real, inverted, and reduced in size. F 1 O Principal axis 3 2 C I F Front Back Cengage Learning/Charles D. Winters a

8 Principal axis Front Examples of Ray Diagrams a Back Cengage Lear When the object is located between the focal point and a concave mirror surface, the image is virtual, upright, and enlarged. 2 3 C F O I 1 Front Back b When the object is in front of a convex mirror, the image is virtual, upright,

9 Front Back Examples of Ray Diagrams (Convex Mirror) b When the object is in front of a convex mirror, the image is virtual, upright, and reduced in size O I F C Front Back c

10 e Formation Concave Mirrors and the Mirror Equation The real image lies at the location at which the reflected rays cross. h O a C a I h' u u V Principal axis q R p Figure 36.9 The image formed by a spherical concave mirror when the object O lies outside the center of curvature C. This geometric construction is used to derive Equation The angles of incidence and reflection are the same magnitude, θ. negative So, sign h h = is introduced q p, and because the image is inverted, so h9 is taken to be negative. Therefore, from Equation 36.1 and M = these q results, we find that the magnification of the image is p

11 Concave Mirrors and the Mirror Equation Looking at the green triangle and the red triangle with angle α: h R q = h p R which rearranges to h h = q R p R Using our magnification expression: q p = q R p R

12 Concave Mirrors and the Mirror Equation q p = q R p R Cross-multiplying and rearranging gives 2 R = 1 p + 1 q However, we already concluded that f = R/2, so 1 f = 1 p + 1 q We have confirmed the mirror equation for spherical concave mirrors, and it follows from simple geometry.

13 not derive any equations for convex spherical mirrors because Eq.4, Convex and 36.6 Mirrors can be used for either concave or convex mirrors if we t sign convention. We will refer to the region in which light rays or e toward the mirror as the front side of the mirror and the other sid We can draw the ray diagrams for convex mirrors in the same way. The image formed by the object is virtual, upright, and behind the mirror. Front Back O I F C q p

14 Mirror Question Quick Quiz Consider the image in the mirror in shown. Based on the appearance of this image, which of the following should you conclude? 1098 Chapter 36 Image (A) the mirror is concave and the image is real (B) the mirror is concave and the image is virtual (C) the mirror is convex and the image is real (D) the mirror is convex and the image is virtual in vi ro ve Q Q NASA 1 Serway & Jewett, page Figure (Quick Quiz 36.3)

15 Mirror Question Quick Quiz Consider the image in the mirror in shown. Based on the appearance of this image, which of the following should you conclude? 1098 Chapter 36 Image (A) the mirror is concave and the image is real (B) the mirror is concave and the image is virtual (C) the mirror is convex and the image is real (D) the mirror is convex and the image is virtual in vi ro ve Q Q NASA 1 Serway & Jewett, page Figure (Quick Quiz 36.3)

16 Sign Conventions for Mirrors! 1 f = 1 p + 1 q Variable is Positive is Negative p object in front of mirror q image in front of mirror image behind mirror (real) (virtual) h and M image upright image inverted f and R concave mirror convex mirror

17 Example 36.4: Convex Mirror Image An automobile rearview mirror shows an image of a truck located 10.0 m from the mirror. The focal length of the mirror is 0.60 m. Find the position of the image of the truck and the magnification of the image. le 36.4) An n in a convex of an autoimage of the frame of the onstrates the same locace.. Bo Zaunders/Corbis

18 Example 36.4: Convex Mirror Image An automobile rearview mirror shows an image of a truck located 10.0 m from the mirror. The focal length of the mirror is 0.60 m. Find the position of the image of the truck. (where does it appear to be?)

19 Example 36.4: Convex Mirror Image An automobile rearview mirror shows an image of a truck located 10.0 m from the mirror. The focal length of the mirror is 0.60 m. Find the position of the image of the truck. (where does it appear to be?) q = 0.57 m Find the magnification of the image.

