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2019-01-16 Prof. Herbert Gross Yi Zhong, Uwe Lippmann Friedrich Schiller University Jena Institute of Applied Physics Albert-Einstein-Str. 15 07745 Jena

Exercise Solutions 10:

Optical Design with Zemax for PhD - Basics

Exercise 10-1: Field lens flattener A field lens near to the image plane can help to correct a system for field curvature. a) Load the f = 50 mm lens AC 300-050-C of the Thorlabs lens catalog. let the incoming light

be a collimated beam at = 0.55 m with a diameter of 10 mm. Define the three field points for initial field angles of 0, 3 and 5 degree. b) Calculate the Seidel bar diagram and the spot diagram. What are the dominant aberrations of the system ? c) Introduce a concave-plane field lens with thickness 3 mm. What is the preferred material and what is the best focal length to correct the system for Petzval curvature? The distance between the achromate and the field lens should be optimized only on axis. Now optimize not the Petzval curvature but the medium image shell and discuss the result. d) What is the result, if the spot is optimized for all field points ? Solution: The achromate with f = 50 mm looks like the following figure:

b) The Seidel diagram and the spot diagramm looks as follows. The dominant aberrations are astigmatism and field curvature. Spherical aberration and coma are only a little bit smaller.

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c) The additional field lens is inserted and the distance between the achromate, the front surface radius and the refractive index are chosen to be variable. In the merit function, the spot only for the axis point is forced to be corrected. The first result gives a model index of n = 1.84. The glass, which is nearest to this value if N-LASF41 with n = 1.83913. After inserting this, the system is re-optimized and becomes free of Petzval curvature. If the field curvature operand FCUR is used, the best image plane is not corrected well. This surprising result comes from the fact, that the Petzval curvature only corrects the best image shell, if the astigmatism is corrected. Since this condition is not fulfilled here, the FCUR operands don't flattens the field.

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Therefore be using Seidel theory, a different approach should be used. If a system suffers from astigmatism and the best image plane in the middle between S and T should be flat, the two commands FCGT and FCGS are used and the sum of them is set to zero in the merit function. To allow for a change in the focal length by the field lens, EFFL should not be required in the merit function. What can be done to guaratee the image plane is to force PARY 0 for the marginal ray.

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d) If the spot is optimized for all field points, the mean curvature is reduced, the astigmatism is nearly symmetrically around the image plane. The field lens looks completly different, the astigmatism is reduced.

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Exercise 10-2: Splitting a lens

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If the requirements of a specification can not be fulfilled by a design approach, generally more degrees of freedom are necessary. To generate more lenses without disturbing the current correction, it can be feasible to split a lens into two. The selection of the lens can be determined by inspection of critical surface contributions. Tais principle should be demonstrated in this exercise. a) Establish a system with the wavelength 587 nm a collimated input beam with diameter 5 mm, a thickness of 3 mm and a focal length of 20 mm and a field angle of 3°. A single lens made of BK7 should serve as initial system. optimize the bending and calculate the spot sizes. The system is not diffraction limited and therefore should be improved. b) Now split the lens by keeping the overall thickness nearly constant. Reoptimize first the radii, then adjust the distances and thicknesses to appropriate values. What is the factor of improvement ? Solution: a) The system looks as follows.

b) Now the material is splitted by 2 additional surfaces with a rather small air thickness and two lenses of half the thickness as before. This looks like the following figures. The performance is not changed by this procedure and therefore the results until gives a good starting system.

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After optimization of the radii, we set the maximum center thickness of glass to be 5 mm. The final result is then as follows and improves the spot diameters by a factor of approximately 2.

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Exercise 10-3: Telephoto lens

A telephoto lens is a system with a focal length, that is considerably larger than the overall length of the system. The general approach is to combine a positive lens group with a negative group. a) Establish a telephoto lens system with paraxial lenses for an imcoming collimated ray bundle of diameter 10 mm. The numerical aperture after the first group should be NA = 0.1 and NA =

0.05 in the image space for = 587 nm. The total focal length should be 3 times larger than the free working distance. Determine the focal length of the system and of both lens groups. b) Look for achromates for both groups, rescale the negative achromate, if the desired focal length is not available. Re-adjust the distances of the system by requiering the image sided numerical aperture. Is the system diffraction limited in performance ? Explain the result. c) Now the wavelength is changed to dFC and a field points of 10° and 7° are introduced. The stop is located at the first surface. The focal length and the free working distance are kept constant. Now optimize the second lens group and allow also for larger diameter. Is the performance diffraction limited? Finally optimize also the radii of the first lens group. What is the obtained performance? What happens, if the second group is now reverted back? What is the remaining dominant aberration type? d) Calculate the polychromatic modulation transfer function of the system on axis. What is the contrast at the spatial frequency 20 LP/mm ? Calculate the MTF behavior versus defocussing for this frequency at the central wavelength 587 nm. Determine the maximum defocussing values, that delivers a contrast not smaller than 50%. Solution: a) The focal lengths of the groups are 50 mm and -33 mm, the system has f = 100 mm. This is obtained by a merit function and two ideal lenses with the conditions on the focal lengths and the numerical apertures.

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b) The negative lens is up-scaled from f = -30 mm to the desired f = -33 mm. The system is diffraction limited, if the negative lens is reverted. In this case, the image sided lower aperture must be oriented towards the negative lens.

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c) If the system is optimized first by the second group and than by the first group requiring a spot optimization and the focal length of f = 100 mm, the performance is not diffraction limted. After reverting the second group, this can be achieved. As can be seen, the remaining dominant aberration is distortion with 2.5%.

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d) The MTF curve looks as follows. From the text output we get the contrast 84% at 20 LP/mm. To get this accurate, the range of spatial frequencies is limited to 200 LP/mm.

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The defocussing MTF for the wavelength 587 nm and 20 LP/mm looks as follows for a defocussing range of 2 mm:

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From the text version of this plot we get the limiting values of -0.35 mm and +0.36 m for 50%

contrast, Therefore the corresponding depth of focus is z = 0.71 mm.

Exercise 10-4: Symmetric system

In a symmetric system, all odd aberrations are completly corrected. This is demonstrated in this exercise. a) Establish an incoming collimated beam with wavelength 500 nm and 10 mm diameter with the field angles 0°, 7° and 10°. It is focussed by two lenses with material SF6, thickness 5 mm and distance 10 mm. The image is located in a distance of 100 mm, the stop lies 5 mm before the first lens vertex. Optimize the system by changing only the radii of curvature. Inspect the quality by calculating the spots, the Seidel aberration contributions, the distortion and the Zernike coefficients for the outer field point. b) Now double the system perfectly symmetric. Exchange the field definition from angle to the equivalent finite object height. What is the correction now ? Change the position of the stop only by a slider option. What kind of changes are seen ? Prepare a universal plot to see the change in coma as a function of the stop location between 0 and 10 mm. c) Now re-optimize the system preserving the symmetry. Is the system now diffraction limited ? Solution: a)

It is seen, that the system is not diffraction limited and there is a larger contribution of coma (c8) and 1% distortion.

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b) The image heights are 0 / 10.5 mm and 15.5 mm.

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The spot is now on axis larger than in the field, the distortion and the coma is gone, the even aberrations are increased.

If the slider is established for a stops location between 0 and 10 mm between the inner lenses, the symmetry is perturbed and coma is growing. The vanishing coma for the middle position can also be demonstrated by a universal plot.

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c) The system is reoptimized by fixing the radii in the second lens group via pick-up. The result gives a quite better performance. Due to the automatic correction of coma, the optimization allows to correct the even aberration better.

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