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2019-01-09 Prof. Herbert Gross Uwe Lippmann, Yi Zhong Friedrich Schiller University Jena Institute of Applied Physics Albert-Einstein-Str 15 07745 Jena

Exercise Solutions 9:

Optical Design with Zemax for PhD - Basics

Exercise 9-1: Fourier filtering and Imaging Set up a telecentric 4f-system with a magnification of m = 4. The system should be arranged by two appropriate achromates from conventional vendor catalogs with the focal lengths f1 = 25 mm and f2 = 100 mm. The wavelength of the application is the e-line, the object size is 3 mm and the numerical aperture should be NA = 0.1. a) Set up the system with telecentric behavior in the object and the image space. In particular, adjust the system and a rectangular stop with appropriate size in the intermediate focal plane. The adjustment should be performed on axis only. b) Check the magnification of the realized lens. Calculate the performance of the system in the center and the edge of the field size. Is the system diffraction limited? What are the dominant aberrations in the field? c) Show, that the resolution of the system in the field is anisotropic due to the elongated shape of the spot by inspecting the MTF of the system. d) Calculate a diffraction based imaging of the complete object field. Take a rectangular grid as an object. Can the anisotropic resolution be seen? Now calculate a geometrical imaging model of the same grid. What is the difference between the results of both models? e) To demonstrate the Fourier filtering in the pupil plane, change the rectangular pupil stop to a slit with dimensions Dx = 5 mm, Dy = 0.15 mm. What changes with the diffraction limited image? What is the consequence for the geometrical image? f) Calculate the Airy radius of the system and constitute an object with size 10 times the Airy radius. Put this small object in the center and at the edge of the field. Discuss the results and compare them. Is the grid resolved? Solution: a) Corresponding achromates are found for example in the catalog of Melles&Griot as: First lens with f = 25 mm: LAO – 25.0 – 12.5 Second lens with f = 100 mm: LAO – 100.0 – 50.0 The first achromate must be reversed. The distances must be adjusted in a way to guarantee a collimated ray bundle between the lenses and to let the chief ray cross the optical axis in the back focal plane of the first lens. The system stop has to be fixed at the intermediate focal point. Correspondingly, the distance between the stop and the second lens must be arranged to get telecentricity in the image space, the final image distance must be adjusted to get a sharp image.

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1st step: find t1, merit function angular radius 2nd step: find t2, not telecentric, initial distance infinity, entrance pupil at lens with diameter D = 3 mm 3rd step: after adding the second lens: 4th step: adjusting the distance behind the stop: chief ray parallel to axis, merit function with REAB = 0 5th step: focusing image plane on axis; merit function with spot criterion The result is (for this choice of lenses):

In the intermediate focal plane, a rectangular diaphragm with half widths 2.5 is placed.

b) The performance of the system can be estimated by a spot diagram as follows. From this diagram or the Prescription data menu we get a magnification of m = -4.004. In the center, the system is nearly diffraction limited, for the field of y = 1.5 mm, it is worse.

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If the spot diagram is considered, mainly coma, and defocus seems to be the reasons for the decrease of the performance. The defocus can be astigmatism or field curvature. To decide about this question, the Seidel diagram is considered. It is seen, that both aberration types are there but the astigmatism dominates by a factor of 2.

c) The MTF shows a large separation of the tangential and sagittal contrast curves. This describes the anisotropic resolution of the coma/elliptical spot shape.

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d) The result look like this picture, the settings are documented on the right. The difference in the resolution cannot be recognized in this coarse representation.

The following figure shows the result of the geometrical calculation for 106 rays.

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Both models shows a comparable sharp imaging of the grid. In the diffraction model, the special effects of diffraction are seen at the crossing points, where a higher intensity is calculated. Furthermore, in the diffraction calculation we have a stronger profile across a bar cross-section. This can be seen in a zoom representation of a single crossing point at higher resolution.

e) If a rectangular slit stop is established in the system, we get the following results:

In diffraction calculation, a filtering and blocking of the higher diffraction orders has nearly completely removed the horizontal grid lines. Therefore, the image is no longer similar to the object. In the geometrical picture, a first sight seem to have an unchanged image. A close look shows that the width of the horizontal and vertical bars is now different. This comes from the fact, that the slit diaphragm truncates those parts of the pupil, where large aberration are present.

