2.710 Optics Spring 09 Solutions to Problem Set #3 Due Wednesday, March 4, 2009

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY.70 Optics Sprig 09 Solutios to Problem Set #3 Due Weesay, March 4, 009 Problem : Waa s worl a) The geometry or this problem is show i Figure. For part (a), the object (Waa) is locate isie the bowl a we are itereste to where the image is orme. We start by usig the matrirmulatio to aalyze the give system, i x i 0 s " ( ) R 0 ( ) R R s s( ) R 0 R # o : o () To the imagig coitio, we ote that all the rays, regarless o their eparture agle o, arrive at the image poit x i o 0), R s 0 ) () s R: We see that the image is orme at the ceter o the bowl a is virtual. Usig this result i equatio, ( ) i R o : (3) 0 The lateral magi catio is, x i a thereore, the image is erect. x i M L ; (4) b) For this part, the object (Olive) is locate outsie the bowl a we are itereste to where the image is orme. Agai, we solve this part usig the matrirmulatio, i x i 0 ( ) s 0 R 0 " s( ) ( R s 0 ss s 0 ( ) 0 R 0 o s # o : ) R s ( ) R (5)

2 From the imagig coitio o 0); we solve or s 0, s 0 ss 0 ( s R ) 0 ) (6) sr s 0 : s ( ) R The lateral magi catio is (or a o-axis ray, o 0), x i s 0 ( ) M L (7) R sr ( ) R : (s ( ) R) R s ( ) R From equatios 6 a 7, the ollowig cases arise:. I R > s( )! s 0 < 0; the image is virtual, erect a is locate at a istace js 0 j outsie the bowl.. I R > s( )! s 0 > 0; the image is real, iverte a is locate at a istace js 0 j isie the bowl. c) I we were to cosier the glass cotaier o thickess t as well as the ier, R ; a outer, R, raii, the matrirmulatio becomes, i x i 0 ( g) s 0 R t 0 g " t( 0 g) P R g s 0 t " P s s 0 t( g) R g 0 g R g t( g ) t( g) R g t s 0 g s P ( g ) R 0 o 0 s # 0 o s P t( g ) 0 R g s P t( g ) R g # o (8) ; where, P ( g ) ( g ) t ( g ) ( g ) : (9) R R g R R For the case o uiorm glass: R R R. The imagig coitio is, s 0 ss 0 ( s R ) s 0 t ( g) t st (g ) ss 0 t ( g ) ( g ) 0 R g g R g R R g

3 Figure : Waa s worl problem s 0 ss 0 ( s R ) g 0; (0) where, g s 0 t ( g ) t st (g ) ss 0 t ( g ) ( g ) : () R g g R g R R g Comparig equatios 0 a 6, we see that i orer to eglect the aquarium walls we require that g << ; 0 s t ( g ) s (g ) ss 0 ( g ) ( g ) << ) R g g R g R R g R g t << : () R ss 0 ( g ) ( g ) s 0 R ( g ) sr ( g ) Problem : Ball les magi er a) I class we erive the composite matrir a thick les surroue by air a is give by, M thick " h R ( ) R R R ( ) R R i # : (3) For a ball les magi er, the les thickess a raii are: R; a R R R: Usig these values o equatio 3 we get, 3

4 M ball ; (4) ( ) R R so the EFL is, EF L P R : (5) ( ) b) To calculate the BFL we cosier a o-axis object source at i ity (i.e. s o ; a o 0) which ocuses by the ball les at the optical axis (x i 0). Usig the matrirmulatio, i 0 0 ( ) R 0 BF L R x o EF L 0 BF L R BF L EF L ) (6) BF L EF L BF L 0 ) (7) R( ) : ( ) By symmetry, the F F L BF L. Sice < < 43, BF L > 0. The locatio o the pricipal plae respect to the back o the les is, x P P BF L EF L R: (8) Agai, by symmetry the locatio o the st pricipal plae is also at the ceter o the sphere as show i Figure. c) As show i Figure 3, a object is locate at a istace to the let o the back surace o the ball les, where The istace o the object to the st pricipal plae is, 4 3 R : (9) 4( ) s o (EF L F F L) R ( ) EF L: (0) 4

5 Figure : Locatio o pricipal plaes: ball les magi er. Figure 3: Imagig with a ball les magi er. Sice the object is locate betwee the pricipal plae a the ocal poit, as iicate by equatio 0, the image orme is virtual. To the image positio we use the imagig coitio, ( s o s i EF L ) ) s i EF L; so the image is orme to the let o the pricipal plae (i.e. to the let o the ceter o the ball les). ) The lateral magi catio is, 5

6 M L s i ; () s o a thereore, the image is virtual a erect. e) The aperture stop (AS) is give by the ite size o the ball les a thus is locate at the ceter o the les. The umerical aperture, NA; is, NA air si ta ; where we have use the paraxial approximatio a is the agle orme by a o-axis poit object a the ege o the AS, R NA 4 : (3) R ) By observig the object through the les, our eye s les is e ectively imagig the image geerate by the sphere. So, i our eye is locate at a istace L behi the les, the ew object istace is s o L R EF L. The crystallie les o the eye accomoates by chagig the eye s ocal legth i orer to still satisy a positive image istace o s i 5 ~ cm. From the imagig coitio, we that the resultig ocal legth ater accomoatio is, ( ) : (4) (L R)( ) R To be able to see a object at a relaxe state (i.e. without accomoatio o the eye), the image geerate by the spherical les shoul be at i ity, so the object shoul be place at a istace equal to EF L to the let o the st pricipal plae. Problem 3: Telephoto les esig a) For this problem, we are give two requiremets o a telephoto les system: i) or a object at i ity, 0 raias a h 5 0 (cm); ii) The locatio o the real image prouce by the system. The telephoto les system is show i Figure 4. By ispectio, we see that the rst les, L, ocuses the object at i ity to a poit o height x i, at a istace o to its right. This image poit ow acts as the object source or the seco les, L. The object a image istaces are, s i s o ; a (5) s i 3 : We use the equatio or the lateral magi catio to the separatio istace, 6

