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Physics 212 Lecture 26: Lenses

Physics 212 Lecture 26: Lenses. Music. Why? Last time we talked about the Brewster angle diagram with “circles & arrows ”. Who is the Artist? Ramblin’ Jack Elliott Arlo Guthrie Pete Seeger Phil Ochs U. Utah Phillips. Your Comments.

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Physics 212 Lecture 26: Lenses

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  1. Physics 212 Lecture 26: Lenses

  2. Music Why? Last time we talked about the Brewster angle diagram with “circles & arrows” • Who is the Artist? • Ramblin’ Jack Elliott • Arlo Guthrie • Pete Seeger • Phil Ochs • U. Utah Phillips

  3. Your Comments “would like to go over the diagrams much more slowly, I’ve never been very good at understanding them.” “i really dont understand what a virtual image is can you explain that in detail please” “It was good and going along at a nice pace, but the lens-maker’s formula must have been on a slide with a smaller index of refraction than the others because it went super fast.” We’ll show the key points with examples “This prelecture could not have better timing (I’m actually doing a project about lenses and mirrors in another class)! I am wondering how a flat lens, such as a fresnel lens, works though.” “This prelecture reminded me of my childhood frying ants in the sunlight. Ahh, memories.” “The demonstraion will be very valuable for this lecture. Choose wisely, or a kitten will die ...” • Hour Exam 3: 1 week from tomorrow (Wed. Apr. 25) • covers L19-26 inclusive (LC circuits to lenses) • Sign up for conflicts, etc. no later than Mon. Apr. 23 at 10:00 p.m.

  4. Refraction Snell’s Law n1sin(q1) = n2sin(q2) q1 n1 n2 q2 That’s all of the physics – everything else is just geometry!

  5. q2 q2 The angle of refraction in BIGGER for the water – glass interface: sin(q2)/sin(q1) = n1/n2 n1sin(q1) = n2sin(q2) Therefore the BEND ANGLE (q1 – q2) is BIGGER for air – glass interface air water qi qi 1.3 glass glass 1.5 1.5 CaseI CaseII In Case I light in air heads toward a piece of glass with incident angleqi 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? I or II or Same (A) (B) (C)

  6. 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 “Under water, the index of refraction is greater and closer to the index of refraction of the lens; this makes the angle of refraction smaller by Snell's law so the focal length will increase.” “The index of refraction will increase so the focal length will decrease.” “The lens makers formula shows focal length only depends on N and R”

  7. Checkpoint 2 We can see this also from Lensmaker’s Formula 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

  8. 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…

  9. Waves from object are focused by lens

  10. Two Different Types of Lenses

  11. f f 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 fbehind the lens:

  12. 1) Draw ray parallel to axis refracted ray goes through focus refracted ray is symmetric 2) Draw ray through center image You now know the position of the same point on the image Recipe for finding image: object f

  13. Example image is: real inverted smaller S > 2f object f image f > 0 S’ > 0 S > 0

  14. Example S = f image is: at infinity… object f f > 0 S > 0

  15. Example 0 < S < f image is: virtual upright bigger image f object f > 0 S > 0 S’ < 0

  16. Checkpoint 1a A B C D “The object is real, therefore its image is real. Real things make real things.” “The screen is to the downstream of the lens, so s prime must be positive and the image must be real. Since both s and s prime are positive, the magnification will be inverted.” “The image is virtual because it is behind the lens. It is upright because it is always upright. ” “converging lens always give the virtual image and if it can be projected on the screen, it is inverted”

  17. Checkpoint 1a A B C D Image on screen MUST BE REAL MUST BE INVERTED

  18. f 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 fin front of the lens:

  19. Example image is: virtual upright smaller object f image f<0 S>0 S’<0

  20. Executive Summary - Lenses: real inverted smaller S > 2f converging f f real inverted bigger 2f > S > f virtual upright bigger f > S > 0 virtual upright smaller S > 0 diverging

  21. It’s always the same: 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

  22. 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 screenB. Only the upper half of the object (i. e. the arrow head) will show on the screenC. The whole object will show on the screen “The light from the bottom half only makes it through the lens.“ “When the top half is covered, the image would cover the lower end of the screen. The image is inverted but real size.” “The entire image will still be projected onto the screen but there will not be as many rays so the intensity will be lower.”

  23. image object Cover top half of lens Light from top of object image object 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 !! 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 screenB. Only the upper half of the object (i. e. the arrow head) will show on the screenC. The whole object will show on the screen

  24. 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

  25. 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? h f h’ h’ h f • Is the image real or virtual? • (A) REAL (B) VIRTUAL A virtual image will be upright A real image would be inverted

  26. 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? Lens equation h’ h f s’ s • How does the object distance compare to the focal length? • (A) (B) (C) Virtual Image s’ < 0 Real object s > 0 Converging lens f > 0

  27. 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? Lens equation: Magnification equation: h’ h f s’ s • What is the magnification M in terms of s and f? • (A) (B) (C) (D)

  28. 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? h’ h f (A) 1.7mm (B) 6mm (C) 8mm (D) 40 mm (E) 60 mm

  29. 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(D) sdiv doesn’t exist EQUATIONS PICTURES Draw the rays: s’ will always be smaller than s Magnification will always be less than 1 s negative  not real object

  30. Follow Up • (A) (B) (C)(D) (E) PICTURES h h h ’ ’ f 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? EQUATIONS

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