Lectures_Lect26

Lectures_Lect26 - Lecture 26: Lenses Physics 212 Physics...

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Unformatted text preview: Lecture 26: Lenses Physics 212 Physics 212 Lecture 26, Slide 1 Refraction Snell's Law n1sin(1) = n2sin(2) 1 n1 2 n2 That's all of the physics everything else is just geometry! Physics 212 Lecture 26, Slide 2 air water i glass i 2 2 Case II 1.3 glass 1.5 1.5 Case I In Case I light in air heads toward a piece of glass with incident angle i 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) (B) (C) I or II or Same sin(2)/sin(1) = n1/n2 Physics 212 Lecture 26, Slide 3 n1sin(1) = n2sin(2) The angle of refraction is BIGGER for the water glass interface: Therefore the BEND ANGLE (1 2) is BIGGER for air glass interface Checkpoint 2 What happens to the focal length of a converging lens when it is placed under water? A. increases B. decreases C. stays the same 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 nlens nair Physics 212 Lecture 26, Slide 4 Light rays from sun bounce off object and go in all directions Some hits your eyes Object Location 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 Two Different Types of Lenses Physics 212 Lecture 26, Slide 7 Converging Lens: Consider the case where the shape of the lens is such that light rays parallel to the axis of the mirror are all "focused" to a common spot a distance f behind the lens: f f Physics 212 Lecture 26, Slide 8 Recipe for finding image: 1) Draw ray parallel to axis 2) Draw ray through center refracted ray goes through focus refracted ray is symmetric object f image You now know the position of the same point on the image Physics 212 Lecture 26, Slide 9 S > 2f image is: real inverted smaller Example 1 1 1 + = S S f S M =- S object f image f > 0 S > 0 S' > 0 Physics 212 Lecture 26, Slide 10 S = f image is: at infinity... Example 1 1 1 + = S S f S M =- S object f f>0 S > 0 Physics 212 Lecture 26, Slide 11 0 < S < f image is: virtual upright bigger image Example 1 1 1 + = S S f S M =- S f object S > 0 S' < 0 f > 0 Physics 212 Lecture 26, Slide 12 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 13 image is: virtual upright smaller Example 1 1 1 + = S S f S M =- S object f image f<0 S>0 S'<0 Physics 212 Lecture 26, Slide 14 Executive Summary Lenses: S > 2f 2f > S > f f > S > 0 real inverted smaller real inverted bigger virtual upright bigger converging f 1 1 1 + = S S f S M =- S S > 0 virtual upright smaller diverging f Physics 212 Lecture 26, Slide 15 It's always the same: 1 1 1 + = S S f S M =- S You just have to keep the signs straight: The sign conventions S: positive if object is "upstream" of lens S' : positive if image is "downstream" of lens f: positive if converging lens Physics 212 Lecture 26, Slide 16 Checkpoint 1a A converging lens is used to project the image of an arrow onto a screen as shown above. The image is A. real and upright B. real and inverted C. virtual and upright D. virtual and inverted Image on screen MUST BE REAL s > 0 1 1 1 + = s s f M =- s s Physics 212 Lecture 26, Slide 17 Checkpoint 1b A converging lens is used to project the image of an arrow onto a screen as shown above. A piece of black tape is now placed over the upper half of the lens. Which of the following is true? A. Only the lower half of the object (i.e. the arrow tail) will show on the screen B. Only the upper half of the object (i. e. the arrow head) will show on the screen C. The whole object will show on the screen Physics 212 Lecture 26, Slide 18 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 !! A converging lens is used to project the image of an arrow onto a screen as shown above. A piece of black tape is now placed over the upper half of the lens. Which of the following is true? A. Only the lower half of the object (i.e. the arrow tail) will show on the screen B. Only the upper half of the object (i. e. the arrow head) will show on the screen Physics 212 Lecture 26, Slide 19 C. The whole object will show on the screen 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? Calculation 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 20 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? (A) REAL Is th e im a g e re a l o r virtua l? (B) VIRTUAL A virtual image will be upright A real image would be inverted h f h' h' h f Physics 212 Lecture 26, Slide 21 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) s< f Lens equation s = f (B) (C) 1 1 1 = - s f s s > f s = Virtual Image s' < 0 Real object s > 0 Converging lens f > 0 fs s- f h' h s' s f s- f < 0 Physics 212 Lecture 26, Slide 22 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 Wh a t is th e m a g nific a tio n M in te rm s o f s a nd f? (A) Lens equation: s- f f -s (B) (C) M= M= f f Magnification equation: -f (D) M= s- f M= f s- f 1 1 1 = - f s s s M =- s fs s = s- f h' M= -f s- f h s' s f Physics 212 Lecture 26, Slide 23 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? -f M= s- f (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) f M h' h s= 4 f = 8 mm 5 s = - sM = - 40 mm f Physics 212 Lecture 26, Slide 24 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 EQ UAT IO NS M= -f s- f s= f M -1 M Follow Up (C) sdiv > sconv (D) sdiv doesn't exist P IC T UR ES h s f h' s' M = +5 f = - 10 mm 4 s = f = -8 mm 5 s ne g a tive no t re a l o b je c t Dra w th e ra ys : s ' will a lwa ys b e s m a lle r th a n s Ma g nific a tio n will a lwa y s b e le s s th a n 1 Physics 212 Lecture 26, Slide 25 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? Follow Up M= -f s- f 1 M= (A) (B) (C) M = 5 2 M= (D) (E) 3 8 M= 5 9 M= 4 5 EQ UAT IO NS M= -f s- f M =- PICTURES h s = 8 mm f = - 10 mm -10 10 5 = = 8 - (-10) 18 9 h' f Physics 212 Lecture 26, Slide 26 ...
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