7 Figure 4: Telephoto les system. M L h (6) x i 0 si 3 s o ) 3 5 ) : We use the imagig coitio to the ocal legth o L; ) ( 7) s o s i o ) o 5 5 o 8 b) Reerrig to Figure 5, we see that, 7

8 Figure 5: Locatig pricipal plae. Figure 6: Locatig st pricipal plae. BC EF AO ) (8) OD CD 3 5 ) OF 5 3 : So the pricipal plae is locate to the let o L at a istace o : To the st pricipal plae, we reverse the propagatio irectio o the rays a cosier a poit at i ity comig rom the right as show i Figure 6. The parallel rays will have a virtual image at poit G prouce by L. This poit image the acts as a poit object or L a we use the imagig coitio to the locatio o the image (poit D), 8

9 ( s o s i ) 9) s i ) 9 8 s i 9: From Figure 6 we see that, BC EF AO 9 DO 5 ) DO 5 ) DF 9 OF 4: ( EF ) 30) So the st pricipal plae is locate to the let o L at a istace o 4: c) From part (b) we kow that, EF L DO 5: (3) Or eve rom the problem we ca kow the EF L without ay calculatios: because the parallel ray bule with agle passig through the system will ocus at the ocal plae with height h, so the e ective ocal legth is EF L h 5: ) This part ca be aswere by usig oly the pricipal plaes a applyig the imagig coitio, s o ) (3) s o s i 5 ) 0 s i ) 3 M L : 3 The image plae is locate to the right o L at a istace o : A alterative way to solve this problem is by usig the matrirmulatio, 9

10 out ut 0 o 0 i x i h i 4 o o (3 ) o (3 ) o o o 3 5 i x i (33) ; ut (3 ) o i (3 ) o xi: o (34) Because all the parallel rays with agle will ocus at the poit at the image plae with h 5, ut is iepeet o x i ; (3 ) 5; (35) o (3 ) 0: (36) o o From equatios 35 a 36 we have, 5 3 o ( ) 37) (3 ) ) ; which is cosistet with the result o equatio 6. Problem 4: Microscope esig a) The geometry o this problem is show i Figure 7. To let the huma s uaccommoate eye to ocus the image o the observer s retia, L3 shoul orm a image at i ity (i.e. or a give poit i the object the correspoig output is a o -axis parallel ray bule). The rst les, L; orms a itermeiate image at the S plae, ( s o 70 0 ) 38) s o 0:65mm. 0

11 Figure 7: Microscope esig problem. b) I we place a small object at s o i rot o L; the istrumet acts as a microscope. The magiyig power, MP, is the ratio o the image size orme o the huma s retia whe usig the istrumet a the image size whe viewe with the ake eye, MP M L M A3 (39) : X, where, M L is the lateral magi catio o the objective (L) a M A3 is the agular magi catio o the eyepiece (L3). c) Accorig to part (a), S is at a itermeiate image plae, so it is ot the aperture stop (we will see later that S is the el stop). Potetial aperture stops are the the rims o either L or L3. The aperture stop shoul be the smallest o the two. As show i Figure 8, the iput agle amitte by L is smaller tha the agle amitte by L3, ; < (40) 5 < ; 70 0 so the aperture stop is the rim o L. Sice there are o optical compoets presiig L, the aperture stop is also the etrace pupil. To the exit pupil the aperture stop through L3 a the locatio o the image usig the imagig coitio,

12 Figure 8: Fiig the aperture stop. ( 90 s i 0 ) 4) s i :35mm, si M L3 s o 0:7: So the exit pupil is locate at a istace o.35mm to the right o L3, a its raius is a ep a M L3 0:588mm. S is the el stop which limits the itermeiate image size. The maximum lateral size o a object that ca be image by this istrumet is restricte by the size o the etrace wiow. The raius o the etrace wiow is, a ew a M L (4) 0:65 6 mm, 70 so the maximum lateral size that ca be viewe is mm (iameter). ) I traitioal microscopes, the aperture stop is locate at the objective s rim; thereore, the subsequet optics create a image o the stop (i.e. the exit pupil) that is locate to the right o the eyepiece. The observer s eye ca be comortably locate such that the eye s pupil coicies with the exit pupil a the image ca be observe without vigettig. I the raius o L becomes a 0mm, the aperture stop woul be the rim o L3 a woul be collocate with the exit pupil. To avoi vigettig, the

13 eye pupil woul have to be ajacet to the eyepiece, which is o course ieasible because (a) the eye pupil is locate behi the corea, a (b) eve i the small istace betwee the corea a pupil coul be eglecte, it woul be really ucomortable or the viewer to place his or her eye i cotact with the eyepiece. Oe remey to this problem is to stop ow the objective (i.e. reuce its raius so that it becomes the aperture stop istea o the eyepiece); that is ot a goo solutio because, as we will see whe we o wave optics, this solutio reuces the overall umerical aperture o the system a, hece the resolutio o the microscope. A better remey is to replace the eyepiece with oe that has larger raius (assumig we ca a or oe). The agai, the objective becomes the aperture stop as esire. 3

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