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2019-11-20 Prof. Herbert Gross Uwe Lippmann, Yi Zhong, Dennis Ochse Friedrich Schiller University Jena Institute of Applied Physics Albert-Einstein-Str 15 07745 Jena

Exercise Solutions 5:

Optical Design with Zemax for PhD - Basics

Exercise 5-1: PSF scaling

To check the Airy diameter formula, we establish a simple system. According to the formula

Dairy = 1.22 / NA we select a system with a perfect lens with the data:

wavelength = 1 m

numerical aperture in the image space of NA = 0.61 = 1.22 / 2

With these numerical values, the Airy diameter must be exactly 2 m. A collimated input beam with 10 mm diameter therefore need a special focal length of the ideal lens to produce this angle. This is calculated by Zemax with optimization and a corresponding merit function with REAB = -0.61. We get a focal length and a final image distance of 6.495 mm.

The corresponding cross section of the point spread function has its first zero exactly at 1 m.

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Now we select the focal length to generate a numerical aperture of NA = 0.5. According to the

Rayleigh length formula Re = / NA2 we get a focusing distance of 2*Re = 8 m to locate a zero of the PSF on axis. The focal length to get this aperture is f = 8.660 mm. Now the slider is used to find the zero point on axis. It is seen, that Zemax is not able to calculate the point spread function exactly: there is no zero point found.

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Exercise 5-2: PSF calculation Establish a symmetrical biconvex lens with a focal length of f = 100 mm made of SF1 for a

wavelength of = 1m. The diameter of the incoming beam should be 15 mm. a) Calculate the spot diagram and the cross section of a point spread function (128 sampling points). Discuss the sizes of these two representations. Calculate and compare the exact Strehl ratio and the Marechal approximation.

b) Fix the plot window size of the PSF to 80 m and produce a plot with the normalized PSF in the best plane and the two planes defocused by +0.5 mm and -0.5 mm respectively. Discuss the result. c) If now the aperture is enlarged to a diameter of 35 mm and the final distance is re-optimized, the PSF with 128 points calculation begins to show sampling problems. How can this be seen in the cross section representation? How can this be seen in the 2D-plot? Solution: System

a) The spot has a RMS radius of 9.1 m; the Airy radius is 8.1 m. The spot diagram is

broadened to a radius of approximately 18 m corresponding to the geometrical radius of the spot.

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The exact Strehl ratio is 56.4% (Huygens PSF), the estimated value 55.9% (Zernike Fringe/Standard coefficients). Therefore, the error is in the range of 1%.

b) The plot looks like the following figure. The PSF is broadened and has a smooth one-peak-shape for the positive defocus +0.5 mm and a ring shaped strongly modulated profile for -0.5 mm.

c) If the cross section PSF is calculated, the energy is spreaded out until the outer points of the grid. This means, that the signal is not bandlimited and shows aliasing.

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In the 2D representation it can be seen,that the rotational symmetry is no longer fulfilled. This comes from the fact, that the effective spectral extend is larger along the diagonal directions and shows, that there is already an inaccuracy in the x- and y-direction.

Exercise 5-3: Strehl ratio and PSF vs spot size

A single lens made of K5 with focal length f = 25 mm and thickness d = 5 mm is illuminated by a diverging beam with numerical aperture NA = 0.1. The wavelength is 𝜆 = 0.546 µm (mercury e-Line). After the lens the light should be collimated. If the collimated beam is refocused without further aberrations, the point spread function is not diffraction limited. a) Calculate the accurate Strehl ratio, the estimated Strehl ratio and the geometrical and diffraction encircled energy inside the ideal Airy diameter. b) If now the numerical aperture is reduced, the Marechal estimation becomes better. Calculate the largest NA, for which the relative error is smaller than 2%. What amount for the geometrical and diffraction encircled energy inside the Airy diameter is obtained here? c) Show the Strehl ratio as a function of the numerical aperture as a universal plot. What is the maximum value for getting a diffraction limited correction with DS > 0.8? Solution: a) System data and layout:

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If the cardinal points of the lens are calculated, the unknown first distance is obtained paraxially as t1 = 25 – 3.280 = 21.72 mm When optimizing with “Quick Adjust” using “Angular Radial” as the criterion after the back surface of the lens, the optimal distance for collimation seems to be 21.55 mm.