20 Example 36.4: Convex Mirror Image An automobile rearview mirror shows an image of a truck located 10.0 m from the mirror. The focal length of the mirror is 0.60 m. Find the position of the image of the truck. (where does it appear to be?) q = 0.57 m Find the magnification of the image. M = q p =

21 Images Formed by Refraction When light rays change media they are bent. This also can form images.

22 sider two transparent m boundary between the Rays making small angles with the We can find the location principal andaxis We assume the object a size diverge of the from image a point formed by object at O and are refracted Let s consider the parax considering paraxial rays. through the image point I. at the spherical surface Figure shows a n n 1 2 law of refraction applie Images Formed by Refraction n 1 n 2 R O I Because u 1 and u 2 are a tion sin u < u (with ang p q We know that an exteri interior angles, so apply Figure An image formed by refraction at a spherical surface. For paraxial rays, Snell s law becomes: n 1 θ 1 = n 2 θ 2 (1)

23 Images Formed by Refraction 36.3 Images Formed by Refraction n 1 u 1 P n 2 Figure G derive Equation 3 n 1, n 2. d u 2 O a b C g I R p q Looking at the diagram: likewise: Combining all three expressions and eliminating u 1 and u 2 gives 180 θ 1 = 180 α β θ 1 = α + β (2) n 1 a 1 n 2 g 5 (n 2 2 n 1 )b (36.7) Figure shows three right triangles that have a common vertical leg of length d. For paraxial rays (unlike the relatively large-angle ray shown in Fig ), the horizontal legs of these triangles are approximately p for the triangle containing angle a, R for the triangle β containing = γ + angle θ 2 b, and (3) q for the triangle containing angle g. In the small-angle approximation, tan u < u, so we can write the approximate relationships from these triangles as follows: Multiply (2) by n 1 and (3) by n 2, add them and subtract (1): tan a < a < d tan b < b < d tan g < g < n d 2 β p= n 1 α + n 1 β R + n 2 γ q Substituting these expressions into Equation 36.7 and dividing through by d gives Rearranging: n 1 n 1 α + n 2 γ = (n 2 n 1 )β p 1 n 2 q 5 n 2 2 n 1 R (36.8) Relation betw image distanc

24 Images Formed by Refraction For small angles (paraxial approx): α tan α = d p Similarly, β d R and γ d q So, n 1 p + n 2 q = n 2 n 1 R

25 Flat Refracting Surfaces (Like a rectangular fish tank.) 1102 Chapter 36 Image Formation The image is virtual and on the same side of the surface as the object. O p n 1 n 2 n 1 n 2 Figure The image formed q Flat Refracting Surfaces If a refracting surface is flat, then R is infinite an n 1 In this case R. p 52n 2 q n 1 p + n 2 q = 0 q 52 n 2 n 1 p From this expression, And so we see that the sign of q according to Table 36.2, the image formed by a same side of the surface q as = the n 2 object p n as illustrate 1 in which the object is in the medium of index n 1 case, a virtual image is formed between the objec n 2, the rays on the back side diverge from one an in Figure As a result, the virtual image is fo Q uick Quiz 36.4 In Figure 36.16, what happens point O is moved to the right from very far awa

26 Flat Refracting Surfaces Example (Problem 30) A cubical block of ice 50.0 cm on a side is placed over a speck of dust on a level floor. Find the location of the image of the speck as viewed from above. The index of refraction of ice is

27 Flat Refracting Surfaces Example (Problem 30) A cubical block of ice 50.0 cm on a side is placed over a speck of dust on a level floor. Find the location of the image of the speck as viewed from above. The index of refraction of ice is n 1 p + n 2 q = n 2 n 1 R, R q = n 2 n 1 p

28 Flat Refracting Surfaces Example (Problem 30) A cubical block of ice 50.0 cm on a side is placed over a speck of dust on a level floor. Find the location of the image of the speck as viewed from above. The index of refraction of ice is n 1 p + n 2 q = n 2 n 1 R, R q = n 2 n 1 p = 1 (50.0 cm) = 38.2 cm

29 Sign Conventions for Refracting Surfaces! n 1 p + n 2 q = n 2 n 1 R Variable is Positive is Negative p object in front of surface [virtual object] 2 q image behind surface image in front of surface (real) (virtual) h (and M) image upright image inverted R object faces convex surf. object faces concave surf. (C behind surface) (C in front of surface) C is the center of curvature. M = h h = n 1q n 2 p 2 Will be useful in derivations.

