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Optics Image Formation from Mirrors and Lenses Lana Sheridan De Anza College June 8, 2016 Last time • refraction • Huygens’ Principle • dispersion Overview • mirrors • image terminology • images formed by spherical mirrors • images formed by refraction • images formed by lenses • images formed by lens combinations Ray Optics and Image Formation Simple geometric ray optics can be used to understand how images are formed by simple optical devices: mirrors and lenses. 1 Wikipedia user Heptagon. Images Formed by Flat Mirrors When we see an object “in the mirror”, we are not actually seeing something that is behind the mirror. We are seeing an image of and object that is placed in front of the mirror. The image seems to be the same distance behind the mirror as the object is in front of it. The image seems to be the same size as the object. This is not true for all optical devices, but we can work out things about how the image will form for many different optical devices. Image Formation Terminology object distance, p The perpendicular (shortest) distance from the object to the device. image distance, q The perpendicular (shortest) distance from where the image appears to be to the device. (lateral) magnification, M The factor by which the image size exceeds the object’s size M= image height object height For mirrors and lenses M=− q p Image Formation Terminology real image An image that can be displayed on a screen formed when the light rays pass through and diverge from the image point. virtual image An image that cannot be displayed on a screen, but can be seen “in the device” because the light rays appear to diverge from the image point. The image in a flat mirror is virtual. Image Formation Terminology upright image An image that appears to be right-side-up with respect to the object. inverted image An image that appears to be upside-down with respect to the object. The image in a flat mirror is upright. focal length, f The distance from the optical device to where a set parallel rays striking the device head on (perpendicularly) will be focused. For a flat mirror, the f is infinite. rays to deterand P'QR are congruent, path perpen| ! | q Mirror | and h ! h". Ray Diagram for a| pFlat ay follows the law of reflected rays back p Q q P P' 9 behind the object would h' R h u object behind Image Object u , so |p | 5 |q |. irror is as far h equals the Ray diagrams have the optical sketched vertically in cross Figure 36.2 device A geometric confollows: section, the object on the left represented by an arrow pointing up. struction that is used to locate the image of an object placed in front of a flat mirror. The(36.1) image is also represented by an arrow, but it may be to the left or right, pointing up or down depending on how the image is formed. a spherical d a concave e R, and its ction, and a 6.6a shows a olid, curved port for the flecting suro a point as Henry Leap and Jim Lehman Images formed by Spherical Mirrors: Concave Mirrors Figure 36.7 focus light. Reflection of paral6.6b,Concave wheremirrorslelcanrays from a concave mirror. ght rays that 6 Images formed by Spherical Mirrors: Concave If the rays diverge from Mirrors O at small angles, they all reflect through the same image point I. Center of curvature Mirror Mirror R C C V O I Principal axis a b (a) A concave mirror of radius R. The center of curvature C is located on th oint object placed at O in front of a concave spherical mirror of radius R, wher principal axis farther than R from the mirror surface, forms a real image at I. Assumption: paraxial rays In studying curved mirrors and thin lenses we assume that all rays are paraxial rays. Paraxial rays are rays close to the principle optical axis of our optical device. (Rays that strike close to the middle.) For a spherical lens, rays that strike further from the axis are not focused to the same point (spherical aberation). 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 1 1 = + f p q Spherical Concave Mirrors Focal Length For a spherical concave mirror of radius R f = R 2 Spherical Concave Mirrors The focal point is where the detectors are place on satellite dishes and radio telescopes. 