Exercise 12 Geometrical and Technical Optics WS 213/214 Slide projector and Köhler illumination In this exercise a simplified slide projector (or LCD projector) will be designed and simulated with ray tracing. The imaging optics as well as the illumination optics will be taken into account. The given data are: Instead of a lens system a simple planoconvex lens made of BK7 with focal length f Obj =12 mm is used as imaging objective lens (diameter 6 mm, center thickne 2 mm). A slide with a size of 36 mm x 24 mm shall be imaged to a size of 9 mm x 6 mm. The light source is approximated as a rectangular surface of size mm x 1 mm with the properties of a Lambert radiator. The maximum radiation angle of the Lambert radiator in the simulation can be 4 o (in practice it can be of course up to 9 o ). The condenser lens, which is designed here as planoconvex lens with aspheric curved surface, shall image the light source with a scaling factor three (exactly -3) to the object-sided principal plane of the objective lens (so called Köhler illumination). The slide (here aumed to be infinitely thin) shall be mm behind the vertex (back side) of the condenser lens. a) Make a small sketch to imagine yourself the setup. Include the distances which are calculated using the imaging equation. Start your calculation with the imaging optics since the paraxial data of the condenser lens are then fixed. Calculate by using the lens equation the radius of curvature of the objective lens (spherical planoconvex lens) and of the condenser lens (aspheric planoconvex lens). The design wavelength is nm and the material of the condenser lens shall be SF1. b) The diameter of the condenser lens has to be at least as large as the diagonal of the slide (43.3 mm) to fully illuminate the slide. Due to the convergent ray path behind the condenser lens and the distance of mm between lens and slide, the diameter of the condenser lens is fixed to 6 mm. Could this diameter be realized for a spherical planoconvex lens taking into account the focal length calculated in a)? Then, calculate the data of the aspheric back surface of the condenser lens via optimization. Use the conical constant and the parameters a_4 and a_6 as optimziation parameters (other higher order coefficients should not be used). For the optimization, the axial point of the light source shall be imaged to the object sided principal plane of the objective lens. Pay attention that the step size of the parameters a_4 and a_6 have to be set to quite small values for the optimization. This is neceary since the parameters a_4 and a_6 are itself quite small since they contribute to the surface equation multiplied with r 4 or r 6, respectively! The center thickne of the condenser lens (material SF1) is fixed to 3 mm in order to have identical results. Pay attention that you have to start the optimization with a lens with small diameter (e.g. 2 mm). Just, in the second or poibly third iteration the diameter can be increased to the final value of 6 mm. c) As object a cro grating with absorbing lines, period 1 mm and width. mm is taken (total aperture like for the slide). Simulate the complete system with the program RAYTRACE. How well is the imaging quality? Which orientation is useful for the objective lens? What happens with the irradiance if the condenser is removed. What happens if the size of the light source is varied?
Solution: To a) Imaging optics: The scaling factor of the imaging optics is Obj =-9/36=-2. The focal length is f Obj =12 mm. Using the imaging equation results in: 1 1 1 1 bobj f ' Obj 1 Obj and gobj f ' Obj 1 bobj gobj f ' Obj Obj In our case the data of the imaging objective lens are: b Obj =312 mm, g Obj = -124.8 mm The lens equation for a planoconvex lens delivers its radius of curvature R Obj using the refractive index n BK7 (= nm)=1.214: R n 1 f ' 62.7 Obj mm BK 7 Obj Condenser optics: Since the light source is imaged to the object-sided principal plane of the objective lens and the slide is d= mm behind the vertex of the condenser lens (vertex=position of image-sided principal plane of the condenser lens), the image distance of the condenser lens b Kond is: bkond gobj d 129.8 mm Using the imaging equation and Kond =-3: bkond f ' Kond 32.4 mm 1 Kond 1 g Kond f ' 1 Kond 43.27 mm Kond The (paraxial) radius of curvature R Kond of the aspheric back surface of the condenser lens is (n SF1 (= nm)=1.7432): R n 1 f ' 24.12 Kond mm SF1 Kond To b) The object-sided principal plane of the planoconvex condenser lens lies due to the center thickne of d m =3 mm in the distance d m /n SF1 =17.21 mm right to the plane front surface. So, the object point/light source lies 43.27 mm -17.21 mm = 26.6 mm in front of the plane front surface of the condenser lens. The image point lies 129.8 mm behind the vertex of the aspheric back surface since for a planoconvex lens vertex plane and principal plane coincide. Optimization in two or three steps (first with a lens diameter of only 2 mm, then 4 mm and finally 6 mm) delivers: K=-.9488683629 a_4=-1.688348e-6 a_6=9.68663e-11 or K=-.829939328 a_4=-4.383291e-7 a_6=1.4889664e-1 The result depends on the starting conditions and intermediate steps (diameter of lens aperture). So, a unique result can only be expected if exactly the same starting conditions are taken. 37
