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Introduction to ray diagrams: CHAPTER 32 OPTICAL IMAGES • Introduction to ray diagrams * object image • Reflection from mirrors • Plane mirrors • Spherical mirrors • Sign convention • Lenses • Converging Lenses • Diverging Lenses • Chromatic aberration * Rays emerge from all points on an object. Only some of those from the top of the head are shown here. Note: the image does not really exist ... you cannot display it on a screen, for example! It is called a virtual image. ** Note that only 2 rays are required to locate the image position ** Geometry of image formation (plane mirror): s′ s B 1 θ A Q θ θ θ θ O Question 32.1: You are 1.62 m tall and want to be able to see your full image in a vertical plane mirror. P (a) What is the minimum height of the mirror that will meet your needs? 2 (b) How far above the floor should the bottom of the ΔOAB and ΔOPQ are congruent triangles, i.e., OA = OP and AB = PQ mirror in (a) be placed, assuming that the top of your head is 14 cm above your eye level? (c) If you move closer to the mirror, describe what Lateral magnification m = PQ ( = 1). AB Also, you would see in the mirror. Use ray diagrams to explain your answers. object distance (OA) = image distance (OP) i.e., s = s′ . Image is virtual and upright. 1.62 m A ray diagram showing rays from your feet and the top of your head reaching your eyes is shown above. (a) Therefore, the mirror must be half your height, i.e., 0.81 m. (c) You move closer to the mirror. A ray diagram showing rays from your feet and the top of your head (b) The top of the minimum height mirror must be halfway between your eyes and the top of your head, i.e., 1.55 m from the floor. Therefore, the bottom of the mirror should be 1.55 m − 0.81 m = 0.74 m, above the floor. reaching your eyes is shown above, with the same size mirror. Notice you can still see all your body; it makes no difference where you stand! Formation of an image with two plane mirrors. P1 P2 1 Reflection in a plane mirror produces: 2 lateral inversion z′ y′ z y x RH set of axes (object) x′ LH set of axes (image) The rays from the object into the eye satisfy the law of reflection. We see that the image at P1 due to mirror 1 acts as an object for mirror 2 . The final image appears at P2. Note that both images are virtual. IMPORTANT CONCEPT: Even though the image at P1 is virtual, it can act as an object for mirror 2 . Question 32.2: When two plane mirrors are parallel, such as on opposite walls in a hairdressing salon, multiple images arise because each image in one mirror serves as Even though a virtual image does not actually exist, you an object for the other mirror. An object is placed can see it and photogragh it because it can act as an between parallel mirrors separated by 3.00 m. The object object for another lens or mirror and produce a real is 1.00 m in front of the left mirror and 2.00 m in front of image. the right mirror. PS: Ever done a selfie in a mirror? (a) What is the distance from the left mirror to the first four images in that mirror? (b) What is the distance from the right mirror to the first four images in that mirror? 