1 Parkes telescope in Australia. Concave Mirrors and the Mirror Equation Simple geometry shows why the mirror equation is true for a concave mirror. ge Formation The real image lies at the location at which the reflected rays cross. h a O C V I u a h' u 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 This this geometric construction is used to derive Equation 36.4. We willC.use ray diagram. 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. ge Formation Concave Mirrors and the Mirror Equation The real image lies at the location at which the reflected rays cross. h a O C V I u a h' u 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. Thisof geometric construction is used to derive Equation The angles incidence and reflection are the same36.4. magnitude, θ. h0 q So,sign − p , and because the image is inverted, so h9 is taken to be neganegative is introduced h = q tive. Therefore, from Equation 36.1 and M =these − results, we find that the magnificap tion of the image is Concave Mirrors and the Mirror Equation Looking at the green triangle and the red triangle with angle α: −h 0 h = R −q p−R which rearranges to p−R −h 0 =− h p−R Using our magnification expression: p−R q =− p p−R Concave Mirrors and the Mirror Equation p−R q =− p p−R Cross-multiplying and rearranging gives 2 1 1 = + R p q However, we already concluded that f = R/2, so 1 1 1 = + f p q We have confirmed the mirror equation for spherical concave mirrors, and it follows from simple geometry. o not derive any equations for convex spherical mirrors because Eq 6.4,Convex and 36.6Mirrors can be used for either concave or convex mirrors if we ct sign convention. We will refer to the region in which light rays or ve 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 p q I F C Examples of Ray Diagrams 36.2 Images Formed by Sphe F 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. T S J 2 3 F C O Principal axis I Front Back a When the object is located between the focal point and a concave mirror surface, the image is virtual, upright, and enlarged. © Cengage Learning/Charles D. Winters 1 Front Examples of Ray Diagrams Back a When the object is located between the focal point and a concave mirror surface, the image is virtual, upright, and enlarged. 2 3 C 1 F O Front I Back b When the object is in front of a convex mirror, the image is virtual, upright, © Cengage Lea Principal axis Front Back Examples of Ray Diagrams b When the object is in front of a convex mirror, the image is virtual, upright, and reduced in size. 1 3 2 O I F Front C Back c r 3BZJTESBXOGSPNUIFUPQPGUIFPCKFDUUPXBSEUIFGPDBMQPJOUPOUIFCBDL Mirror Question Quick Quiz 36.31 Consider the image in the mirror in shown. Based on the appearance of this image, which of the following 1098 Chapter 36 Image should you conclude? in vi ro ve (A) the mirror is concave and the image is real (B) the mirror is concave and the image is virtual Q (C) the mirror is convex and the image is real (D) the mirror is convex and the image is virtual NASA Q 1 Serway & Jewett, page 1098. Figure 36.14 (Quick Quiz 36.3) Mirror Question Quick Quiz 36.31 Consider the image in the mirror in shown. Based on the appearance of this image, which of the following 1098 Chapter 36 Image should you conclude? in vi ro ve (A) the mirror is concave and the image is real (B) the mirror is concave and the image is virtual ← Q (C) the mirror is convex and the image is real (D) the mirror is convex and the image is virtual NASA Q 1 Serway & Jewett, page 1098. Figure 36.14 (Quick Quiz 36.3) Sign Conventions for Mirrors! 1 1 1 = + f p q Variable is Positive is Negative p object in front of mirror — q image in front of mirror (real) image behind mirror (virtual) h 0 and M image upright image inverted f and R concave mirror convex mirror 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. ple 36.4) An en in a convex e of an autoimage of the e frame of the monstrates the same locaace. . Bo Zaunders/Corbis Find the position of the image of the truck and the magnification of the image. 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?) 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. 