Lect26 - Lecture 26: Lenses Physics 212 35 30 25 20 15 10 5...

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Unformatted text preview: Lecture 26: Lenses Physics 212 35 30 25 20 15 10 5 0 Confused Avg = 3.3 Confident Physics 212 Lecture 26, Slide 1 Physics Refraction Snell’s Law n1sin(θ1) = n2sin(θ2) θ1 n1 n2 θ2 That’s all of the physics – everything else is just geometry! Physics 212 Lecture 26, Slide 2 Physics 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 3 Physics Waves from object are focused by lens Physics 212 Lecture 26, Slide 4 Physics Ray Tracing can be used to determine Image Physics 212 Lecture 26, Slide 5 Physics Two Different Types of Lenses Physics 212 Lecture 26, Slide 6 Physics Converging Lens: Consider the case where the shape of the lens is such that light rays parallel to the axis of the lens are all “focused” to a common spot a distance f behind the lens: f f Physics 212 Lecture 26, Slide 7 Physics 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 8 Physics 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 9 Physics 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 10 Physics 2f > S > f image is: real inverted bigger 111 += S S′ f S′ M =− S object f image f S S’ Physics 212 Lecture 26, Slide 11 Physics S=f image is: at infinity… 111 += S S′ f S′ M =− S object f f S Physics 212 Lecture 26, Slide 12 Physics 0<S<f image is: virtual upright bigger image 111 += S S′ f S′ M =− S f object S S’<0 f Physics 212 Lecture 26, Slide 13 Physics Diverging Lens: Consider the case where the shape of the lens is such that light rays parallel to the axis of the lens all diverge but appear to come from a common spot a distance f in front of the lens: f Physics 212 Lecture 26, Slide 14 Physics 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 15 Physics Executive Summary - Lenses: S > 2f 2f 2f > S > f f >S>0 real inverted smaller real real inverted bigger virtual upright bigger converging f 111 += S S′ f S′ M =− S S >0 virtual upright smaller diverging f Physics 212 Lecture 26, Slide 16 Physics 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 if image is “downstream” of lens positive if converging lens Physics 212 Lecture 26, Slide 17 Physics Preflight 2 By definition, an image on the screen (downstream of the lens) MUST BE REAL remember, though: 100 80 60 111 += s s′ f 111 =− ′fs s If s < f, then s’ < 0: virtual image 40 20 0 Physics 212 Lecture 26, Slide 18 Physics 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 19 Physics Preflight 5 BB light from the object can not pass through the top of the lens, which is the light that forms the inverted arrow head (MRA) Some of the upper light rays won't pass through the lense and the inverted image will result in the upper half of the object being visible. (GEC) Light rays will still pass through from all points, the image will just be dimmer since the rays that pass through the top are not present anymore. (GPD) 50 40 30 20 10 0 Physics 212 Lecture 26, Slide 20 Physics Cover top half of lens Light from top of object object image Cover top half of lens Light from bottom of object object image 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 21 Physics Preflight 5 Physics 212 Lecture 26, Slide 22 Physics air water θi θi 1.3 ? Case I glass 1.5 ? Case II glass 1.5 BB In Case I light in air heads toward a piece of glass with incident angle θi In Case II, light in water heads toward a piece of glass at the same angle. In which case is the light bent most as it enters the glass? (A) I or II or Same (B) (C) Physics 212 Lecture 26, Slide 23 Physics air water θi glass θi θ2 θ2 Case II 1.3 glass 1.5 BB 1.5 Case I In Case I light in air heads toward a piece of glass with incident angle θi In Case II, light in water heads toward a piece of glass at the same angle. In which case is the light bent most as it enters the glass? (A) I or II or Same (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 24 Physics 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 25 Physics 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 26 Physics 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 A real image would be inverted (B) VIRTUAL A virtual image will be upright h f h’ h’ h f Physics 212 Lecture 26, Slide 27 Physics 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? How does the object distance compare to the focal length? (A) BB s< f (B) s= f (C) s> f Lens equation 111 =− s′ f s s′ = Virtual Image fl s’ < 0 Real object fl s > 0 Converging lens fl f > 0 fs s− f h’ h s’ s f s− f <0 Physics 212 Lecture 26, Slide 28 Physics 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 equation: s− f f (B) M = f −s f (C) M = −f s− f (D) M = f s− f Magnification equation: 111 =− s′ f s s′ M =− s fs ′= s s− f h’ M= −f s− f h s’ s f Physics 212 Lecture 26, Slide 29 Physics 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 ( M −1) s= f M h’ h s= 4 f = 8 mm 5 s′ = − sM = −40 mm f Physics 212 Lecture 26, Slide 30 Physics 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 EQUATIONS M= −f s− f s= f M −1 M (B) sdiv = sconv (C) sdiv > sconv (D) sdiv doesn’t exist PICTURES h s f BB h’ s’ M = +5 f = −10 mm 4 s = f = −8 mm 5 s negative fl not real object negative Draw the rays: s’ will always be smaller than s Magnification will always be less than 1 Physics 212 Lecture 26, Slide 31 Physics Follow Up Suppose we replace the converging lens with a diverging lens with focal length of 10mm. What is the magnification if we place the object at s = 8mm? M= −f s− f (A) M = 1 2 (B) M = 5 (C) M= 3 8 (D) M= 5 9 (E) M= 4 5 EQUATIONS EQUATIONS M= −f s− f PICTURES h h’ f s = 8 mm f = −10 mm −10 10 5 = = M =− 8 − ( −10) 18 9 BB Physics 212 Lecture 26, Slide 32 Physics ...
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