Lect26 - Physics 212 Lecture 26: Lenses Lecture Lenses 50...

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Unformatted text preview: Physics 212 Lecture 26: Lenses Lecture Lenses 50 40 30 20 10 0 Confused Avg = 3.0 Confident Physics 212 Lecture 26, Slide 1 Music Who is the Artist? BB A) B) C) D) E) Pete Fountain Dr. Michael White Al Hirt Al Hirt George Lewis Dr. John Why?? I heard “Horn Man Blues” on the radio coming in this morning! heard Master of traditional New Orleans jazz !! Theme of the week: Classic Jazz Coltrane and Dr. Michael White Physics 212 Lecture 26, Slide 2 Physics 212 Lecture 26: Lenses Lecture Lenses 50 40 30 20 10 0 Confused Avg = 3.0 Confident Physics 212 Lecture 26, Slide 3 Refraction Snell’s Law n1sin(θ1) = n2sin(θ2) θ1 n1 n2 θ2 That’s all of the physics – all everything else is just geometry! Physics 212 Lecture 26, Slide 4 Object Location • Light rays from sun bounce off object and go in all directions – Some hits your eyes We know object’s location by where rays come from. We will discuss eyes in lecture 28… Physics 212 Lecture 26, Slide 5 Waves from object are focused by lens Physics 212 Lecture 26, Slide 6 Ray Tracing can be used to determine Image Physics 212 Lecture 26, Slide 7 Two Different Types of Lenses Physics 212 Lecture 26, Slide 8 Converging Lens: Consider the case where the shape of the lens is Consider such that light rays parallel to the axis of the mirror are all “focused” to a common spot a distance f behind the lens: to f f Physics 212 Lecture 26, Slide 9 Recipe for finding image: 1) Draw ray parallel to axis refracted ray goes through focus 2) Draw ray through center refracted ray is symmetric object f image You now know the position of the same point on the image Physics 212 Lecture 26, Slide 10 10 S > 2f image is: real inverted smaller 111 += S S′ f S′ M =− S object f image f S S’ Physics 212 Lecture 26, Slide 11 11 S = 2f image is: real inverted same size 111 += S S′ f S′ M =− S object f image f S S’ Physics 212 Lecture 26, Slide 12 12 2f > S > f 111 += S S′ f image is: real inverted bigger S′ M =− S object f image f S S’ Physics 212 Lecture 26, Slide 13 13 S=f 111 += S S′ f image is: at infinity… S′ M =− S object f f S Physics 212 Lecture 26, Slide 14 14 0<S<f 111 += S S′ f image is: virtual upright bigger S′ M =− S image f object S S’<0 f Physics 212 Lecture 26, Slide 15 15 Diverging Lens: Consider the case where the shape of the lens is Consider such that light rays parallel to the axis of the lens all diverge but such but appear to come from a common spot a distance f in front of the lens: f Physics 212 Lecture 26, Slide 16 16 image is: virtual upright smaller 111 += S S′ f S′ M =− S object f image f<0 S S’<0 Physics 212 Lecture 26, Slide 17 17 Executive Summary - Lenses: Executive S > 2f 2f real inverted smaller 2f > S > f real real inverted bigger f >S>0 virtual upright bigger S >0 virtual upright smaller converging 111 += S S′ f diverging f S′ M =− S f Physics 212 Lecture 26, Slide 18 18 It’s always the same: 111 += S S′ f S′ M =− S You just have to keep the signs straight: The sign conventions S: S’ : f: positive if object is “upstream” of lens positive of positive if image is “downstream” of lens positive positive if converging lens Physics 212 Lecture 26, Slide 19 19 Preflight 2 By definition, an image on the screen (downstream of the lens) MUST BE REAL 60 remember, though: 111 += s s′ f 80 111 =− ′fs s If s < f, then s’ < 0: virtual image 40 20 0 Physics 212 Lecture 26, Slide 20 20 Preflight 3 M =− s′ s s’ > 0 (real) M < 0 (inverted) s’ < 0 (virtual) M > 0 (upright) 100 80 60 40 20 0 Physics 212 Lecture 26, Slide 21 21 Preflight 5 BB Refraction cannot occur for the top half so the only thing seen will be that of the lower half of the object 50 Only the top part of the arrow will not be seen as that part of the lens is responsible for the arrow head. 30 The tape will block some of the light rays going onto the screen, but the other half of the lens will still refract the entire arrow onto the screen; just a little dimmer. 10 40 20 0 Physics 212 Lecture 26, Slide 22 22 image object Cover top half of lens Light from top of object object image Cover top half of lens Light from bottom of object What’s the Point? The rays from the bottom half still focus The image is there, but it will be dimmer !! Physics 212 Lecture 26, Slide 23 23 Preflight 5 Physics 212 Lecture 26, Slide 24 24 air water θi θi ? glass 1.3 ? 