Lectures_Lect28

Lectures_Lect28 - Physics 212 Optical Devices Lecture 28...

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Unformatted text preview: Physics 212 Optical Devices Lecture 28 Physics 212 Lecture 28, Slide 1 Executive Summary Mirrors & Lenses: S > 2f 2f > S > f f > S > 0 real inverted smaller real inverted bigger virtual upright bigger f > 0 c o nc ave ( c o nve r g ing ) f c o nve r g ing f 1 1 1 + = S S f f < 0 S M =- S S > 0 virtual upright smaller ( d ive r g ing ) c o nve x f d ive r g ing f Physics 212 Lecture 28, Slide 2 It's always the same: 1 1 1 + = S S f S M =- S You just have to keep the signs straight: s' is positive for a real image f is positive when it can produce a real image Lens sign conventions S: positive if object is "upstream" of lens S' : positive if image is "downstream" of lens f: positive if converging lens Mirrors sign conventions S: positive if object is "upstream" of mirror S' : positive if image is "upstream" of mirror f: positive if converging mirror (concave) Physics 212 Lecture 28, Slide 3 System of Lenses Trace rays through lenses, beginning with most upstream lens Image from first lens Becomes object for second lens Physics 212 Lecture 28, Slide 4 System of Lenses Virtual Objects are Possible !! Object Distance is Negative !! Image from first lens Becomes object for second lens Physics 212 Lecture 28, Slide 5 A parallel laser beam of width w1 is incident on a two lens system as shown below. Checkpoint 3 Each lens is converging. The second lens has a larger focal length than the first (f 2 > f1). What does the beam look like when it emerges from the second lens? A B A. The beam is converging B. The beam is diverging C C. The beam is parallel to the axis with a width < w1 D E D. The beam is parallel to the axis with a width = w1 E. The beam is parallel to the axis with a width > w1 1. 2. 3. Parallel rays are transmitted and pass through focal point (f1) Those rays also pass through focal point of second lens (f2) and therefore are transmitted parallel to the axis. f2 > f1 implies that the width > w1 Physics 212 Lecture 28, Slide 6 Normal Eye Physics 212 Lecture 28, Slide 7 Checkpoint 2 A person with normal vision (near point 28 cm) is standing in front of a plane mirror. What is the closest distance to the mirror the person can stand and still see herself in focus? A. 14 cm B. 28 cm C. 56 cm r i d The image is formed an equal distance BEHIND the mirror Therefore, if you stand a distance = of your near point, the distance to the image will be the near point distance. d Physics 212 Lecture 28, Slide 8 Farsighted Converging Lens creates virtual image at person's near point Physics 212 Lecture 28, Slide 9 Nearsighted Fix with diverging lens that creates virtual image at far point. Physics 212 Lecture 28, Slide 10 Checkpoint 1 Two people who wear glasses are camping. One of them is nearsighted and the other is farsighted. Which person's glasses will be useful in starting a fire with the sun's rays? A. The farsighted person's glasses B. The nearsighted person's glasses Farsighted = Converging Lens Only Converging Lens can produce a REAL IMAGE !! Physics 212 Lecture 28, Slide 11 T wo c o nve rg in g le n s e s a re s e t u p a s s h o wn . T h e fo c a l le ng th o f e a c h le ns is 4 7 c m . T h e o b je c t is a lig h t b u lb lo c a te d 7 0 c m in fro nt o f th e firs t le ns . Multiple Lenses Exercises s 1 =7 0 c m f=4 7 c m What is the nature of the image from the first lens alone? (A) R EAL UP R IG HT (B) R EAL INVER T ED (C ) VIR T UAL UP R IG HT (D) VIR T UAL INVER T ED EQUATIONS 1 1 1 = - s f s s > f M =- s s s ' > 0 M < 0 s = fs s- f PICTURES Dra w R a y s a s a b o ve re a l im a g e inve rte d im a g e P h y s ic s 2 1 2 Le c tu re 2 8 , S lid e 1 