Chapter 3 –Geometrical Optics

Chapter 3 – Geometrical OpticsGabriel PopescuUniversityy of Illinois at Urbana‐Champaignp gBeckman InstituteQuantitative Light Imaging Laboratoryhttp://light.ece.uiuc.eduPrinciples of Optical ImagingElectrical and Computer Engineering, UIUC

ECE 460 – Optical ImagingObjectives Introduction to ggeometrical opticspand Fourier opticsp–precedes MicroscopyChapter 3: Imaging2

ECE 460 – Optical Imaging3.1 Geometrical Optics If the objectsjencountered byy lightg are largeg comparedptowavelength, the equations of propagation can be greatlysimplified (λ 0) i.e.i ththe wave‐phenomenah( tt i interference,(scattering,i t fetc)t ) areneglected In homogeneousgmedia,, lightg travels in straightg lines raysy G.O. deals with ray propagation trough optical media (eg.Imaging systems)ObChapter 3: Imagingblack boximagingsystemImOptical Axis3

ECE 460 – Optical Imaging3.1 Geometrical Optics G.O. ppredicts imageg location troughg complicatedpsystems;y;accuracy is fairly good Nowadays there are software programs that can run “raypropagation”ti ” troughth arbitrarybitmaterialst i l So, what are the laws of G.O.?Chapter 3: Imaging4

ECE 460 – Optical Imaging3.2 FermatFermat’ss principlea)) n constantb)) n n((r ) function ofpositionBdsBLAAcv nL 1Time:t AB v c nL straight lineChapter 3: Imagingcn( r )ds 1dt n( s )dsvB c1 t AB c n( s )dsAv(r ) (3.1)5

ECE 460 – Optical Imaging3.2 FermatFermat’ss principle Fermat’s Principlep is reminiscent of the followingg pproblem thatyou might have seen in highschool: Someone is drowning in the Ocean at point (x,y) The lifeguard at point (u,w)pv1 and in the water at speedpv2. What is hiscan travel across the beach at speedbest possible path?(x,y)v2v1(u,w)Chapter 3: Imaging6

ECE 460 – Optical Imaging3.2 FermatFermat’ss principle Definition:(3.2)S ct n( s )ds optical path length How can we predict ray bending (eg. mirage)? Fermat’s Principle: Light connects any two points by a path of minimum time(the least time principle)B B n( S )dS 0 A (3.3)A If n constant in space, AB line, of courseChapter 3: Imaging7

ECE 460 – Optical Imaging3.3 SnellSnell’ss Law Consider an interface between 2 media:yθ1n1Bθ2XXxn2A The rays are “bent” such that:n1 sin 1 n2 sin 2(3.4) SnellSnell’ss law (3.4)(3 4) can be easily derived from Fermat’sFermat s principle,principleby minimizing:S n1 AO n2 OB total path‐length Take it as an exerciseChapter 3: Imagingdemo available8

ECE 460 – Optical Imaging3.3 SnellSnell’ss Lawn2 n1 Consequences of Snell’s Law:a)) Iff n2 n1 Ѳ2 Ѳ1 (ray( gets closerlto normal)l)b)If n2 n1 quite interesting! 1 n1 2 sin sin 1 (3.5) n2 Ray gets away from normalyn 1n2 n1n21k2θ1k1 2θ2xk n1 i 1 1 2 NO TRANSMISSION So, if sin2 n2 Chapter 3: Imagingdemo available2 k 9

ECE 460 – Optical Imaging3.3 SnellSnell’ss Law The angleg of incidence c for whichn1 sin c n2(3.6)is called critical angle This is total internal reflectionn1c)) law of reflectionn2 n1 Snell’s law is: n1 sin 1 n2 sin 2 1 2 (reflection law) Energy conservation: Pt Pr PiChapter 3: Imagingyrn2tθ2θ1xi(3.7)demo available10

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.O Efficient wayy of ppropagatingp gg raysy throughg opticalpsystemsyyθ1Optical Systemy1y2θ2OxOA Optical Axis Any given ray is completely determined at a certain plane bythe angle with OA,OA Ѳ1 , and height w.r.tw r t OA,OA y1 Let’s propagate (y1 , Ѳ1), assume small angles GaussianapproximationChapter 3: Imaging11

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.Oa)) Translationy1y2θOAd 2 1y2 y1 d tan 1Small angles:y2 y1 d 1 2 0 y1 1 1Chapter 3: Imaging(3.8)12

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.Oa)) Translation We can re‐write in compact form: y2 1 d y1 01 1 2 Chapter 3: Imaging(3.9)13