3.0 m 1.0 m 5.0 m 2.0 m 4.0 m See the notes I wrote on 7.0 m 8.0 m “retroreflection” 10.0 m 11.0 m available on the useful notes link on the web-site. (a) For the left hand mirror (red) the images are 1.0 m; 5.0 m; 7.0 m; 11.0 m behind the mirror. (b) For the right hand mirror (blue) the images are 2.0 m; 4.0 m; 8.0 m; 10.0 m behind the mirror. REFLECTION FROM CURVED SURFACES 1. Concave mirror optical axis object 2 1 θ α φ •C β ℓ θ γ image r s′ optical axis object • image r s′ s s NOTE: The image produced in this example is real; it Using simple geometry we have: actually exists, i.e., we could see it on a screen. β = α + θ and γ = α + 2θ. Eliminate θ and we get: α + γ = 2β If α, β and γ are small then: ℓ ℓ ℓ α≈ β≈ γ≈ s r s′ 1 1 2 i.e., + = s s′ r My convention is that if the rays forming the image are solid lines, it is a real image. Dashed lines are rays that do not actually exist ... they show paths that rays appear to follow, for example, behind a mirror. Beam parallel to the optical axis (i.e., from a very distant object) •C • F r s= ∞ • s′ C 1 1 2 + = s s′ r r If s = ∞ then s′ = = f , 2 where F is the principal focus and f is the focal length. 1 1 1 ∴ + = . s s′ f Note, it is a real focus. Can you think of an example? s= f s′ = ∞ To produce a parallel beam of light put the source at the principal focus. Can you think of an example? Image formation in a concave mirror Tee, hee, hee ... there’s a sign convention for mirrors and lenses! I. Object distance: s > r s object y •F y′ •C image • light, s > 0. Image distance ( s′ ): if the image is on the same side of the reflecting or refracting surface as the outgoing light, s′ > 0. • 1 Object distance ( s): if the object is on the same side of the reflecting or refracting surface as the incoming • 2 Radius of curvature (r): if the center of curvature is on the same side as the outgoing light, r > 0. s′ s>0 s′ > 0 r>0 The lateral (or linear) magnification is defined as m = y′ y . But y s = − y ′ s′ . ∴m = − s′ s (⇒ negative). Definition of magnification. Note in this case that m < 0 and m < 1. The image is real, inverted and reduced. Image formation in a concave mirror Image formation in a concave mirror II. Object distance: r 2 < s < r (i.e., f < s < r) III. Object distance: s < f (i.e., s < r 2) s image object •F y′ y object y •F •C y′ s •C s′ image 1 2 s′ s>0 2 1 s′ > 0 r>0 s>0 s′ < 0 r>0 The lateral (or linear) magnification is m = y ′ y = − s′ s (⇒ negative). The lateral (or linear) magnification is m = y ′ y = s′ s (⇒ positive). Note that m < 0 and m > 1. The image is real, inverted Note that m > 1. The image is virtual, upright and and enlarged. enlarged. Here’s a summary! For a concave mirror ... F C r f >0 m = − s′ s . Question 32.3: A concave mirror has a radius of curvature of 24.0 cm. Find the image position for objects that are placed • When s > r: then m < 0 and m < 1. The image is real, inverted and reduced. • When f < s < r: then m < 0 and m > 1. The image is real, inverted and enlarged. • When s < f : then m > 1. The image is virtual, upright and enlarged. positive m ( m > 0) ⇔ virtual image (upright) negative m ( m < 0) ⇔ real image (inverted) (a) 55.0 cm, (b) 24.0 cm, (c) 12.0 cm, and (d) 8.00 cm from the mirror. For each case give the magnification and state whether the image is real or virtual; upright or inverted. Given r = 0.24 m. Since the mirror is concave then r and f ( = r 2 = 0.12 m) are both positive. 1 1 1 (a) For s = 0.55 m and using + = , we have: s s′ f 1 1 1 = − = 6.52 m −1, i.e., s′ = 0.15 m. s′ 0.12 m 0.55 m ∴m = − s′ s = −0.15 m 0.55 m = −0.27. Since s′ > 0, m < 0 and m < 1, the image is on the same side as the object, it is real, inverted and reduced. (b) For s = 0.24 m: 1 1 1 = − = 0.417 m −1, i.e., s′ = 0.24 m. s′ 0.12 m 0.24 m ∴m = − s′ s = −0.24 m 0.24 m = −1.00. Since s′ > 0, m < 0 and m = 1, the image is on the same side as the object, it is real, inverted and the same size as the object. (c) s = 0.12 m: 1 1 1 1 = − = 0, i.e., s′ = − = ∞. s′ 0.12 m 0.12 m 0 So, the emerging rays are parallel and do not form an image. (d) s = 0.08m: 1 1 1 = − = −4.17 m −1, i.e., s′ = −0.24 m. s′ 0.12 m 0.08 m ∴m = − s′ s = 0.24 m 0.08 m = 3.0. Since s′ < 0, m > 0 and m > 1, the image is on the opposite side to the object, it is virtual, upright and enlarged. 