To c) If the vertex of the condenser lens is put to z=- mm, the object (slide) is at z= mm. The vertex and the object-sided principal plane of the objective lens are then both at z=124.8 mm. The image-sided principal plane lies at z=131.6 mm (d m /n BK7 =13. mm in front of the plane back surface of the lens, i.e. 2 mm-13. mm = 6.8 mm right to the vertex). Finally, the image plane is at z=312 mm + 131.6 mm=321.6 mm. It results that the imaging quality of the cro grating is quite well in the center, but deteriorates towards the rim. At the rim the single squares can just be resolved and for a real slide with much more pixels the rim would be blurred. If the planoconvex imaging lens is rotated in such a way that the plane surface points towards the object, the imaging quality even deteriorates (although the difference between both orientations is not very large). Without the condenser lens the slide cannot be illuminated fully and homogeneously and the imaging quality is worse due to large angles. If the light source has a larger distance the slide can be fully illuminated and le imaging errors, but a lot of energy is lost and the image is distorted. If the size of the light source decreases, the image becomes sharper. Simulation results: In the following some simulation results are shown. To do so, a radiation angle of the Lambert radiator of up to 36 degree is taken (i.e. rays with higher angles are not simulated, but the radiance is that of a Lambert radiator) and 8196769 rays are taken for the simulation. It appears that only the system with Köhler illumination gives good results. There, the illuminated area of the imaging lens is quite small and similar for all field points of the slide. A smaller light source leads of course to better results since then the used area of the imaging lens and its aberrations are even smaller. However, for a very small light source (point source) diffraction effects would have to be taken into account due to the small numerical aperture. If the light is just collimated with the help of the condenser lens rays coming from the rim of the slide also hit the imaging lens at the rim. This generates larger aberrations (point aberrations as well as distortion). Without condenser lens the illumination of the image is quite bad since rays hitting the slide at the rim do not hit the imaging lens. Therefore, the image is cut, inhomogeneous, distorted and blurred, especially at the rim. For a larger distance of the light source the image field would be complete, but the intensity decreases since only part of the emitted light is used. However, to obtain the full image field, the distance would be so large, that the intensity is really small. Besides, also with very large distance the image quality would be at best as good as with collimated light and a small light source. So, still outer parts of the imaging lens would be used which leads to higher aberrations as for the Köhler illumination. Summary: The Köhler illumination with magnified imaging of the light source into the entrance pupil of the imaging lens gives the best result concerning intensity, homogenity, sharpne and distortion. In a real slide projector the imaging lens is of course a lens system consisting of several lenses. Then, several aberrations are corrected. But, also there, the Köhler illumination is used since it reduces the requirements on the aberration correction (better correction means in practice more lenses and higher costs). 38
Köhler illumination as described in the exercise, light source size: mm x 1 mm - 1 3 3 3 2 2 1 - -1 - -2-2 -3-3 - 1 RAYTRACE Copyright 28 Mittelwert 11 RMS 21 P-V 1e+2 Max 1e+2 Min 1e+2 8 6 4 2-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 3 2 2 1 - -1 - -2-2 -3-3 23.1.29 14:19:17-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 RAYTRACE Copyright 28 Left: rim region upper side left; right: central region 39
Köhler illumination as described in exercise, but smaller light source size: 1 mm x 1 mm - 1 3 3 3 2 2 1 - -1 - -2-2 -3-3 - 1 RAYTRACE Copyright 28 Mittelwert 11 RMS 26 P-V 1,3e+2 Max 1,3e+2 Min 1,2e+2 1e+2 8 6 4 2-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 3 2 2 1 - -1 - -2-2 -3-3 23.1.29 14:24:39-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 RAYTRACE Copyright 28-6 -4-2 2 4 6 8 Left: rim region upper side left; right: central region 4
Köhler illumination as described in exercise, larger light source size: 1 mm x 1 mm - 1 3 3 3 2 2 1 - -1 - -2-2 -3-3 - 1 RAYTRACE Copyright 28 Mittelwert 11 RMS 11 P-V 61 Max 61 Min 6 4 3 2 1-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 3 2 2 1 - -1 - -2-2 -3-3 23.1.29 14:29:2-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 RAYTRACE Copyright 28 Left: rim region upper side left; right: central region 41
Light collimated by condenser lens, light source size: mm x 1 mm - 1 3 3 3 2 2 1 - -1 - -2-2 -3-3 - 1 RAYTRACE Copyright 28 Mittelwert 14 RMS 21 P-V 1,4e+2 Max 1,4e+2 Min 1,2e+2 1e+2 8 6 4 2-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 3 2 2 1 - -1 - -2-2 -3-3 23.1.29 14:37:6-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 RAYTRACE Copyright 28 Left: rim region upper side left; right: central region 42
Without condenser lens, distance Lambert radiator to slide: 48 mm - 1 3 3 2 2 1 - -1 - -2-2 -3-3 - 1 RAYTRACE Copyright 28 Mittelwert 1,6 RMS 3,7 P-V 9 Max 9 Min 4 3 2 1-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 3 3 2 2 1 - -1 - -2-2 -3-3 23.1.29 14:4:7-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 RAYTRACE Copyright 28 Zoom of central region 43
Without condenser lens, distance Lambert radiator to slide: 2 mm - 1 3 3 2 2 1 - -1 - -2-2 -3-3 - 1 RAYTRACE Copyright 28 Mittelwert 2,2 RMS 7,3 P-V 1,8e+2 Max 1,8e+2 Min 1,e+2 1e+2-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 3 3 2 2 1 - -1 - -2-2 -3-3 23.1.29 14:44:28-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 RAYTRACE Copyright 28 Zoom of central region 44
Without condenser lens, distance Lambert radiator to slide: 1 mm - 1 3 3 2 2 1 - -1 - -2-2 -3-3 - 1 RAYTRACE Copyright 28 Mittelwert,93 RMS 1,9 P-V 21 Max 21 Min 2 1-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 3 3 2 2 1 - -1 - -2-2 -3-3 23.1.29 14:48:39-4e+2-3e+2-2e+2-1e+2 1e+22e+23e+24e+2 RAYTRACE Copyright 28 Zoom of central region 4