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.Ob)Refraction‐sphericaldielectric interface)pyn1α1 θ1θ2y1 xα2n2OACR Snell’s law: n1 1 n2 2 Geometry: 1 1 2 2 y1 y2 R Rn1n21n1 1 y1 n2 2 y2 RRn2Chapter 3: Imaging(3 10)(3.10)14

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.Ob)Refraction‐spherical)pdieletric interface So: y2 y1 0 1n1y1 n1 2 ( 1) 1n2R n2 Chapter 3: Imaging 1 y2 n1 n2 2 n R 20 y1 n1 1 n2 (3.11)15

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.Ob)Refraction‐spherical)pdieletric interface Important: To avoid confusion between Ѳ and – Ѳ angles,use “sign convention”1. angle convention ‐OA Counter clock‐wise positive22. distance conventionLeft negativeRight positiveAOAB‐Chapter 3: Imaging 16

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.Ob)Refraction‐spherical)pdieletric interface Example:RR 1We found n1 n2 nR 2‐0 1n1 and n2 n1 nRn2 20 n1 n2 Same /‐/ convention applies to spherical mirrors. Withoutsign convention, it’s easy to get the wrong numbers.Chapter 3: Imaging17

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.Oc)) Dielectric interface – pparticular case of R n1θ1θ2n2OA 1 n nlim2R 1 nR 2Chapter 3: Imaging0 1n1 0 n2 0 n1 n2 (3.12)18

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.O The nice thingg is that cascadingg multiplep opticalpcomponentspreduces to multiplying matrices (linear systems) Example:n2n1n3n4ABT1R1T2R2T3R3 yB yA T4 R3 T3 R2 T2 R1 T1 B A T4(3.13) Note the reverse order multiplication (chronological order)Chapter 3: Imaging19

ECE 460 – Optical Imaging3.4 Propagation Matrices in G.O Note the reverse order multiplicationp(chronological(gorder)) 1 d T Translation matrix 01 R refraction matrix Chapter 3: Imaging 1 n n 2 1 nR 20 n1 n2 20

ECE 460 – Optical Imaging3.5 The thick Lenstn1 1n2 1BAR1R2n Typical glass: n 1.5 Basic optical component: typically ‐ 2 spherical surfaces yB yA RB Tt RA . B A 0 0 1 1 yA 1 t n 1 1 n 1 . A n 0 1 R 2 nR1 n MChapter 3: Imaging21

ECE 460 – Optical Imaging3.5 The thick Lens After some algebra:gtt 1C 1 R1n M tt (C1 C2 C1C2 n ) 1 C2 nR 2 In general C (3 14)(3.14)n2 n1 convergence of spherical surfaceR R1 0, R2 0 C1 0 & C2 0 convergent Note [C] m‐1 dioptriesChapter 3: Imaging22

ECE 460 – Optical Imaging3.5 The thick Lens1t Definition: C1 C2 C1C2fn((3.15)) f is the focal distance of lens Eq (3.15) is the “lens makers equation”Chapter 3: Imagingdemo available23

ECE 460 – Optical Imaging3.6 Cardinal pointsImage Formation Ray TracingyOO’Fy’F’z O object; O’ image ; O‐O’ conjugate pointsF’ focal point image (image of objects from ‐ )F focal point objectTransverse magnification:y'M (3.16)yChapter 3: Imaging24

ECE 460 – Optical Imaging3.6 Cardinal points Definition: pprincipalp planespare the conjugatej g pplanes for whichM 1F’ff’FHChapter 3: ImagingH’H, H’ principal planesf,f f’ focal distances! f, f’ measured from H25

ECE 460 – Optical Imaging3.7 Thin lens Particular use: t 0 Transfer matrix for thin lens:tt 1C1 R1n10 limt 0 tt (C1 C2 ) 1 (C1 C2 C1C2 n ) 1 C2 nR 2 11 1 C1 C2 (n 1)( ) SincefR1 R2 (Note R1 0,0 R2 0) 1 Mthin lens 1 f Chapter 3: Imaging0 1 (3 17)(3.17)26

ECE 460 – Optical Imaging3.7 Thin lens Remember other matrices: 1 d Translation: T 01 1 Refraction‐spherical surface: R n2 n1 nR 2 1 0 Spherical mirror: M 2 1 (f R/2) R Chapter 3: Imaging(3.18)0 n1 n2 (3.19)(3.20)27

ECE 460 – Optical Imaging3.7 Spherical MirrorsConvergentCfRf 2DivergentChapter 3: Imaging28

ECE 460 – Optical Imaging3.8 Ray Tracing – thin lensesLBθyAf’F’A’fFy’θ’xx’ B’ convergent lens; f 0 divergent lens; f 0Chapter 3: Imaging29

