2015-11-24 Mateusz Oleszko, Chang Liu Friedrich Schiller University Jena Institute of Applied Physics Albert-Einstein-Str 15 07745 Jena

Solution: Lens Design II – Part 3

Exercise 3-1: Alvarez Lens An Alvarez lens pair is a special system to change the focal position by transverse translating two lenses. The first lens is a plate of thickness 1 mm made of BK7. The front surface is plane, the rear surface has an aspherical shape, which is described by the polynomial

23 302.0),( xyxyxz

The second lens has the same thickness and material, the same aspherical surface on the front side and is plane on the rear side. The air distance between the two components is 1 mm. The system is illuminated by a collimated beam with diameter 2 mm and wavelength 550 nm. After the pair of plates an ideal lens with focal length f = 20 mm is places 1 mm behind. The final distance to the image plane should be 19.5 mm.

a) Establish the system. The aspherical surfaces should be modelled by an extended polynomial surface type with 8 terms and a normalization radius of 2 mm. The data of the aspherical surfaces can be seen in the extra data editor. The system should be prepared for a later movement of the plates in the x-direction by the same amount in different directions (first plate in +x and 2nd plate in -x).

b) Calculate the spot vs defocus option with a relative delta focus of 500 m. Where is the focus point?

c) Now shift the first plate by 0.24 mm in the positive x direction and calculate the spot vs defocus again. What is observed?

Solution: a)

The second coordinate break surface is established by a pickup with the factor -2 relative to the offset of the 1st coordinate break. The 3rd coordinate break is the same as the 1st one.

In the 3d layout, a 90° rotation around the z-axis must be performed to see the x-z-section for the later movement.

b) The figure below is obtained. The smallest spot and therefore the focus lies at +0.5 mm at an absolute distance of 20 mm from the perfect lens.

c) The focus point is shifted by approximately 1 mm to the left as proposed. The spot remains nearly circular symmetric.

Exercise 3-2: TMA mirror system TMA (three mirror anastigmates) are pure reflective systems without central obscuration and with corrected astigmatism. According to the Coddington equations, a mirror generates large astigmatism, if it is used under a finite incidence angle. Due to the fact, that the astigmatism is caused by the splitting of the effective radius of curvature according to R cos(I) and R/cos(I), it is necessary to use at least one negative convex mirror to correct astigmatism. Typically a setup this a plus-minus-plus power distribution is used.

a) Establish first a centered system with 3 mirror with the radii -1000, -160, -400 mm. The incoming

ray bundle has a diameter of 100 mm, the wavelength is = 1 m. The distances between the mirrors are 250 and 300 mm. Determine the optimal final image distance, the focal length and the F-number of the setup. The stop is located at the second mirror (in this example we don't consider a field, therefore this is not of relevance).

b) Now introduce every mirror as a bending mirror to avoid the central obscuration. The tilt angles should be --12°, 22° and -10°. Is the system now anastigmatic? What are the dominating residual aberrations (inspect the Zernikes)?

c) Now optimize the system by maintaining the focal length and the F-number. Due to the complicated interaction of the possible variables, the control of the optimization is not trivial and unique. Especially the overall setup should not get strange sizes and further overcrossings of rays and mirrors. Therefore the following steps are recommended:

1. First allow only the final image distance and the tilt angles at mirror 2 and 3 to be changed. 2. The result shows overcrossing, therefore the angles are fixed to be -16° and -5° to avoid unphysical solutions. Now the radii of curvature are optimized 3. Now the distances and the first tilt angle are optimized 4. Finally, the tilt angles 2 and 3 are reset by hand to obtain the critical case of smallest angles without obscuration. Solution: a) The system is seen here. The focal length is f = 216.22 mm, the F-number 2.178.

b) The system now looks like this:

From the Zernike coefficients, tilt, astigmatism and coma are identified to be the dominant contributions.

c) 1st step:

c) 2nd step

c) 3rd step

It is seen, that the dominating residual aberration is astigmatism of 3rd and 5th order.

c) 4th step The angles can be reduced step by step to -11° and -2°. The result is not diffraction limited but a Strehl ratio of 64% are achieved.

Exercise 3-3: Aspherical IR relay system Courtesy K. Uhlendorf Aspheres are suitable for correction of spherical aberration as well as field aberrations, but don’t help with the correction of color. Aspheres are widely used in infrared systems, because the IR material is expensive and the turning of the lenses makes no difference cost between spherical, aspherical and even diffractive surfaces. Load example: Ex 10-13 aspherical IR relay (1).zmx The system is a 1:1 relay for the MWIR range (3.6 – 4.9 µm), starting from the detector (including a cold stop). There are requirements on the performance and manufacturing defined in the Merit function. Do you understand all definitions? a) Analyze, which are the dominant aberrations. Use the Zernike-over-field representation. b) Which is the most suitable surface for correction of spherical aberration? Apply the asphere and optimize the system. c) Which surface can be suitable for the correction of field aberrations? Apply the asphere and optimize the system. Why is the image performance only slightly improved? d) Use a binary 2 – surface description on surface 10 for color correction. Solution:

a) The Zernike Coefficient vs. field analysis is very useful to understand which aberrations are still remaining for the optical system. In the diagram below the Standard Zernikes (Z4-Z11) are shown.

It can be seen that the spherical aberration (Z11), field curvature (Z4) and coma (Z7) are the dominant aberrations of the system. b) To correct spherical aberration a surface close to the pupil should be chosen to be aspherical. In our case it is surface 6. After optimization the image performance is clearly improved. Besides the correction of the spherical aberration also the field aberrations are reduced, because the system used the newly gained degrees of freedom also for field correction.

c) For further field correction a surface with good field separation should be chosen. In our case it is surface 11. The improvement is only small, because the color aberrations (longitudinal as well as lateral) are now dominant, which can’t be corrected by an asphere.

d) Now surface 10 is defined as a Binary 2 surface (diffraction order 1, 2 coefficients, and normalization radius 12). After optimization the image performance is strongly improved. The drawback of using

diffractive surfaces is the worsened signal-noise ratio by the change of diffraction efficiency over the wavelengths and the unwanted side orders.

Exercise 3-4: P-N-P triplet field flattening A plus-minus-plus triplet approach is feasible to flatten the curvature of field.

a) Establisch a symmetrical systetem with outer plano-convex lenses made of SF66 and a symmetrical inner lens made of K5. The incoming beam is collimated with a diameter of 10 mm at

= 632 nm. The outgoing beam should have the same size and the total length of the system is fixed to be L = 50 mm. A field of 5 ° is defined for a stop position in the center of the system. The ratio of the marginal ray heights at the center (negative lens) and the initial ray bundle is denoted by k = h2 / h1. To get a finite image location, an ideal lens of f = 100 mm is added. Optimize the system for k = 0.8 and look at the field curvature.

b) Optimize the system for k = 0.7 and 0.6 3 and compare the effect on the field curvature.

c) What happens, if the diameter at the negative lens is further decreased ? Solution: a) The system for k = 0.8 looks as follows.,.

The remaining field curvature is 0.31 mm (operator mean of FCGS and FCGT in the merit function with weight 0). It is noticed, that the curvature is positive ! b) For the stronger ratio of the heigths we get a stronger effect:

No k h2 Sag of field curvature in [mm]

1 0.8 4.0 0.31 mm

2 0.7 3.5 0.56 mm

3 0.6 3.0 0.63

c) If the diameter at the negative lens is further decreased, the astigmatism and field curvature grow tremendously.