Light Ray Tracing in GRIN Lens

Light Ray Tracing in GRIN Lens 1:Distribution of reflective index in GRIN Lens Imaging condition: n(r)ds,const ………………….(1) ,s Following equations can be got according to Fig.1(Light ray tracing in Grin lens) ds , dr s rBy Light reflection Law: 0 r 2a ,0 z Fig.1 ………………….(2) n(0)cos,,n(r)cos,,n(r)00 is the reflective index at axial of Grin lens. n(r) is the reflective n(0)0 index at amplitude. drsin,,Use following equation can be calculated: ds n(r)drds, ………………….(3) 222n(r),n(0)cos,0 If light ray tracing is sinusoid: r,rsin(Az) ………………….(4) 0 222ds,dr，dzUse ………………….(5) dr,rAcos(Az) ………………….(6) 0dz when z=0, 1 dr ………………….(7) ,tg,,rA00dz Using (3),(4),(5) and (7),we can get following equation: 2222 ………………….(8) n(r),n(0)(1,Acos,r)0 (1) can be calculated by (8) and (3): r220,(1,cos)rA0(),4(0)nrdsndr ………………….(9) ,,222sin,,cos,rA0s00 Using transform integral variable,(9) can be calculated: sin,,rActg,0 ,12,(1sin),20,nn4(0)2(0)222,,,n(r)ds(1sinsin)d ,,,0,,cos,AcosA,0s00 ………………….(10) If near axial light ray is considered,,(10) can be cos,,1,sin,,000 simplified: 2n(0),n(r)ds,,n(0)l ………………….(11) ,As l is the periodic length.Formula (11) show that the rays of giving off from a location at axial will gather together at another location of axial. The distance between two location at axial is equal to the periodic llength . For near axial light ray,the distribution of reflective index can be expressed: 11222n(r),n(0)(1,Ar),n(0)(1,Ar) ………………….(12) 2 2: the light ray tracing in GRIN Lens: 2 If ,(12) is : n(r),n,n(0),n0 12 ………………….(13) n,n(1,Ar)02 Now the near axial meridian light ray is considered only,according to light ray equation: ,ddr(n()),,n ………………….(14) dsds x dx2ds,dz1，(),dz ….(15) P dzs (14) can be calculated: o z Fig.2 2dx,n,ndxn,,, ………………….(15) 2,x,zdzdz 12n(x),n(1,Ax)and: ………………….(16) 02 1,n2,0Ax,,1Using ,and if ,we can get following simplified light ray ,z2 equation: 2dx,,Ax ………………….(17) 2dz Light ray tracing is: ………………….(18) x,Bcos(Az)，Csin(Az) dxP,,,BAsin(Az)，CAcos(Az) ………………….(19) dz Integred constant B and C can be obtained by following beginning x,x,Bcondition:when z=0, ,P,P,CA,we have following 00 equation: P0 ………………….(20) x,xcos(Az)，sin(Az)0A 3 ………………….(21) P,,xAsin(Az)，Pcos(Az)00 1,,xxcos(Az)sin(Az),,,,,,0 ………………….(22) ,,,,,A,,,,,,PP,,0,,,,Asin(Az)cos(Az),,, Two special cases: (1):When light ray input Grin lens at axial with ,in Grin lens the , ,angle between light ray and z axial is : so , x,B,0P,P,CA,tg,00 tg, ………………….(23) x,sin(Az) A ………………….(24) P,tg,cos(Az) ml,. (23) show the light ray is sinusoid.whenz,,m,m,1,2,3......,x,02A , , l Fig.3 (2):When light ray input Grin lens,the light ray is paralleled to the axial of Grin lens: x,BSo, , P,P,CA,000 x,xcos(Az) ………………….(25) 0 P,,xAsin(Az) ………………….(26) 0 ml,z,,m,m,1,2,3......,P,0(25) show light ray is cosine curve. when. 2A 4 P 0.75P 0.5P 0.25P Fig.4 3: Imaging of Grin lens: (1):image: M a ’ Mb x 0 ,xxb , ’ xa O O LzL0 Fig.5 For a light ray in Grin-lens: ………………….(27) x,xcos(Az)a0 ………………….(28) P,,xAsin(Az)a0 a Light ray output Grin lens,the slope is ,P,nP,,nxAsin(Az) ………………….(29) a0a00 a light ray output Grin lens is a beeline,the beeline equation is: ,,,x,x,Pz ………………….(30) aa For b light ray in Grin lens: 5 xtg,0 ………………….(31) x,sin(Az),,sin(Az)bAnLA00 x0 ………………….(32) P,tg,cos(Az),,cos(Az)bnL00 b Light ray output Grin lens,the slope is x,0 ………………….(33) P,nP,,cos(Az)0bbL0 b light ray output Grin lens is a beeline,the beeline equation is: ,,,x,x,Pz ………………….(34) bb ,,From (27)~(34), the intersection point () of a light ray (z,x),(L,x) output Grin lens and b light ray output Grin lens can be calculated: nLAcos(Az)，sin(Az)100() ………………….(35) L, nAnLAsin(Az),cos(Az)000 x0x,, ………………….(36) nLAsin(Az),cos(Az)00 LIn the above expressions,is object distance, is image distance; is Lx00 object height, is image height. Transverse magnification is defined as x ratio of image height and object by: x,1m,, ………………….(37) xnLAsin(Az),cos(Az)000 L,0L,0m,0m,0If ,real image, ,virtual image;erect image,,inverted m,1m,1image,,enlarged image ,,lessen image. f (2): effective focal length f: x s Reference to Fig.6: x,xcos(Az) 0, x0 , x z 6 F z Fig.6 tg, P,,xAsin(Az),,tg,,,0n0 So,the distance between focal point F and rear surface of Grin Lens s can be calculated by: xcos(Az)x10 ………………….(38) s,,,ctg(Az)tg,nxAsin(Az)nA000 xx100f,,, ………………….(39) tg,nxAsin(Az)nAsin(Az)000 Discussion: (1): if input light ray is paralleled to principal axial of Grin lens: we have formulae (38) and (39). For 0.25pitch Grin ,2*0.25Az,,lens,, ,2 s=0, 1f, ………………….(40) nA0 It show that the focal point is at rear surface of grin lens. L,,(2): If the image is infinite,according to (35),,So: nLAsin(Az),cos(Az),000 cos(Az)1L,,ctg(Az),s ………………….(40) 0nAsin(Az)nA00 Object should be at front focal point. 7