2. Convex mirror: 2 2 θ θ θ θ 1 γ α object s s′ θ r β image 1 •C θ = α + β and 2θ = α + γ Eliminate θ and we get: γ − α = 2β ℓ ℓ 2ℓ 1 1 2 i.e., − = so, − = . s′ s r s′ s r But, using the sign convention we find: s′ < 0 1 1 2 ∴ + = , s′ s r s s′ θ r β image •C 1 1 2 + = . s′ s r Using simple geometry: s>0 object γ α r < 0, which is the same result obtained for a concave mirror! When s = ∞, i.e., an incoming parallel beam of light, we 1 2 1 have: = = s′ r f r i.e., f = (virtual focus), 2 1 1 1 so + = . s′ s f ** Note that by the sign convention, r < 0, and so f < 0 for a convex lens ** Image formation in a convex mirror 1 • object 2 image s′ F C • Question 32.4: A convex mirror has a radius of curvature of 24.0 cm. Find the image position for objects that are placed s (a) 55.0 cm, By simple geometry the lateral magnification of a (b) 24.0 cm, convex mirror is: (c) 12.0 cm, and m = y ′ y = − s′ s , but s′ < 0, so m > 0. (d) 8.00 cm from the mirror. For each case give the magnification and state whether the image is real or virtual; upright or inverted. Note also, 0 < m < 1 always and so the image is upright, virtual and reduced in size. Can you think of an example ? Given r = 0.24 m. Since the mirror is convex then r and f ( = r 2 = 0.12 m) are negative. (a) s = 0.55 m: 1 1 1 =− − = −10.2 m −1, s′ 0.12 m 0.55 m i.e., s′ = −0.10 m. ∴m = − s′ s = 0.10 m 0.55 m = +0.18. Since s′ < 0, m > 0 and m < 1, the image is on the opposite side as the object, it is virtual, upright and reduced. (b) s = 0.24 m: 1 1 1 =− − = −12.5 m −1, s′ 0.12 m 0.24 m i.e., s′ = −0.08 m. ∴m = − s′ s = 0.08 m 0.24 m = +0.33. Since s′ < 0, m > 0 and m < 1, the image is on the opposite side as the object, it is virtual, upright and reduced. (c) s = 0.12 m: 1 1 1 =− − = −16.7 m −1, s′ 0.12 m 0.12 m i.e., s′ = −0.06 m. ∴m = − s′ s = 0.06 m 0.12 m = +0.50. Since s′ < 0, m > 0 and m < 1, the image is on the opposite as the object, it is virtual, upright and reduced. (d) s = 0.08 m: 1 1 1 =− − = −20.83 m −1, s′ 0.12 m 0.08 m i.e., s′ = −0.048 m. ∴m = − s′ s = 0.048 m 0.08 m = +0.60. Since s′ < 0, m > 0 and m > 1, the image is on the opposite side to the object, it is virtual, upright and reduced. (b) The mirror cannot be convex because a convex mirror always produces a diminished virtual image. Therefore, it must be concave. s = 2.10 cm. (a) Question 32.5: A dentist wants a small mirror that will s produce an upright image that has a magnification of s′ The image is upright and so it is virtual, i.e., behind 5.50 when the mirror is located 2.10 cm from a tooth. the mirror, so s′ < 0. (a) Should the mirror be concave or convex? (b) What should the radius of curvature of the mirror be? The magnification: m = − s′ s = 5.50 ∴ s′ = −5.50 × 2.10 cm = −11.6 cm. 1 1 1 2 1 1 ∴ + = = = − = 0.39 cm−1 s s′ f r 2.10 cm 11.6 cm 2 i.e., r = = 5.13 cm. 0.39 cm−1 ∴f = r 2 = 2.56 cm, so s < f . Types of Lenses ... Converging Lenses Planoconvex Biconvex** Positive Meniscus See the notes on the web-site for chapter 32 on how to solve mirror problems using a scale drawing ... Diverging Lenses Planoconcave Biconcave** Negative Meniscus Cross-sections of various types of lenses. The upper group are converging lenses, the lower group are diverging lenses. Convex lens ... Converging lens (convex) 5.0 cm Principal focus Parallel beam Real focus Diverging lens (concave) Focal length (f) Lens-maker’s formula: n1 Principal