VS203B Lecture Notes Spring, 2011 Topic: Diffraction

Transcription

1 Diffracion Diffracion escribes he enency for ligh o ben aroun corners. Huygens principle All poins on a wavefron can be consiere as poin sources for he proucion of seconary waveles, an a a laer ime he new wavefron posiion is he envelope (or surface of angency) o hese seconary waveles. Fresnel s Aiion For ligh in he same bunle (ie. waves from he same poin source) all waves are muually coheren an all inerfere wih each oher. Fresnel s iffracion occurs in he near fiel when he screen is hel close o he aperure. Fraunhofer iffracion occurs in he far fiel or near he focal plane creae by converging he ligh wih a posiive lens. These iffracion effecs are relevan for imaging sysems like he eye, elescopes, cameras ec. Each wave of ligh across he aperure ravels a ifferen pah lengh o reach he screen. You nee o consier how all waves combine o generae he iffracion paern. Along he ais, he inensiy coul be high, or coul be zero, epening on he isance. Away from he ais, we again have o a all he ampliues across he wavefron o ge an esimae of wheher he iffracion paern has a peak or a valley. When you go very far off he ais, here is an equal amoun of in an ou of phase waves so he inensiy rops o zero. The paerns ha you observe across he screen are calle Fresnel Diffracion aerns. These are observe when he screen is hel near o he aperure. 44 Ausin Roora

2 In hse figures he aperure is on he righ an he screen is on he lef. You can see how he inensiy isribuion on he screen will vary as he screen is move owar an aperure. (images from Davi C. Banks, John T Foley, Kiril N. Viimce an Ming-Hoe Kiu, "Insrucional Sofware for Visualizing Opical henomena," 1997 IEEE Visualizaion Conference.) Now we move very far away from he aperure. The effec is ha he isance for each ray across he aperure o opical ais a he screen is very nearly he same. This calle he far fiel. In he far fiel, here is always a peak a he cener because all he ligh is in phase Now if we look off he ais, we inrouce a phase ifference across he rays. When he phase ifference a he eges is equal o 1 wavelengh. One half of he waves are ou of phase wih he oher half an we ge he firs low inensiy poin in he iffracion paern. The iffracion paern is calle he Fraunhofer Diffracion aern.noe ha his is opposie o he case of he ouble sli, because now we have o inclue he phase of all he rays. a In his iffracion paern, for small angles, he posiions of he ima occur a: asin() = m, where m =± 1,±.an a is he wih of he sli. For a screen a a isance s, m sin ima occur when = m = ± 1, ±... noe ha he equaion is he same as for he ouble sli, ecep i is use o locae he ima, an here is no m=zero 45 Ausin Roora

3 Eample (eample.1a from Keaing p 498) A sli of wih 0.5 mm is illuae wih 633 nm ligh. A wha angular locaion is he firs imum observe in he single sli iffracion paern on a isan screen? 0.5 sin = sin = = using a small angle approimaion = 1.7 raians m Where is he locaion of he imum if he screen is m away from he sli? = 1.7 raians = an = 1.7 cm Inensiy profile of he iffracion paern The cenral peak is he mos inense of all. This is because all of he ligh is in phase a ha poin. As you move furher off-ais, he presence of ou of phase zones reuce he ampliue an also here is an overall fall-off in inensiy ue o he fac ha you are geing furher from he slis. An infiniely small sli will prouce a uniform illuaion wih he epece fall-off ue o isance from he sli. The ampliue an inensiy profiles are given by: The zeros occur when: π sin π sin sin sin E = I = E = π sin π sin sin() his form,, is calle he sinc funcion π sin = m π, where m = ± 1, ±... mπ m sin = = π for small angles, m sin = The iffracion paern is sprea perpenicular o he irecion of he sli (ie if he sli is verical, he iffracion paern is horizonal). 46 Ausin Roora

