Exercise 1 - Lens bending

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The object field has a diameter of 30 mm.

1 Exercise 1 - Lens bending Most of the aberrations change with the bending of a lens. This is demonstrated in this exercise. a) Establish a lens with focal length f = 100 mm made of BK7 with thickness 5 mm for nm for an object space numerical aperture of NA = The object distance is 170 mm. The object field has a diameter of 30 mm. The stop is located at the lens b) Generate a universal plot for coma, spherical aberration, astigmatism and Petzval curvature, if the curvature of the first lens surface is varied between Explain the results. c) Now modify the setup by placing the system stop 30 mm in front of the lens. What is changing? d) It is obvious that the stop position and the bending have influence on the aberrations. Therefore it makes sense to look at the combined effect. For getting this, generate a 2D universal plot, where the stop position and the bending are changed. Formulate the second thickness as a pickup to keep the object distance constant and change the distance between object and stop from mm. Plot the 2D-dependence for spherical aberration, coma and astigmatism. Interpret the results. What is the optimal bending for a distance of 100 mm? Is it possible to correct all three aberrations simultaneously? Solution: a) The data are as follows, the second lens surface is fixed as a solve with the power of b) In the merit function, the lines with SPHA, COMA, ASTI, PTZC are established. The petzval curvature is multiplied by a factor to get similar numbers.

2 The universal plot with the various lines are generated with constant y-interval, then all the 4 plots are overlayed. As expected, the spherical aberration changes quadratically, the coma nearly linear, the astigmatism nearly not and the petzval curvature also very weak. c) As can be seen, the spherical aberration and the Petzval curvature is not changing. The variation of the coma now is also quadratic and the astigmatism is now stronger changed. d) The corresponding plots look as follows.

It is seen in the diagrams: 1. spherical aberration does not depend on the stop position 2. the minima for all 3 aberrations are at different bending values 3.

3 It is seen in the diagrams: 1. spherical aberration does not depend on the stop position 2. the minima for all 3 aberrations are at different bending values 3. coma and astigmatism become large, of large stop distances produce large chief ray heights 4. coma and astigmatism has a zero correction option for a certain stop distance, these are lying at different values Minima of the bending / first surface curvature of the quadratic dependencies for the distance 100: SPH Coma Astigmatism No of the aberrations can be corrected to zero, the minima occur at different bendings and therefor only a compromise is possible.

Exercise 2 - Aplanatic lens Consider a collimated incoming beam with wavelength 500 nm and diameter 10 mm. This bundle should be focussed by a perfect lens of focal length f = 50 mm.

Show at least two different methods to find the best image position. b) Show that the spherical aberration of this setup is exactly zero for all orders.

What is the largest present aberration? Solution: The initial focussing lens is established as follows: a) A lens with thickness 1 mm is placed 1 mm behid the lens.

5 Exercise 2 - Aplanatic lens Consider a collimated incoming beam with wavelength 500 nm and diameter 10 mm. This bundle should be focussed by a perfect lens of focal length f = 50 mm. a) Place an aplanatic-concentric lens shortly behind the ideal lens with the material SF57. What is the resulting numerical aperture in the image space? Show at least two different methods to find the best image position. b) Show that the spherical aberration of this setup is exactly zero for all orders. c) Aplanatic means, that the linear coma vanishes and the imaging is free of coma for a small but finite field size. Show this property by using a small field of 2 for the current system. What is the largest present aberration? Solution: The initial focussing lens is established as follows: a) A lens with thickness 1 mm is placed 1 mm behid the lens. The first surface is made aplanatic by a solve, the second surface is mad concentric by choosing the solve 'marginal ray normal' to force the marginal ray to be concentric. If a single ray trace is performed, we get the direction cosine of the marginal ray to be It can also be seen, that the marginal ray is concentric at the surface 5.

Optimizing the last thickness as a variable with minimal spot size b) If the Zernike polynomials are calculated, they are

6 The best image position can be obtained by 1. Quick focus option 2. Solve at the last surface with marginal ray height 0 3. Pick up on the last (concentric) surface radius 4. Optimizing the last thickness as a variable with minimal spot size b) If the Zernike polynomials are calculated, they are exactly zero for all orders. c) If a field of 2 is introduced and the Zernike coefficients are calculated for the field point in the image and behind the 4th surface (the aplanatic), we get the following picture:

7 In the image, defocus (which is here in field the field curvature) and astigmatism are the dominating aberrations. Directly behind the aplanatic surface, only defocus has a considerable amount. This shows, that the concentric surface limits the system performance by astigmatism and field curvature. The change in the coma of the aplanatic surface is extremly small.