ECE 460 – Optical Imaging3.8 Ray Tracing – thin lenses y ' y ' Tx ' M f Tx 1 0 1 x ' 1 x y 1 0 1 f 1 0 1 x' 1 f 1 f Chapter 3: Imagingxx ' x x ' f y x 1 f (3.21)30

ECE 460 – Optical Imaging3.8 Ray Tracing – thin lenses y ' A B y ' C D ; yy’ can be found as: y ' Ay B ((3.22)) Condition for conjugate planes:yOAy’(Figure page III‐13) For conjugate planes, y’ should be independent of angle Ѳ B 0 i.e.i e stigmatism condition (points are imaged into points) We neglect geometric/chromatic aberrationsChapter 3: Imaging31

QuizyOAy’Explain how FermatFermat’ss principle works here.

S l tiSolutionn Sn 1S n( s )dsBecause all of the rays leaving a given point converge again in the image, we knowfrom Fermat’s principle that their paths must all take the same amount of time.Another way to say this is that all the paths have the same optical path length.length Thisis because those paths that travel further in the air, have a “shorter” distance totravel in the more time‐expensive glass. If the optical path lengths were not thesame, the image would not be in focus because rays from a single point would bemapped to several points.

ECE 460 – Optical Imaging3.8 Ray Tracing – thin lensesxx ' 0 So,So B 0 x x ' f1 1 1 x' x f(3.23) Eq above is the conjugate points equation (thin lens) Eq 3.22 becomes: y’ yAx' M A 1 Transverse magnificationfChapter 3: Imaging(3.24)34

ECE 460 – Optical Imaging3.8 Ray Tracing – thin lenses Use Eq 3.23:3 23:x' 1 1 M 1 1 x ' f x' x x' 0 (inverted image)xx'M xy' ? 1 2f(3.25) If object and image space have different refractive indices,3.23 has the more general form:n' n 1(3.26) x' x fChapter 3: Imaging35

ECE 460 – Optical Imaging3.8 Ray Tracing – thin lenses x x' n' ff focal distance in air x ' x nf Let’s differentiate (3.26) for air, n’ n 1:dx 'dx 22x'x2 x' dx ' dx x dx ' M 2 dx(3.27) Eq 3.27 says that if the object gets closer to lens, the imagemoves away!Chapter 3: Imaging36

ECE 460 – Optical Imaging3.8 Ray Tracing – thin lensesOAyFF’xx’Δ’yOAΔChapter 3: Imagingy’FF’y’demo available37

ECE 460 – Optical Imaging3.8 Ray Tracing – thin lenses What happensppwhen x f ?1 1 11 1 1 0 ?x' x fx' f xyFy’x’F This image is formed by continuations of rays Sometimes called “virtual images” These images cannot be recorded directly(need re‐imaging)Chapter 3: Imagingdemo available38

ECE 460 – Optical Imaging3.8 Ray Tracing – thin lenses Other useful formulas in G.O ((figuregabove: Δ,, Δ’)) ' f 2 (Newton’s formula)y' ' f (“lens formula”) yf Chapter 3: Imaging(3.28)demo available39

ECE 460 – Optical Imaging3.9 System of lenses The image through one lens becomes object for the next ��F1y1’ y2 Apply lens equation repeteatly. Or, use matricesL2L1BA’AT1 AA, A’ conjugateconj gate throthroughgh L1 B,B’ conjugate through L2Chapter 3: ImagingB’T2demo available40

ECE 460 – Optical Imaging3.9 System of lenses Use T T2.T1 ; T matrix from 3.21 x1 ' 10 f1 T1 x 11 ff 1((3.29)) Note: 1 x1 1 x1 1 1 f1 x1 x1 ' x1 1 1 1 magnificationx1 ' M 1Chapter 3: Imaging41

ECE 460 – Optical Imaging3.9 System of lensesx11 1 f1 M 1also: 1 (3.30)x1 ' M1f1y fdet(T1) 1 T T2 .T1 Transverse Magnification0 M 10 M2 1 111 f MfM 22 11 0 M 1M 2 M111 f fM MM 21 21 2 Chapter 3: Imaging 'f'y'y'M y ' y 1 y' M(3.31)42

ECE 460 – Optical Imaging3.9 System of lenses 2‐lens system is equivalent to:1 M11 ff2f1M 2M M1 M 2 Microscopes achieve M 10‐100 easily Can be reduced to 2‐lens system Question: cascading many lenses such that M 106, would web ablebebl tot see atoms?t? Well, G.O can’t answer that. So,backSo back to wave opticsChapter 3: Imaging43

3.1 Geometrical Optics ECE 460 –Optical Imaging If the objects encountered by light are large compared t