2 T wo c o nve rg ing le n s e s a re s e t u p a s s h o wn . T h e fo c a l le n g th o f e a c h le ns is 4 7 c m . T h e o b je c t is a lig h t b u lb lo c a te d 7 0 c m in fro n t o f th e firs t le n s . Le ns s e p a ra tio n = 2 m Multiple Lenses Exercises What is the object distance s2 for lens 2? (A) s 1 = 7 0 c m f = 4 7 c m s 1 ' = 1 .4 3 m s 2 = 1 .4 3 m (B) s 2 = +1 .4 3 m (C ) s 2 = 0 .5 7 m (D) s 2 = +0 .5 7 m (E) s 2 = +2 .7 m THE OBJECT FOR THE SECOND LENS IS THE IMAGE OF THE FIRST LENS s 2 = 0 .5 7 OR s 2 = +0 .5 7 Im a g e o f firs t le ns is a R EAL o b je c t fo r th e s e c o nd le ns P h y s ic s 2 1 2 Le c tu re 2 8 , S lid e 1 3 T wo c o nve rg in g le n s e s a re s e t u p a s s h o wn . T h e fo c a l le ng th o f e a c h le ns is 4 7 c m . T h e o b je c t is a lig h t b u lb lo c a te d 7 0 c m in fro nt o f th e firs t le ns . Le ns s e p a ra tio n = 2 m Multiple Lenses Exercises s 1 =7 0 c m f=4 7 c m (A) R EAL UP R IG HT (B) R EAL INVER T ED (C ) VIR T UAL UP R IG HT s 1 ' = 1 .4 3 m s 2 = 0 .5 7 m (D) VIR T UAL INVER T ED What is the nature of the FINAL image in terms of the ORIGINAL object? EQUATIONS s2 = fs2 s2 - f PICTURES Dra w R a y s a s a b o ve s 2 > f M2 = - s2 s2 s 2 ' > 0 M2 < 0 re a l im a g e M = M1 M2 > 0 up rig h t im a g e R ES ULT S s 2 ' = 2 .6 9 m M = 9 .6 P h y s ic s 2 1 2 Le c tu re 2 8 , S lid e 1 4 S up p o s e we inc re a s e th e in itia l o b je c t d is ta n c e to 7 4 c m Le ns s e p a ra tio n = 2 m Multiple Lenses Exercises s 1 =7 4 c m f=4 7 c m R ES ULT S s 1 ' : 1 .4 3 m 1 .2 9 m s 2 ' : 2 .6 9 m 1 .3 8 m How does the L, the distance to the FINAL image, change? (A) L inc re a s e s (B) L d e c re a s e s (C ) L re m a ins th e s a m e Step through images, one at a time WORDS Inc re a s ing s 1 will d e c re a s e s 1 ' (m o ving c lo s e r to fo c a l p o int wo uld inc re a s e th e im a g e d is ta nc e ) De c re a s ing s 1 ' will inc re a s e s 2 Inc re a s ing s 2 will d e c re a s e s 2 ' EQUATIONS 1 1 1 = - s1 f s1 1 inc re a s e s s1 s2 = 2m - s1 1 1 1 = - s2 f s2 s 2 inc re a s e s 1 inc re a s e s s2 P h y s ic s 2 1 2 Le c tu re 2 8 , S lid e 1 5 S up p o s e we no w d e c re a s e th e in itia l o b je c t d is ta nc e to 5 8 c m . Ap p lying th e le ns e q ua tio n, we fin d s 1' = 2.48m Le ns s e p a ra tio n = 2 m Multiple Lenses Exercises What is the object distance s2 for lens 2? (A) s 1 = 5 8 c m f = 4 7 c m s 1 ' = 2 .4 8 m s 2 = 0 .4 8 m (B) s 2 = +0 .4 8 m (C ) s 2 = 2 .4 8 m (D) s 2 = +2 .4 8 m (E) s 2 = +2 .5 8 m THE OBJECT FOR THE SECOND LENS IS THE IMAGE OF THE FIRST LENS s 2 = 0 .4 8 OR s 2 = +0 .4 8 Im a g e o f firs t le ns is a VIR T UAL o b je c t fo r th e s e c o nd le ns P h y s ic s 2 1 2 Le c tu re 2 8 , S lid e 1 6 S up p o s e we no w d e c re a s e th e in itia l o b je c t d is ta nc e to 5 8 c m . Ap p lying th e le n s e q u a tio n , we fin d s 1' = 2.48m Lens separation = 2 m s 1 =5 8 c m f=4 7 c m s 1 ' = 2 .4 8 m s2 = 0.48 m (D) VIR T UAL INVER T ED Multiple Lenses Exercises What is the nature of the FINAL image in terms of the ORIGINAL object? (A) R EAL UP R IG HT (B) R EAL INVER T ED (C ) VIR T UAL UP R IG HT EQUATIONS s2 = fs2 s2 - f PICTURES Dra w R a y s a s a b o ve s 2 < 0 M2 = - s2 s2 s 2 ' > 0 M2 > 0 re a l im a g e M = M1 M2 < 0 inve rte d im a g e RESULTS s2' = 0.24 m M = 2.1 Physics 212 Lecture 28, Slide 17 0 < S < f image is: virtual upright bigger image Virtual object 1 1 1 + = S S f S M =- S f object We could just as well take the rays to be going right to left. In that case the "image" shown above becomes a virtual object and the "object" shown above becomes a real image. S > 0 f > 0 Physics 212 Lecture 26, Slide 18 S' < 0 ...
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