focus Parallel beam n2 r1 Real focus r2 Focal length (f) 1 ⎛ n2 ⎞ ⎛ 1 1 ⎞ = ⎜ − 1⎟ ⎜ − ⎟ f ⎝ n1 ⎠ ⎝ r1 r2 ⎠ Virtual focus Optometrists who prescribe lenses and opticians who Principal focus make them do not normally specify the focal length of a Parallel beam lens but its refractive power, in units called diopters. Refractive power = Focal length (f) 1 (diopters) f (in meters) Converging lenses have f > 0 but diverging lenses have f < 0. Sign convention: If the center of curvature is on the same side as the outgoing light r > 0 . Incoming light • • r2 r1 > 0 Incoming light r2 < 0 Question 32.6: Place the following in order of increasing focal length, i.e., from shortest to longest: r1 Outgoing light r2 • • r1 Incoming light Outgoing light r1 r1 < 0 r2 > 0 Outgoing light r1 • • r2 r1 > 0 r2 > 0 1 2 r1 r1 r1 r2 3 r2 > r1 r1 1 r1 > 0 r2 = ∞ r1 r1 r1 r2 2 both positive 3 r2 > r1 both positive 1 1 r 1: = (n − 1) . ∴f1 = 1 . f1 r1 (n − 1) 2: 3: ⎛ 1 1⎞ 1 = (n − 1)⎜ − ⎟ = 0. ∴f 2 = ∞. f2 ⎝ r1 r1 ⎠ 1 ⎛ 1 1⎞ ⎛r −r ⎞ = (n − 1)⎜ − ⎟ = (n − 1)⎜ 2 1 ⎟ . f3 ⎝ r1 r2 ⎠ ⎝ r1r2 ⎠ r r2 ∴f 3 = 1 . (n − 1) (r2 − r1 ) >1 ∴f1 < f 3 < f 2 . Note, if n1 = 1 (air), the lens maker’s formula becomes: ⎛ 1 1⎞ 1 = (n 2 − 1)⎜ − ⎟ . f ⎝ r1 r2 ⎠ But n 2 varies with wavelength (λ) ... dispersion. Therefore, the focal length, f, varies with light color. Since, for glass, n 2 increases for decreasing λ (i.e., from red light to blue light ), the focal length for blue light is less than for red light. So, blue light is “bent” more than red light. blurred focus “chromatic aberration” ** See the articles on the web-site I wrote about aberration ** ** See also lens-maker’s formula under “useful notes” ** Incoming light Outgoing light r1 • • r1 • • r2 Question 32.7: Find the focal length of a lens that has an index of refraction of 1.62, a convex surface with a radius of curvature of r1 = 0.40 m and a concave surface with a radius of curvature of r2 = 1.00 m. r2 Using the sign convention r1 > 0 and r2 > 0. We have 1 ⎛ n2 ⎞ ⎛ 1 1 ⎞ = ⎜ − 1⎟ ⎜ − ⎟ , f ⎝ n1 ⎠ ⎝ r1 r2 ⎠ where n 2 = 1.62 and n1 = 1.00. 1 1 1 ⎞ ∴ = (1.62 −1)⎛ − = 0.93 m −1, ⎝ 0.40 m 1.00 m ⎠ f i.e., f = 1.08 m. Convex lens ... f s′ s f F F s y y′ virtual upright image Image Object f f If s > 2f then m < 1, i.e., an inverted and reduced image. s′ < 0 0 For a convex lens we have (see “useful notes”): 1 1 1 + = s s′ f and m = y ′ y = − s′ s . s′ − f f (Also, by substitution: m = − =− .) f s−f f s−f s>0 f >0 m= − real inverted image s′ > 0 object distance (s) s<f s= f s>f When: • s>f • s = 2f then m = 1 (real, inverted) • s=f then m= ∞ • s<f then m > 1 (virtual, upright) then m < 0 (real, inverted) Interesting case !! s′ s Object Question 32.8: An object that is 0.03 m high is placed Image f 0.25 m in front of a thin lens that has a power equal to f 10.0 D. m= − f s− f If s < f , then m > 1. Also, 1 1 1 since > , then < 0, s f s′ i.e., s′ < 0. So, to use a convex lens as a magnifying glass make s < f , i.e., object distance less than the focal length of the lens. The image is on the same side as the object and it is enlarged, upright and virtual. (a) Find the position and the size of the image. (b) Repeat part (a) if the object is placed 0.20 m in front of the lens. (c) Repeat part (a) if the object is placed 0.05 m in front of the lens. 1 The power of a lens is given by P = . f ∴f = P −1 = 0.10 m. 1 1 1 (a) With s = 0.25 m. The lens formula is + = . s s′ f 1 1 1 s−f fs ∴ = − = , i.e., s′ = . s′ f s fs s−f With s = 0.25 m and f = 0.10 m, we find (0.10 m)(0.25 m) s′ = = 0.167 m. (0.25 m − 0.10 m) ∴m = − s′ 0.167 m =− = −0.667. s 0.25 m The object height