30 Refracting Surface Question Quick Quiz 36.4 In the figure, what happens to the image point I Rays making small angles with the as the object point O is moved to the rightprincipal from very axis diverge far away from a point to very close to the refracting surface? (A) It is always to the right of the surface. (B) It is always to the left of the surface. (C) It starts off to the left, and at some position of O, I moves to the right of the surface. (D) It starts off to the right, and at some position of O, I moves to the left of the surface. object at O and are refracted through the image point I. O p n 1 n 2 n 1 n 2 Figure An image formed by refraction at a spherical surface. q R I 36 In t und side bou We Let at th F law Bec tion We inte 2 Serway & Jewett, page 1102.

31 Refracting Surface Question Quick Quiz 36.4 In the figure, what happens to the image point I Rays making small angles with the as the object point O is moved to the rightprincipal from very axis diverge far away from a point to very close to the refracting surface? (A) It is always to the right of the surface. (B) It is always to the left of the surface. (C) It starts off to the left, and at some position of O, I moves to the right of the surface. (D) It starts off to the right, and at some position of O, I moves to the left of the surface. object at O and are refracted through the image point I. O p n 1 n 2 n 1 n 2 Figure An image formed by refraction at a spherical surface. q R I 36 In t und side bou We Let at th F law Bec tion We inte 2 Serway & Jewett, page 1102.

32 Refracting Surface Question 1102 Chapter 36 Image For Quick Quiz 36.5 In the figure, what happens The image to the is virtual image and on point I the same side of the surface as the object point O moves toward the right-hand as the object. surface of the material of index of refraction n 1? (A) It always remains between O and the surface, arriving at the surface just as O does. (B) It moves toward the surface more slowly than O so that eventually O passes I. (C) It approaches the surface and then moves to the right of the surface. 2 Serway & Jewett, page O p n 1 n 2 n 1 n 2 Figure The image formed by a flat refracting surface. All rays are assumed to be paraxial. q Flat If a r From acco same in wh case, n 2, t in Fi Q ui po su th rig

33 Refracting Surface Question 1102 Chapter 36 Image For Quick Quiz 36.5 In the figure, what happens The image to the is virtual image and on point I the same side of the surface as the object point O moves toward the right-hand as the object. surface of the material of index of refraction n 1? (A) It always remains between O and the surface, arriving at the surface just as O does. (B) It moves toward the surface more slowly than O so that eventually O passes I. (C) It approaches the surface and then moves to the right of the surface. 2 Serway & Jewett, page O p n 1 n 2 n 1 n 2 Figure The image formed by a flat refracting surface. All rays are assumed to be paraxial. q Flat If a r From acco same in wh case, n 2, t in Fi Q ui po su th rig

34 t of me, is flat yes. lowin car rva- 15). ind our oes changes according to Snell s law, n 1 sin u 1 5 n 2 sin u 2. For paraxial rays, we assume u 1 and u 2 are small, so we A curved may refracting write n surface separates a material with index of 1 tan u 1 5 n 2 tan u 2. The magnification is refraction defined n 1 from as M a material 5 h9/h. with Prove index that n 2 the. Prove magnification that the is magnification is given by M = n 1q given by M 5 2n 1 q/n 2 p. Refracting Surface Example (Problem 34) prolose an tual h O n 2 p. n 1 n 2 u 1 p u 2 q C Figure P A glass sphere (n ) with a radius of 15.0 cm has a M tiny air bubble 5.00 cm above its center. The sphere is viewed looking down along the extended radius containing the bubble. What is the apparent depth of the bubble below the surface of the sphere? I h

35 t of me, is flat yes. lowin car rva- 15). ind our oes changes according to Snell s law, n 1 sin u 1 5 n 2 sin u 2. For paraxial rays, we assume u 1 and u 2 are small, so we A curved may refracting write n surface separates a material with index of 1 tan u 1 5 n 2 tan u 2. The magnification is refraction defined n 1 from as M a material 5 h9/h. with Prove index that n 2 the. Prove magnification that the is magnification is given by M = n 1q given by M 5 2n 1 q/n 2 p. Refracting Surface Example (Problem 34) prolose an tual h O n 2 p. n 1 n 2 u 1 p Hint: For paraxial rays, wefigure assumep36.34 θ 1 and θ 2 are small, so we may write 35. Snell s A glass Law sphere as n 1 (n tan 5 θ1.50) 1 = n 2 with tan θa 2 radius. of 15.0 cm has a M tiny air bubble 5.00 cm above its center. The sphere is viewed looking down along the extended radius containing the bubble. What is the apparent depth of the bubble below the surface of the sphere? u 2 q C I h