Exercise 3 - Aberrations and Performance of an Diode Collimator Load the lens GLC-6.5-8.0-830 DIODE LASER COLLIMATING LENS form the catalog of CVI Melles Griot.

8 Exercise 3 - Aberrations and Performance of an Diode Collimator Load the lens GLC DIODE LASER COLLIMATING LENS form the catalog of CVI Melles Griot. a) What is the numerical aperture of the lens in the image? Determine the Strehl ratio and compare the estimated and the exact value. b) What are the two surfaces with the largest contribution to the spherical aberration? The fisrt lens groups looks like an achromate. Corresponds the correction of this part of the system to an achromate? c) Show, that all intersection points of the spot diagram are inside the Airy diameter. If the spot is analysed it is seen, that there is a bright kernel with a surrounding halo. What is the relative power content of the inner kernel region? What is the diameter of this inner part? d) What is the contrast of a grating image with 100 linepairs per millimeter of this system? Determine the MTF of the lens for defocussing. What is the depth of focus, inside which the cointrast is larger than 50%? Explain the asymmetry of the curve. Solution: a) The system looks as follows. The numerical aperture is NA = in the image space.

The Strehl ratio in Marechal approximation is D = 0.972, the exact value is D = 0.975.

the system have the largest contributions to the residual aberrations.

spherical aberration is not corrected. Therefore, this part does not correspond to a conventional achromate.

9 The Strehl ratio in Marechal approximation is D = 0.972, the exact value is D = b) The Seidel surface contributions show, that the 2nd and the 7th surface (without the first help surface) of the system have the largest contributions to the residual aberrations. If the Seidel diagram is calculated only for the 3 surfaces of the first cemented group it is seen, that the spherical aberration is not corrected. Therefore, this part does not correspond to a conventional achromate. c) The spot diagram has a geometrical radius of µm, the Airy radius is µm and therefore all spot points are inside. The diameter of this inner part is D = 0.2 µm.

$If the geometrical encircled energy function is considered (without 'multiply by diffraction limit) and with 256x256 sampling points), it is seen, that inside this region we have approximately 60% of$

10 If the geometrical encircled energy function is considered (without 'multiply by diffraction limit) and with 256x256 sampling points), it is seen, that inside this region we have approximately 60% of the beam power. d) The calculation of MTF versus focus gives in the nominal image plane a contrast of 80%. From the text version of this plot we get the limiting values of mm and µmm for 50% contrast, Therefore the corresponding depth of focus is z = 20 µm. The asymmetric behavior is a result of the residual spherical aberration of the system.

If in particular in a front region there are large residual aberrations, they generate macroscopic changes in the ray height at the following rear surfaces and therefore the reference of the linear

12 Exercise 4 - Induced aberrations If a systems suffers from higher order aberrations, we get the so called induced aberrations. The reason is the combination of 3 rd order aberrations to higher order contributions, which are not additive. If in particular in a front region there are large residual aberrations, they generate macroscopic changes in the ray height at the following rear surfaces and therefore the reference of the linear paraxial ray is no longer meaningful. The change of the ray bundle diameter at the later surfaces and the corresponding changes of the surface contributions are called induced aberrations. In particular it is seen in these cases, that the performance of the system is different, if it is reverted. These effects can only be seen in the 5th or higher orders of aberrations. This is not computed in Zemax explicitely. a) Establish a collimating aspherical lens up to the 6th order for a focal length of 50 mm at a wavelength of nm and an initial numerical aperture of NA = 0.2. The lens should have a thickness of 6 mm and is made of BK7. b) Now set a single spherical plano convex lens of the same paraxial data a distance of 100 mm behind the asphere. The final image distance should be also mm. Calculate for this system the diameter of the ray bundle at the last lens, the Zernike coefficients c9 and c16, the marginal ray angle and the rms of the wavefront. c) Invert the system and calculate and compare the same data as in b) Solution: a) The data are as follows: b) The system data are: c) Reverted system:

13 The desired values are collected in the following table: asphere in front spherical lens in front diameter c c Wrms sinu' It is seen, that in the configuration with the asphere in front, the diameter at the singlet is 1mm larger and therefore the normalization radius and the Zernike aberrations are different in comparison to the other setup. The macroscopic change in diameter and angle therefore produces a nonsymmetrical setup with considerably different numerical aperture, which is decreased by 10%.

Lens Design I Seminar 5

Y. Sekman, X. Lu, H. Gross Friedrich Schiller University Jena Institute of Applied Physics Albert-Einstein-Str 15 07745 Jena Lens Design I Seminar 5 Exercise 5-1: PSF scaling (Homework) To check the Airy