1.5 glass BB 1.5 Case I Case II In Case I light in air heads toward a piece of glass with incident angle θi In Case light air In Case II, light in water heads toward a piece of glass at the same angle. In Case light water heads same In which case is the light bent most as it enters the glass? I or II or Same or or (A) OK (B) (C) or NOT OK Physics 212 Lecture 26, Slide 25 25 air water θi θi glass θ2 1.3 θ2 1.5 glass BB 1.5 Case I Case II In Case I light in air heads toward a piece of glass with incident angle θi In Case light air In Case II, light in water heads toward a piece of glass at the same angle. In Case light water heads same In which case is the angle of refraction the largest ? I or II or Same or or (A) n1sin(θ1) = n2sin(θ2) (B) (C) sin(θ2)/sin(θ1) = n1/n2 Physics 212 Lecture 26, Slide 26 26 air water θi θi glass θ2 1.3 θ2 1.5 glass BB 1.5 Case I Case II In Case I light in air heads toward a piece of glass with incident angle θi In Case light air In Case II, light in water heads toward a piece of glass at the same angle. In Case light water heads same In which case is the light bent most as it enters the glass? I or II or Same or or (A) (B) (C) The angle of refraction in BIGGER for the water – glass interface: n1sin(θ1) = n2sin(θ2) sin(θ2)/sin(θ1) = n1/n2 Therefore the BEND ANGLE (θ1 – θ2) is BIGGER for air – glass interface Physics 212 Lecture 26, Slide 27 27 Preflight 7 BB The rays are bent more from air to glass than from water to glass Therefore, the focal length in air is less than the focal length in water We can see this also from Lensmaker’s Formula 50 40 30 20 10 0 nlens nair Physics 212 Lecture 26, Slide 28 28 Calculation A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted? • Conceptual Analysis • Lens Equation: 1/s + 1/s’ = 1/f • Magnification: M = -s’/s • Strategic Analysis • Consider nature of image (real or virtual?) to determine relation between object position and focal point • Use magnification to determine object position Physics 212 Lecture 26, Slide 29 29 A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted? BB Is the image real or virtual? (A) REAL (B) VIRTUAL A virtual image will be upright A real image would be inverted h h’ f h’ h f Physics 212 Lecture 26, Slide 30 30 A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted? BB How does the object distance compare to the focal length? (A) s< f Lens Lens equation equation (B) s= f s> f (C) 111 =− s′ f s s′ = fs s− f Virtual Image fl s’ < 0 Real object fl s > 0 Converging lens fl f > 0 h’ s’ h s f s− f <0 Physics 212 Lecture 26, Slide 31 31 A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted? s′ = fs s− f BB What is the magnification M in terms of s and f? (A) M = Lens Lens equation: equation: 111 =− s′ f s fs ′= s s− f s− f f (B) M = f −s f (C) M = −f s− f (D) M = f s− f Magnification Magnification equation: equation: s′ M =− s h’ M= −f s− f s’ h s f Physics 212 Lecture 26, Slide 32 32 A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted? M= −f s− f BB (A) 1.7mm (B) 6mm (C) 8mm (D) 40 mm (E) 60 mm M = +5 f = +10 mm M= −f s− f s= ( M −1) s= f M h’ 4 f = 8 mm 5 h s′ = − sM = −40 mm f Physics 212 Lecture 26, Slide 33 33 Follow Up Suppose we replace the converging lens with a diverging lens with focal length of 10mm. If we still want to get an image magnified by a factor of 5 that is NOT inverted, how does the object sdiv compare to the original object distance sconv? (A) sdiv < sconv (B) sdiv = sconv (C) sdiv > sconv (C) EQUATIONS M= −f s− f (D) sdiv doesn’t exist (D) PICTURES s= f M −1 M BB h h’ M = +5 f = −10 mm 4 s = f = −8 mm 5 s negative fl not real object s f s’ Draw the rays: s’ will always be smaller than s Magnification will always be less than 1 Physics 212 Lecture 26, Slide 34 34 Follow Up Suppose we replace the converging lens with a diverging lens with focal length of 10mm. M= What is the magnification if we place the object at s = 8mm? (A) M = 1 2 (B) M = 5 EQUATIONS EQUATIONS M= −f s− f s = 8 mm f = −10 mm (C) (C) M= 3 8 (D) M= 5 9 (E) −f s− f M= 4 5 PICTURES −10 10 5 M =− = = 8 − (−10) 18 9 h BB h’ f Physics 212 Lecture 26, Slide 35 35 ...
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