is y = 0.03 m, so the image height is y ′ = my = (−0.667)(0.03 m) = −0.02 m. Since s′ > 0 the image is real. Since y ′ < y and y ′ < 0, the image is reduced and inverted. ∴m = − s′ 0.20 m =− = −1.00, s 0.20 m and the image height is y ′ = my = (−1.00)(0.03 m) = −0.03 m. Since s′ > 0 the image is real. Since y ′ = y and y ′ < 0, the image is the same size as the object and inverted. (c) With s = 0.05 m, we find (0.10 m)(0.05 m) s′ = = −0.10 m. (0.05 m − 0.10 m) ∴m = − s′ 0.10 m = = +2.00, s 0.05 m and the image height is y ′ = my = (2.00)(0.03 m) = 0.06 m Since s′ < 0 the image is virtual. Since y ′ > y and (b) With s = 0.20 m, we find (0.10 m)(0.20 m) s′ = = 0.02 m. (0.20 m − 0.10 m) y ′ > 0, the image is enlarged and upright. Remember Fermat’s principle? It states that a light ray traveling between two points will take the path with the least time. So, if one of A A B C C B these light ray paths took less time than any of the others, by Fermat’s principle they would all take that path! But we Question 32.9: If the rays all start from the object at the same time, which arrives at the image first or do they know that light travel along A, B and C ... so they must arrive at the same time? all take the same time (even though the distances traveled are not the same)! Although path C is the shortest, the ray travels further through glass where the velocity is less. Concave lens ... s Object A Image F A Question 32.10: A double concave lens* that has an s′ index of refraction of 1.435 has radii whose magnitudes f are 0.03 m and 0.25 m. An object is positioned 0.80 m to the left of the lens. Find Note for a concave lens: s > 0, s′ < 0, f < 0 and s′ < s . Since m = − s′ s then 0 < m < 1, always. (a) the focal length of the lens, (b) the position of the image, and (c) the magnification of the image. (d) Is the image real or virtual, upright or inverted? So, a diverging lens always produces a virtual image of a real object no matter where it is positioned. The image is upright and smaller relative to the object ABCDEF ABCDEF * Both sides of the lens are concave. Incoming light r1 r2 Outgoing light (c) m = − s′ −0.22 m =− = +0.275. s 0.80 m (d) Since m > 0 the image is virtual and upright. Since r1 = −0.30 m (opposite side of outgoing light). r2 = 0.25 m (same side of outgoing light). n1 = 1.00 and n 2 = 1.45 ⎞⎛ 1 1 ⎞ 1 ⎛n (a) = ⎜ 2 − 1⎟ ⎜ − ⎟ f ⎝ n1 ⎠ ⎝ r1 r2 ⎠ 1.45 ⎞ ⎛ 1 1 ⎞ =⎛ −1 − = −3.30 m −1 ⎝ 1.00 ⎠ ⎝ −0.30 m 0.25 m ⎠ ∴f = −0.303 m m > 1 the image is diminished. See useful notes on web-site to see how to solve this problem using a scale drawing. Object (diverging lens). 1 1 1 (b) s = 0.80 m and + = . s s′ f 1 1 1 ∴ + = 0.80 m s′ −0.303 m 1 1 1 i.e., = − = −4.55 m−1 s′ −0.303 m 0.80 m ∴ s′ = −0.22 m (opposite side to outgoing light) Image Incoming light Outgoing light Note: the image is not necessarily on the same side as the outgoing light! FRESNEL LENS: Only the radii of the front and back surfaces of a lens determine the focal length, not the thickness in between! So, much weight can be saved with this plano-convex lens by removing the “unused” part of a lens! Compared with regular lenses, Fresnel lenses produce a poorer image quality, so they tend to be used only where quality is not critical or where the bulk of a solid lens would be prohibitive. Cheap Fresnel lenses can be stamped or moulded out of transparent plastic and are used in overhead projectors, projection televisions, and hand-held sheet magnifying glasses. Fresnel lenses have been used to increase the visual size of CRT displays in pocket televisions, and they are also used in traffic lights. Lighthouse lens