b) If the numerical aperture is changed, the following steps are performed: 1. Reduce NA 2. Determine the Airy diameter out of the spot diagram window 3. Set the aperture in the image plane exactly to the Airy value 4. Calculate the estimated Strehl ratio from the Zernike window 5. Calculate the accurate Strehl ratio from the Huygens PSF window with appropriate sampling 6. Calculate the geometrical encircled energy by the footprint diagram (with option: delete vignetted) 7. Calculate the diffraction encircled energy by the text output of the EE window. Then the following table is obtained:

NA Strehl exact

Strehl estimated

relative error

Airy radius geometrical EE inside Airy

diffraction EE inside Airy

0.1 0.019 0 0 0.003299 0.0389 0.0394

0.08 0.058 0 0 0.004123 0.0889 0.0889

0.07 0.053 0 0 0.004712 0.1449 0.1443

0.06 0.172 0.2158 0.255 0.005497 0.2514 0.3403

0.055 0.342 0.3662 0.0661 0.005996 0.3414 0.4611

0.051 0.486 0.4958 0.0198 0.006466 0.4467 0.5534

0.05 0.520 0.5277 0.0146 0.006600 0.4780 0.5650

0.045 0.676 0.6753 0.00104 0.007328 0.6887 0.6607

0.04 0.798 0.7940 0.00501 0.008244 1.0 0.7281

The relative error of the estimated Strehl ratio is smaller than 2% for NA < 0.051. Here the geometrical encircled energy is 45%, the diffraction calculated encircled energy 55%.

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d) The universal plot is obtained with the following setting:

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From the corresponding text window, we also get a limiting value of approximately NA = 0.04 for the diffraction limit.

Exercise 5-4: Transfer function

a) Load the Cooke triplet 40° from the sample files of Zemax. Restrict to the center wavelength and reduce the field to 12° at one off axis field point only. Locate the stop 20 mm in front of the system. What is the residual power transmitted in the field position? Fix the vignetting in the field menu. Calculate the distortion. b) What is the Airy radius in the field point? Estimate the diameter at 10 % intensity of the PSF in the field in x and y respectively. Calculate the modulation transfer function and discuss the

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three curves. Estimate the contrast in the case that the PSF diameter is approximately of half the width of a grating period. c) Determine the location and the size of the exit pupil. Calculate the phase transfer function in the field position. Discuss the green curve of the tangential orientation. If the green curve is extrapolated to a contrast reversal of 180°, what is the corresponding spatial frequency? What is the corresponding feature size? Explain the difference of this number with the distortion calculated above. Solution: a)

Only 40% of the energy is transmitted. Footprint at the pupil position without / with setting the vignetting:

Due to the remote pupil position, there is a distortion of 0.711 % corresponding to a height

difference of 75.5 m.

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b) The Airy radius on axis is 3.34 m, the diameter therefore on axis 6.7 m. For the field point with vignetting factors set, the Airy disc is calculated as an ellipse with axis diameters 15 µm and 8 µm.

If the MTF is calculated, a strong separation between tangential and sagittal is seen for the field position due to coma and vignetting.

If the cross section of the point spread function is calculated for the field position, we get a large difference in x- and y-direction respectively. A rough estimate of the diameters at intensity 10 % are:

Dx = 12 m , Dy = 60 m.

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A corresponding width of the grating bars is obtained for the spatial frequencies

x = 0.5 / 0.012 = 42 Lp/mm and y = 0.5 / 0.060 = 8.3 Lp/mm. According to the zoomed MTF curve we get a corresponding behavior with a contrast of 60 % - 70 %.

c) The exit pupil is located at -143.8 mm with a diameter of 28.83 mm. The phase transfer function looks as follows. The green T-curve describes the distortion with growing field height.

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If the green curve is extrapolated, a contrast reversal with 180° is obtained approximately at a

spatial frequency of 80 Lp/mm corresponding to a period of 1/80 = 0.0125 m and a bar width

of 6.25 m. The reason for this discrepancy is the existence of coma and other nonsymmetrical aberrations. If the spot diagram in the field is calculated with respect to the chief ray and to the

centroid respectively, we see a shift of 6.3 m corresponding to the value above.