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 = +0.057 p Images Formed by Refraction When light rays change media they are bent. This also can form images. sider two transparent m boundary between the We assume the object a principal axis diverge from a point We can find the location and size of the image formed by Let’s consider the parax object at O and are refracted considering paraxial rays. through the image point I. at the spherical surface Figure 36.17 shows a n1 ! n2 law of refraction applie Images Formed byRaysRefraction making small angles with the n1 n2 R O I p q Because u1 and u2 are a tion sin u < u (with ang We know that an exteri interior angles, so apply Figure 36.16 An image formed by refraction at a spherical surface. For paraxial rays, Snell’s law becomes: n1 θ1 = n2 θ2 (1) Images Formed by Refraction n1 u1 36.3 Images Formed by Refraction P Figure 36.17 G derive Equation 3 n 1 , n 2. n2 d u2 b a g C O I R p q Looking at the diagram: Combining all three expressions and eliminating u1 and u2 gives a 1−n 2β g 5 (n⇒ 2 2 n 1)bθ1 = α + β 180◦ − θ1 = 180◦ −n 1α (36.7) (2) Figure 36.17 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. 36.17), the likewise: horizontal legs of these triangles are approximately p for the triangle containing = γ + angle θ2 b, and(3) angle a, R for the triangleβ containing q for the triangle containing angle g. In the small-angle approximation, tan u < u, so we can write the approximate(2) relationships follows:them and subtract (1): Multiply by n1 from andthese (3)triangles by n2as , add tan a < a < d d d b<b< tan g < g < n2 β p= ntan q R + n2 γ 1 α + n1 β Substituting these expressions into Equation 36.7 and dividing through by d gives Rearranging: n n n 2n 2 2 1 5 (n2 − n1 )β n1 α + pn1 1 2 γq = R (36.8) Relation betw image distanc Images Formed by Refraction For small angles (paraxial approx): α ≈ tan α = d p Similarly, β≈ d d and β ≈ R q So, n1 n2 n2 − n1 + = p q R Flat Refracting Surfaces (Like a rectangular 1102 Chapterfish 36 tank.) Image Formation The image is virtual and on the same side of the surface as the object. n1 ! n2 n1 n2 Flat Refracting Surfaces If a refracting surface is flat, then R is infinite an In this case R → ∞.n 1 5 2 n 2 q p n1 n2 n2 + =0 q52 p p q n 1 O q p Figure 36.18 The image formed And so we see that the sign of q From this expression, according to Table 36.2, the image formed by a n2 − object p as illustrate same side of the surfaceqas=the n1 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 36.18. 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 Sign Conventions for Refracting Surfaces! n2 − n1 n1 n2 + = p q R Variable is Positive is Negative p object in front of surface [virtual object]2 q image behind surface (real) image in front of surface (virtual) h 0 (and M) image upright image inverted R object faces convex surf. (C behind surface) object faces concave surf. (C in front of surface) C is the center of curvature. 0 M = hh = − nn12 qp (proof for homework). 2 Will be useful in derivations. 36 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 far away to axis diverge from a point object at O and are refracted very close to the refracting surface? through the image point I. (A) It is always to the right of the surface. (B) It is always to the left of the surface. n1 ! n2 n1 n2 R O I (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. 2 Serway & Jewett, page 1102. p In t und side bou We Let’ at th F law q Figure 36.16 An image formed by refraction at a spherical surface. Bec tion We inte 36 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 far away to axis diverge from a point object at O and are refracted very close to the refracting surface? through the image point I. (A) It is always to the right of the surface. (B) It is always to the left of the surface. n1 ! n2 n1 n2 R O I (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. ← 2 Serway & Jewett, page 1102. p In t und side bou We Let’ at th F law q Figure 36.16 An image formed by refraction at a spherical surface. Bec tion We inte Refracting Surface Question 1102 Chapter 36 Image For The image is virtual and on Quick Quiz 36.5 In the figure, what happens to the image 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 n1 ? (C) It approaches the surface and then moves to the right of the surface. O q p Figure 36.18 The image formed by a flat refracting surface. All rays are assumed to be paraxial. 