2012-10-20Prof.Herbert GrossFriedrich Schiller University JenaInstitute of Applied PhysicsAlbert-Einstein-Str 1507745 JenaSolution of ExercisesLecture Optical design with Zemax-Part 11.1Singlet imaging setup..1.2System layout with ideal lenses..1.3Symmetrical 4f-system...........1.461.1 Singlet imaging setupSetup a system with a single lens to image an object.The wavelength should be =632.8 nm,thelens is made of the glass N-BK7 with radii R1 30 mm,R2 =-50 mm and a thickness t =5 mm.Theobject distance is 50 mm.The numerical aperture in the object space is NA 0.1.a)Establish the system and determine the approximate image distance by three methods:1)distance,where the marginal ray intersects the optical axis or 2)distance,where the spot of the axialobject point has a minimal size or 3)determine the location in the layout,where the ray caustic has aminimal size.b)What is the focal length of the system Determine the magnification and the numerical aperture inthe image space.What is the reason for the differences in the direction cosines for real and paraxialraytracingc)What is the relative enlargement of the spot size in the image for an object with diameter 10 mm atthe edge in comparison to an object point on axisSolution:a)System data:Lens Data EditorCommentOBJ StandardInfinity50.0000000Standard30.00000005.0000000N-BKStandard-50,000000121,294000Standard1/82The image distance can be found by determining the distance,where is diameter of the marginal rayheight is zero(in Lens Data Manager)or by calculating the marginal ray trace and deviding the heightat the lens by the cosine of the ray behing the lens.Real Ray Trace Data:X-coordinateY-coordinateZ-coordinateX-cosineY-cosineOBJ0.0000000000E+0000.0000000000E+0000.0000000000E+0000.00000000000.1000000000120.0000000000E+0005,0685330379E+0004.3126697263E-0010.00000000000,00716345130,0000000000E+0005.0993910862E+000-2,60717938902-0010,0000000000-0,04204160820,0000000000E+000-1.5484000638E-0020.0000000000E+0000.0000000000-0.0420416082This gives approximately 121.3 mm.Other possibilities are minimizing the spot diameter(this gives124.8mm).5:RMS vs.Focus5If the menue 'longitudinal aberration'is used for an image distance zero in the data,the exactintersection lengths of the paraxial and the marginal ray can also be taken from the text outputnumerically.1,238E+0021.236建+0020.97001214E+0021.212+0021.209+002b)The data are found in the menue Prescription data:Focal length f 37.19 mm.The numericalapertur is NA'=0.0380.3Temperature (C)1.00000E+000Adjuat Index Data To Environment off37.19172 (in air at syatenEffectiveFocal Length37.19172(in image space)Back Focal Length35,08436129.8081Image Space F/3.458545Paraxial orking F/13.1479111,89298Image Space NA0.03800140.1Stop Radius5,0721270Entranc世Pupil10.7535810.14425Exit Pup1-124.8081Field TypePrimary Wavelength0.6328=Millimeters0The magnification of m=-2.64 is only seen,if a finite object height is inserted.The image sided numerical aperture can also be seen in the raytrace data table.Due to sphericalaberration,the real ray has a significantly(10%)larger numerical aperture value in the image spaceX-ccordinateI-cocxdinateY-cosine0.10000000000,00000000002Y-coordinate24.5945413703-001c)If the object size is defined and the spot diagram is calculated,we get the following figure:According to the numbers,the spot is grown by a factor of 2.2...5 depending on the criterion.Sincealso the shape is changed,an approximate factor of 3 seems to be reasonalbe