36 t of me, is flat yes. lowin car rva- 15). ind our oes changes according to Snell s law, n 1 sin u 1 5 n 2 sin u 2. For paraxial rays, we assume u 1 and u 2 are small, so we A curved may refracting write n surface separates a material with index of 1 tan u 1 5 n 2 tan u 2. The magnification is refraction defined n 1 from as M a material 5 h9/h. with Prove index that n 2 the. Prove magnification that the is magnification is given by M = n 1q given by M 5 2n 1 q/n 2 p. Refracting Surface Example (Problem 34) prolose an tual h O n 2 p. n 1 n 2 u 1 p u 2 Hint: For paraxial rays, wefigure assumep36.34 θ 1 and θ 2 are small, so we may write 35. Snell s A glass Law sphere as n 1 (n tan 5 θ1.50) 1 = n 2 with tan θa 2 radius. of 15.0 cm has a ForMsmall tiny angles: air bubble 5.00 cm above its center. The sphere is viewed looking down along the extended radius containing 1 tan θ n the 1 = n bubble. 2 tan θ What 2 is the apparent depth of the bubble below h the surface ( h ) n 1 = n 2 of the M sphere? = h p q h = n 1q n 2 p q C I h

37 Images Formed by Thin Lenses We will derive the thin lens equation 1 f = 1 p + 1 q (Notice it is the same as the mirror equation!) And the lens maker s equation 1 f ( 1 = (n 1) 1 ) R 1 R 2 We do this by considering each side of the lens as a refracting surface.

38 Images Formed by Thin Lenses A thick lens: The image due to surface 1 is virtual, so I 1 is to the left of the surface. I 1 Surface 1 Surface 2 n O q 1 R 1 p 1 n 1 1 t R2 C 1 refraction a the notion t object for th let the thick Consider faces with r R 1 is the rad reaches first lens.) An ob Let s beg and assumin the image I 1 a p 2 We can imagine that there first surface forms an image I 1, and that image becomes where q 1 is The the object image due for the to surface second1 surface. is real, so I formed by su 1 is to the right of the surface. The pink arrow shows I 1 if the image

39 First image formed The image due to surface 1 is virtual, Let n 1 = 1, n 2 so = I 1 n. is to the left of the surface. I 1 Surface 1 Surface 2 n O p 1 q 1 n 1 1 R 1 R2 t C 1 the image fo refraction at the notion th object for the let the thickn Consider a faces with rad R 1 is the radi reaches first a lens.) An obje Let s begin and assuming the image I 1 f p 2 a 1 The image + n due = n 1 to surface 1 is real, p so I 1 is 1 q to the 1 R right of 1 the surface. where q 1 is th formed by sur if the image i

40 Second image formed n + 1 = 1 n p 2 q 2 R 2 We must relate p 2 to q 1. The new object is the image from the previous surface so: p 2 = q 1 + t where t is the lens thickness.

41 p 2 could be positive (real object)... If q 1 is negative The image (virtual due image to surface 1)... 1 is virtual, so I 1 is to the left of the surface. n 1 1 I 1 Surface 1 Surface 2 n O p 1 q 1 R 1 R2 t C 1 the image refraction the notion object for let the thi Consid faces with R 1 is the r reaches fi lens.) An Let s be and assum the image p 2 a... p 2 is positive (object in front of surface). where q 1

42 a p 2 could be negative (virtual object)... If q 1 is positive (real The image image 1)... due to surface 1 is real, so I 1 is to the right of the surface. R 1 p 1 p 2 q 1 R 2 Surface 1 Surface 2 n O n 1 1 b... p 2 is negative (object behind surface). t I 1 C 1 Figure To locate the image formed where q 1 formed b if the ima Now le (We mak 2 are in t the objec We now i face acts is virtual related to q 1 is nega image fro

43 a p 2 could be negative (virtual object)... If q 1 is positive (real The image image 1)... due to surface 1 is real, so I 1 is to the right of the surface. R 1 R 2 Surface 1 Surface 2 n O n 1 1 t I 1 C 1 where q 1 formed b if the ima Now le (We mak 2 are in t the objec p 1 p 2 b... p 2 is negative (object behind surface). Either way, the sign is taken care of in the expression p 2 = q 1 + t. Figure To locate the image formed q 1 We now i face acts is virtual related to q 1 is nega image fro