4 Fraunhofer iffracion paerns for oher aperures. Recangular aperure: A square aperure can be consiere as wo sli aperures in he verical an horizonal irecions. The sinc funcions for each sli aperure are simply muliplie ogeher. a b Circular aperure: ( α ) sin( β ) sin πa sin πbsinφ E =, whereα =, β = α β This is he mos relevan an imporan aperure shape since i is he shape of mos lenses an aperures. The iffracion paern for a circular aperure is efine as: π sin π sin J1 J1 E = I = E = π sin π sin Where J 1 is he Bessel funcion. The shape ha his equaion efines is calle he Airy isc. The raius o he firs imum in his funcion is: 1. = also, since 1. = compare wih = for firs imum hrough a sli aperure I have been saing ha he Fraunhofer iffracion paern can be obaine by moving he screen far from he aperure. A Fraunhofer iffracion paern can also be generae a he focal poin of a lens. So Fraunhofer iffracion paerns are also observe in he image plane. In orer o generae a paern, you se he lens agains he aperure an simply use he focal lengh of he lens as he aperure o screen isance (ie replace s wih f ) Everyhing else is he same. Diffracion is anoher facor ha limis he image qualiy in opical sysems. Recall ha he size of he iffracion paern is inversely proporional o he iameer of he aperure. Therefore, in a lens, he bigger he aperure, he smaller he Airy isc. 47 Ausin Roora

5 Two oin Resoluion Consier wo muually incoheren poin sources. When he sources are muually incoheren, you simply a he inensiies. Each source prouces an Airy isc in he image plane. When he sources ge close ogeher, he wo Airy isc paerns begin o overlap unil hey can no longer be resolve as isinc objecs. The smalles value of he angle for which images of wo poins can be eece as wo isinc objecs is he limi of resoluion. We have o aop an arbirary sanar o come up wih a number for he resoluion limi so we efine he limi as when he peak from one Airy isc sis aop he firs imum of he oher isribuion. This is calle he Rayleigh resoluion limi. The equaion ha efines his coniion is simply he isance from he cener o he firs imum which I showe earlier: 1. = also, since 1. = This shows he images of wo poin sources ha are consiere o be resolve a he Rayleigh resoluion limi peak of n Airy isk 0% conras firs imum of 1 s Airy isk Recall ha aberraions en o blur ieal images. Bu even if here were no aberraions, rays woul no be image o a poin source bu o a iffracion paern. erfec opical sysems ha have no aberraions are sill limie by iffracion. A sysem ha is free from aberraions is calle iffracion-limie (ie is image qualiy is limie solely by iffracion) 48 Ausin Roora

6 imum angle of resoluion (ues of arc 500 nm ligh) = The graph plos he imum angle separaion ha wo poin objecs have angle subene a he noal poin o be separae by o be resolve in a iffracion-limie sysem. The smaller he imum angle of resoluion, he beer he vision. wavelengh of he ligh pupil iameer pupil iameer (mm) Eample. A iffracion-limie eye wih a 6 mm pupil is looking a an approaching car whose healighs have a wavelengh of 550 nm an are separae by 1.5 m. Wha is he imum angular resoluion of he eye? A wha isance can you no longer resolve wo isinc healighs? = = = raians = eg = ues of arc = 1.5 m = = 13, 41.8 m = km = 8.4 miles 6 3 Two poin resoluion Image qualiy in he eye: For 630 nm ligh 6 mm pupil, angular resoluion is abou 0.44 ues of arc. 3 mm pupil, angular resoluion is abou ues of arc. How oes his relae o vision? The 0/0 leers on a Snellen Char suben 5 ues of arc. The cener lines on he wo arms of he E are separae by ues of arc on he reina. The gap beween wo arms on he E is one ue of arc. The efiniion of wha resoluion is require o resolve an E is no clear cu, bu you shoul epec o be able o resolve 0/0 leers if you are well-correce. 0/ leers are half he size an resolving hem may be ifficul for pupil sizes smaller han 3 mm. Recall ha aberraions reuce image qualiy when he pupil ges larger, so even hough iffracion heory ells us ha he image shoul ge sharper, i oes no. If he paraial opics are free from aberraions, why on we jus use a iny pupil? Diffracion akes over for small pupils an makes he image qualiy worse. We are beaen a boh ens of our pupil range. Given he aberraions ha are presen in he eye, he bes poin sprea funcion is obaine for a pupil size of abou 3 mm. 49 Ausin Roora