2 Serway & Jewett, page 1102. If a r n1 ! n2 n1 n2 (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. Flat From acco same in wh case, n 2, t in Fi Q ui po su th rig Refracting Surface Question 1102 Chapter 36 Image For The image is virtual and on Quick Quiz 36.5 In the figure, what happens to the image 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 n1 ? (C) It approaches the surface and then moves to the right of the surface. O q p Figure 36.18 The image formed by a flat refracting surface. All rays are assumed to be paraxial. 2 Serway & Jewett, page 1102. If a r n1 ! n2 n1 n2 (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. Flat From acco same in wh case, n 2, t in Fi Q ui po su th rig Images Formed by Thin Lenses We will derive the thin lens equation 1 1 1 = + f p q (Notice it is the same as the mirror equation!) And the lens maker’s equation 1 = (n − 1) f 1 1 − R1 R2 We do this by considering each side of the lens as a refracting surface. Images Formed by Thin Lenses The image due to surface 1 is virtual, A thick lens: so I1 is to the left of the surface. n1 ! 1 R1 I1 Surface 1 R2 Surface 2 n O p1 t C1 q1 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 I1 p2 a We can imagine that there first surface forms an image I1 , and where q 1 is that image becomes The the object for the second1 surface. image due to surface is real, so I is to the right of the surface. The pink arrow shows I11 formed by su if the image First image formed The image due to surface 1 is virtual, Let n1 = 1, n2so=I1n.is to the left of the surface. n1 ! 1 R1 I1 Surface 1 R2 Surface 2 n O t p1 C1 q1 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 I1 f p2 a 1 n n−1 + due=to surface 1 is real, The image p q R so I1 is1 to the1 right of1 the surface. where q 1 is th formed by sur if the image i Second image formed n 1 1−n + = p2 q2 R2 We must relate p2 to q1 . The new object is the image from the previous surface so: p2 = −q1 + t where t is the lens thickness. p2 could be positive (real object)... The image due to surface ... if q1 is negative (virtual image 1). 1 is virtual, so I1 is to the left of the surface. n1 ! 1 R1 I1 Surface 1 R2 Surface 2 n O p1 q1 p2 a t C1 just learn 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 a p2 could be negative (virtual object)... The image image 1). due to surface 1 is real, ... if q1 is positive (real so I1 is to the right of the surface. n1 ! 1 R1 Surface 1 R2 Surface 2 n where q 1 formed b if the ima Now le (We mak 2 are in t the objec I1 O p1 t C1 p2 q1 b Figure 36.21 To locate the image formed We now i face acts is virtual related to q 1 is nega a p2 could be negative (virtual object)... The image image 1). due to surface 1 is real, ... if q1 is positive (real so I1 is to the right of the surface. n1 ! 1 R1 Surface 1 R2 Surface 2 n where q 1 formed b if the ima Now le (We mak 2 are in t the objec I1 O p1 t C1 p2 We now i face acts q1 is virtual related to b Either way, the sign is taken care of in the expression p2 = −q1 + t. q 1 is nega Figure 36.21 To locate the image formed Relate the original object to the final image surface 1 1 n n−1 + = p1 q1 R1 n surface 2 (−q1 + t) + (2) 1 1−n = q2 R2 For a thin lens, t is negligible, so take it to be zero. − n 1 1−n + = q1 q2 R2 Add equation (2) to equation (3) to get 1 1 1 1 + = (n − 1) − p1 q2 R1 R2 (3) Thin Lens Equation Lastly, measure the object distance to be p = p1 + t/2 and q = q2 + t/2. Since t is effectively zero, we have: 1 1 + = (n − 1) p q 1 1 − R1 R2 The RHS is the inverse of the focal length (lens maker’s equation) 1 = (n − 1) f 1 1 − R1 R2 This gives the thin lens equation 1 1 1 + = p q f Magnification with a Lens By definition, M= h0 h And it follows from simple trigonometry that M=− Same as for a mirror! q p Sign Conventions for Lenses! 1 1 1 + = p q f Variable is Positive is Negative p object in front of surface [virtual object]3 q image behind lens (real) image in front of lens (virtual) h 0 (and M) image upright image inverted R1 and R2 object faces convex surf. (C behind surface) object faces concave surf. (C in front of surface) f lens is converging lens is diverging 3 Useful in derivations. Understanding the Sign of f 1 = (n − 1) f 1 1 − R1 R2 Always n > 1, so (n − 1) is positive (if zero, there’s no lens!). 