44 Relate the original object to the final image surface 1 surface n = n 1 (2) p 1 q 1 R 1 n ( q 1 + t) + 1 q 2 = 1 n R 2 For a thin lens, t is negligible, so take it to be zero. n q q 2 = 1 n R 2 (3) Add equation (2) to equation (3) to get ( 1 = (n 1) 1 ) p 1 q 2 R 1 R 2

45 Thin Lens Equation Lastly, measure the object distance to be p = p 1 + t/2 and q = q 2 + t/2. Since t is effectively zero, we have: 1 p + 1 ( 1 = (n 1) 1 ) q R 1 R 2 p and q can vary, but since n, R 1, and R 2 are constant for a fixed lens, we know 1 p + 1 q = const.

46 Thin Lens Equation Lastly, measure the object distance to be p = p 1 + t/2 and q = q 2 + t/2. Since t is effectively zero, we have: 1 p + 1 ( 1 = (n 1) 1 ) q R 1 R 2 p and q can vary, but since n, R 1, and R 2 are constant for a fixed lens, we know 1 p + 1 q = const. If p, q f, by the definition of the focal length. This gives the thin lens equation 1 p + 1 q = 1 f

47 Lens Maker s Equation Thin lens equation 1 p + 1 q = 1 f and 1 p + 1 q ( 1 = (n 1) 1 ) R 1 R 2 The LHS is the inverse of the focal length, giving the lens maker s equation: 1 f ( 1 = (n 1) 1 ) R 1 R 2

48 Magnification with a Lens By definition, M = h h And it follows from simple trigonometry that M = q p Same as for a mirror!

49 Sign Conventions for Lenses! 1 p + 1 q = 1 f Variable is Positive is Negative p object in front of surface [virtual object] 3 q image behind lens image in front of lens (real) (virtual) h (and M) image upright image inverted R 1 and R 2 object faces convex surf. object faces concave surf. (C behind surface) (C in front of surface) f lens is converging lens is diverging 3 Useful in derivations.

50 Understanding the Sign of f 1 f ( 1 = (n 1) 1 ) R 1 R 2 Always n 1, so (n 1) is positive (if zero, there s no lens!). ( ) 1 R 1 1 R 2 could be positive or negative, depending on the signs and magnitudes of R 1 and R 2.

51 applies to a refracting surface.) Sign Examples: Converging Lenses Biconvex Ra 1 R 2 R 1 R 2 R 1 R 2 Planoconvex Convexconcave R 1 +ve R 1 +ve R 1 +ve Biconcave R 2 ve RConvex- concave concave 2 +ve Plano- R 2 R 1 < R 2 f +ve f +ve f +ve Figure 36 sign conven refracting s Various l thicker at t center than Magnific Consider a (Eq. 36.2), image is

52 a Sign Examples: Diverging Lenses Biconcave Convexconcave Planoconcave Consider a (Eq. 36.2) image is b R 1 ve R 1 +ve R 1 Figure Various lens shapes. R 2 +ve (a) Converging R 2 +ve lenses R 2 ve have a positive Rfocal 1 > R 2 length and are fthickest ve at the f ve middle. f ve (b) Diverging lenses have a From this e the same s and on the Ray Diag Ray diagra tems of len diagrams f

53 Focal Length Question Quick Quiz 36.6 What is the focal length of a pane of window glass? (A) zero (B) infinity (C) the thickness of the glass (D) impossible to determine

54 Focal Length Question Quick Quiz 36.6 What is the focal length of a pane of window glass? (A) zero (B) infinity (C) the thickness of the glass (D) impossible to determine

55 Ray Diagrams for Converging Lenses Again, draw rays whose behavior we know. Rays parallel to the principle axis are refracted through the focal point. Rays that travel through the center of the lens are (effectively) not refracted. For Converging Lenses: 1 Draw a ray from the top of the object parallel to the principle axis refracted through the focal point F. 2 Draw a ray from the top of the object through the focal point (or back to the focal point) and refracted parallel to the principal axis. 3 Draw a ray from the top of the object through the center of the lens, and continuing in a straight line.