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The large number of blocked rays can also be recognized from the reduced computational time.

f) The Airy radius is D = 13.4 m. Therefore, the field size in the image calculation is chosen to be 0.134 mm. The results of the imaging calculations in the center and at the edge are as follow:

It is seen, that the grid bars are resolved on axis, but the degradation is seen clearly. In the field, the resolution of the grid bars is lost with a bit more residual contrast in the sagittal direction.

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Exercise 9-2: Incoherent Imaging Load the file “Double Gauss 28 degree field.zmx” from the folder {Zemax Documents}\Samples\Sequential\Objectives. Fix all diameters to the values, which are given. For this system, the nominal field is angle 14° (half) corresponding to an image height of 25.8 mm. a) Determine the largest field angle, which can be transmitted through the system. Add this field point to the data and check the vignetting behavior of the system. b) Calculate a geometrical image of a line grid with 105 rays for a field size of 10 (degree) in the center of the field. Calculate the corresponding images for the new field point and the field size 47.2°. What are the differences?

On axis, the spot size is ~16 m. The bar size 25 m corresponds to an image 0.025 x 20 = 0.5 mm, corresponding to the field angle 0.27°. Explain the image in this case. c) Now, calculate the geometrical image of the “Demo picture – 640 x 480.bmp” bitmap for the full field size, 400x300 pixel with width 0.125 Solution: a) Layout:

Fix the diameters by the command: Tools / Convert Semi-Diameters to circular apertures The largest field angle is ~ 23.6°. If the relative illumination is calculated, it drops down to nearly zero.

b) Geometric imaging with ray trace

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Here the object has a size of 10 degrees, is centered at field 1, e.g. on-axis, and the image size is 20 mm. The field 23.6° corresponds to image size 41.7 mm, 10° to 17.4 mm. For field 4 the vignetting is visible

With Field: 47.2 the whole image is filled

With Field: 0.27 the limited resolution is nearly reached.

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c) Geometrical image of a bitmap with 640x480 pixel

For a good quality, 100 Rays per pixel have to be traced. Since the image is rectangular and the field size corresponds to the full image height in Y the edges in X direction and the corners are outside of the designed FOV of the system. Especially for the corners vignetting is clearly visible.

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Exercise 8-3: Physical Image simulation (Courtesy Moritz Eßlinger) Open the file “{Zemax Documents}\Samples\Sequential\Objectives\Apochromat.zmx”. Is the system diffraction limited? Simulate two images with image field sizes 0.1 mm and 50 mm. Solution:

- Spot shows: system is diffraction limited on axis (Airy = 6 um at primary wavelength

479 nm)

- “Image simulation”

- Grid of lines.bmp

- Image size 0.1 mm

- Oversampling None (object pixel doubling to increase pixel resolution)

- Guard band None (object zero padding, required for large PSFs)

- Geometric

- Input bmp has 201x201 pixel

- This means a pixel size of 0,5um, 1 pixel is below diffraction limit

- “Diffraction” mode required

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In convolution algorithms, the image automatically has the same pixel-size as the object and automatically fits the magnification of the objective.

- Image size 50 mm (250 µm/pixel)

- Diffraction limit cannot be seen by pixel discretization

- Field dependent, geometric aberrations dominate

For the large field of view, no difference is observable between diffraction mode (left) and geometric aberrations only (right). Oversampling increases the object number of pixels by coping the same brightness value in 2,4, … pixels.

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More details can be seen in this 4x oversampled (804x804 pixel) grid_of_lines.bmp. Distortion becomes evident in the edges that was previously unresolved (see above). In fast Fourier transforms, a periodic continuation of the sample is assumed. For aberration calculation, this imposes a problem as it implies that the PSF from the right end of the image leaks into the left end of the image. This is a problem of the numeric method used.

This can be avoided by choosing a guard band. The object field is extended and filled with zeros (zero padding). As a result, the image is zero at the edge and no artifacts from the periodicity occur.