1 R1 1 R2 − could be positive or negative, depending on the signs and magnitudes of R1 and R2 . applies to a refracting surface.) Sign Examples: Converging Lenses Biconvex Convexconcave Planoconvex Figure 36 sign conven refracting s Various l thicker at t center than Magnific Ra1 R2 R1 +ve Biconcave R2 −ve f +ve R1 R2 R1 +ve RConvex2 +ve concave R1 < R2 f +ve R1 R2 R1 +ve PlanoR2 ∞ concave f +ve Consider a (Eq. 36.2), image is a Sign Examples: Diverging Lenses Biconcave Convexconcave Planoconcave Consider a (Eq. 36.2) image is From this e the same s and on the b R −ve R +ve R ∞ 1 Figure 36.25 1Various lens1 R2 +ve R2 +ve lenses R2 −ve shapes. (a) Converging have a positiveRfocal 1 > R2length and are fthickest at the middle. f −ve −ve f −ve (b) Diverging lenses have a Ray Diag Ray diagra tems of len diagrams f 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 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 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. Wh foca the tha side Ray Diagrams Converging Whenfor the object is in frontLenses of and outside the focal point of a converging lens, the image is real, inverted, and on the back side of the lens. 1 2 O 3 F2 I I F 1 F1 Front Back Fron b a Figure 36.26 Ray diag Ray Diagrams for Converging Lenses Wh foca the tha side When the object is in front of and outside the cases focal of point of a converging lens, the There are two interest: image is real, inverted, and on the back side of the lens. Object is beyond focal point 1 2 O 3 F2 I I F 1 F1 Front Back b a Image is real and inverted. Fron Figure 36.26 Ray diag Object Beyond Focal Point The object is the Sun. 1 Photo from http://www.mahalo.com/how-to-start-a-fire-with-a-magnifying-glass 6 than the and on the front Ray Diagrams forobject, Converging Lenses o l side of the lens. Object is closer than the focal point 2 1 I F 1 O O 3 Front b F2 Image is virtual, upright, and magnified. Back c Ray diagrams for locating the image formed by a Object Beyond Focal Point The object is the stamp. 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. Converging Lens Example A converging lens has a focal length of 10.0 cm. mation (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. d by a Converging Lens 10.0 cm. he lens. Constance, and verging, the focal e expect the poses. F2 I O 30.0 cm a I, F1 O F1 10.0 cm Figure 36.28 (Example 36.8) An image is formed by a converging lens. The object is closer to the lens than the focal point. The object is farther from the lens than the focal point. F2 10.0 cm 5.00 cm 15.0 cm 10.0 cm b Converging Lens Example (a) An object is placed 30.0 cm from the lens. q = +15.0cm M = −0.500 (b) An object is placed 5.00 cm from the lens. q = −10.0cm M = +2.00 (c) An object is placed 10.0 cm from the lens. Find the image distance and describe the image. Converging Lens Example (a) An object is placed 30.0 cm from the lens. q = +15.0cm M = −0.500 (b) An object is placed 5.00 cm from the lens. q = −10.0cm M = +2.00 (c) An object is placed 10.0 cm from the lens. Find the image distance and describe the image. q=∞ No image is formed! 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. object, for andDiverging on the front side of the Ray Diagrams Lenses lens. 1 2 O F1 Front I F2 3 Back c The image formed is upright, virtual, and smaller in size. med by a thin lens. Diverging Lenses The image formed is upright, virtual, and smaller in size. 3 Photo from the Capilano University Physics Dept. webpage. 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 1 1 = + f f1 f2 ual focal lengths by the expression Combinations of Lenses Example 36.10 (36.19) Focal length for a cm and Two thin converging lenses of focal lengths f1 = 10.0 combination of two thin f2 = 20.0 cm are separated by 20.0 cm as illustrated. An object is lenses in contact her areplaced equivalent to to a single 30.0 cm the leftthin of lens 1. 9. Find the position and the magnification of the final image. e? .0 cm rated e left of the h the f ) in these Lens 1 Lens 2 I2 I1 O1 6.67 cm 10.0 cm 30.0 cm Figure 36.30 15.0 cm 20.0 cm (Example 36.10) A combination of two converging lenses. The ray diagram shows the location of the final image Summary • image terminology • images formed by spherical mirrors • images formed by refraction • images formed by lenses • images formed by lens combinations Collected Homework! due Monday, June 13. Homework Serway & Jewett: • Carefully read all of Chapter 36. • Ch 36, onward from page 1123. 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)