56 Ray Diagrams for Converging Lenses When the object is in front of and outside the focal point of a converging lens, the image is real, inverted, and on the back side of the lens. Wh foca the tha side O F 1 F 2 I I F1 Front Back Fron a b Figure Ray diag

57 Ray Diagrams for Converging Lenses When the object is in front of and outside There are two the cases focal of point interest: of a converging lens, the image is real, inverted, and on the back side of the lens. Object is beyond focal point Wh foca the tha side O F 1 F 2 I I F1 Front Back Fron a Image is real and inverted. b Figure Ray diag

58 Object Beyond Focal Point The object is the Sun. 1 Photo from

59 Ray Diagrams than the forobject, Converging and on Lenses the front side of the lens. Object is closer than the focal point o l 2 1 F 2 I F 1 O 3 Front Back O b Image is virtual, upright, and magnified. c Ray diagrams for locating the image formed by a

60 Object Closer Than the Focal Point The object is the stamp.

61 Converging Lens Example A converging lens has a focal length of 10.0 cm. (a) An object is placed 30.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. (b) An object is placed 5.00 cm from the lens. Find the image distance and describe the image.

62 Converging Lens Example A converging lens has a focal length of 10.0 cm. ation (a) An object is placed 30.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. (b) An object is placed 5.00 cm from the lens. Find the image by a Converging Lens distance and describe the image cm. he lens. Contance, and The object is farther from the lens than the focal point. The object is closer to the lens than the focal point. erging, the focal expect the poss. Figure (Example 36.8) An image is formed by a converging lens. F 2 I 10.0 cm 15.0 cm 30.0 cm a I, F 1 O F 2 b 10.0 cm 10.0 cm 5.00 cm

63 Converging Lens Example (a) An object is placed 30.0 cm from the lens. q = +15.0cm M = (b) An object is placed 5.00 cm from the lens. q = 10.0cm M = (c) An object is placed 10.0 cm from the lens. Find the image distance and describe the image.

64 Converging Lens Example (a) An object is placed 30.0 cm from the lens. q = +15.0cm M = (b) An object is placed 5.00 cm from the lens. q = 10.0cm M = (c) An object is placed 10.0 cm from the lens. Find the image distance and describe the image. No image is formed! q =

65 Ray Diagrams for Diverging Lenses For Diverging Lenses: 1 Draw a ray from the top of the object parallel to the principle axis refracted outward directly away focal point F on the front side of the lens. 2 Draw a ray from the top of the object toward the focal point on the back side of the lens and refracted parallel to the principal axis. 3 Draw a ray from the top of the object through the center of the lens, and continuing in a straight line.

66 Ray Diagrams object, for and Diverging on the front Lenses side of the lens. 1 2 F 2 O F 1 I 3 Front Back c The image formed is upright, virtual, and smaller in size. ed by a thin lens.

67 Diverging Lenses The image formed is upright, virtual, and smaller in size. 3 Photo from the Capilano University Physics Dept. webpage.

68 Combinations of Lenses Two or more lenses can be used in series to produce and image. This is used in eyeglasses (your eyeball lens is the second lens) refracting telescopes microscopes camera zoom lenses Lenses can also be used together with curved mirrors, for example in reflecting telescopes. Two thin lenses placed right up against each other (touching) will act like one lens with a focal length 1 f = 1 f f 2

69 al focal lengths by the expression Combinations of Lenses Example Two thin converging lenses (36.19) of focal lengths Focal length f 1 = 10.0 for a cm and combination of two thin f 2 = 20.0 cm are separated by 20.0 cm as illustrated. An object is er are placed equivalent 30.0 cmto to a single the left thin lenses in contact of lens 1.? 0 cm ated left f the Find the position and the magnification of the final image. O 1 Lens 1 Lens 2 I 2 I cm 15.0 cm 6.67 cm the ) in hese 30.0 cm 20.0 cm Figure (Example 36.10) A combination of two converging lenses. The ray diagram shows the location of the final image

70 Summary images formed by spherical mirrors images formed by refraction images formed by lenses images formed by lens combinations Collected Homework! due Monday, June 19. Final Exam 6:15-8:15pm, Monday, June 26. Homework Serway & Jewett: Carefully read all of Chapter 36. Ch 36, onward from page OQs: 1, 3, 5, 11, 13; CQs: 5, 9, 11; Probs: 1, 7, 9, 19, 21, 29, 37, 39, 43, 53, 55, 71, 73, 87, 89, (93)