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2020-03-04 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 15:

Optical Design with Zemax for PhD - Advanced

Exercise 15-1: Illumination system with non-sequential raytrace a) In non-sequential mode, define an ellipsoidal volume source with dimensions of diameters 1 mm in x and y and 2 mm in z-direction. A parabolic reflector is located 10 mm to the left of the source center. The radial aperture is selected to be 40 mm. A rectangular detector with size 10 mm x 10 mm is established in a distance 200 mm. Optimize the curvature of the paraboloid for the best efficiency of light detection under these conditions. Now increase the detector distance to 500 mm and optimize again. b) Now an aspheric lens is located in front of the mirror. The diameter is 80 mm, the thickness 20 mm and the material BK7. The shape of the mirror is now fixed with a radius of curvature of 20 mm. Optimize the lens shape to get the best efficiency on the detector, if this is now at a location z=150 mm. What is the overall efficiency of the illumination system now? Solution: a) System: A radius of 20 mm is used for R as initial value. The conic constant is set to -1.

In the merit function, first the detector is reset to zero by operand NSDD with all parameters set to zero. After that, a ray trace is performed with NSTR. After the ray trace detector data can be retrieved with NSDD. Pixel number -3 corresponds to the ray count on the detector. This number is maximized by minimizing the inverse with RECI.

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After optimization, the radius is 17.9 mm.

b) For the optimization, only the radius of curvature and the conic constant is changed. The efficiency is in the range of 77 %.

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Exercise 15-2: RXI collimator a) Establish a point source, which radiates into the 180° half space. A parabolic reflector nearly collimates the rays. The reflector vertex point is -8 mm to the left of the source. Put a perfect lens with focal length f = 40 mm in a distance 40 mm from the source point. A rectangular detector of the size 10 mm x 10 mm is located in the focal plane of the perfect lens. Furthermore, put a thin aspherical lens in the distance of 8 mm to the right of the source. The diameter of the lens is 20 mm, the thickness 10 mm and the material BK7. First optimize the mirror data and the aspherical shape parameters R and kappa to maximize the efficiency of power transfer onto the detector and minimize the spot size. b) Insert an annular aspherical lens outside the inner asphere and optimize its shape together with the mirror data to optimize the transfer efficiency. The outer diameter of the annular lens should be 40 mm. Solution: a) System data:

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Exercise 15-3: Projector illumination system For a given projection lens an illumination system is to be designed. After laying out the illumination optics in sequential mode, the system shall be simulated in non-sequential mode. The projection system shall project an image of 2 m width in a distance of 5 m from the projection lens. The light modulator is a micro display with dimensions 20 x 15 mm² and a resolution of 1280 x 960 pixels. We assume a thin transmissive display for sake of simplicity.

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The light source shall be a Lambertian LED with emitter size 2 x 2 mm². The light from the LED shall be homogenized with a tapered hollow integration rod. a) Load the lens “Double Gauss 28 degree field.zmx” from the samples folder. Save it in a different folder before changing anything. Remove all variables and adapt the system to the projection task. Is the lens suitable for projecting the image? b) Design an appropriate illumination system for the projection lens. This shall image the exit of the integrator rod onto the display. Behind the display there is a minimum distance of 20 mm to the illumination system. Start with a paraxial layout of the illumination system and turn the paraxial lenses into real ones after determining the focal lengths and distances. What has to be cared about when transferring light between optical systems? Can the illumination system be made from one element/group? c) Convert the system to non-sequential mode (not mixed mode). Add an absorbing “Detector Color” for the projection screen. Add the integration rod and the LED. Choose a blackbody spectrum with a color temperature of 6000 K for the LED. In place of the display add a “Slide” surface with a bitmap image. Run a simulation of the projection. What can be seen in the projected image? How can the image be improved? Solution: a) The system has to be scaled to the correct focal length. The magnification we want to have

is 𝛽′ = 2000 mm / -20 mm = -100. With the projection distance of 5 m we calculate a focal length of 50 mm, the lens has thus to be scaled by a factor of 0.5x.

After scaling the object distance is changed to 5000 mm and the field definition is changed to “object height” with the edge at 1000 mm (for simplicity, we ignore the fact that we actually need to define the diagonal of the field in this example). A “quick focus” refocuses the image plane to smallest spot radius.

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The shortest possible spatial wavelength of patterns on the display is two pixels. Thus the

highest spatial frequency that can be displayed is 𝑅𝑚𝑎𝑥 =1280 𝑃𝑖𝑥𝑒𝑙𝑠

2 × 20 𝑚𝑚= 32 𝑚𝑚−1. With an MTF

of about 0.8 the resolution of the system should be more than adequate for this purpose.

The projection lens has an effective F-number of 3.0 while projecting a display of 20 x 15 mm².

The corresponding entendue is 𝐸𝑡 = 𝜋 𝐴 𝑛2 sin2 𝑢 = 26.2 𝑠𝑟 𝑚𝑚2. The etendue of the light

source is 𝐸𝑡 = 𝜋 𝑛2 𝐴 = 12.5 𝑠𝑟 𝑚𝑚2. Thus, the lens is capable of carrying the light from the light source. b) When connecting optical systems the intermediate image and pupil locations and sizes must match. Designing the system in reverse direction from screen to illumination, the chief rays coming from the display are diverging, thus our illumination lens must be hypercentric on the display side to match the pupil of the projection optics. On the rod side the illumination system must be telecentric since every point of the rod exit emits light into a cone with its axis parallel

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to the optical axis. The conjugate of the display plane must be on the rod exit with matching image size. The pupil and image requirements cannot be fulfilled at the same time with a single lens group. Thus, we add two paraxial lenses after the display surface. The first will pick up the light from the projection lens (we are still designing the system in reverse direction) and form an intermediate pupil. The second lens will produce an image of the display on the rod exit and make the image space telecentric. The magnification of the illumination system is somewhat arbitrary – we choose -0.75x to get 15 mm width for the exit of the integrator rod.

Most of the system can be calculated by solves:

A solve for chief ray height zero calculates the distance from the first lens to the intermediate pupil.

The distance from the intermediate pupil to the second lens is equal to the focal length of the second lens, ensuring telecentric image space.

A solve for marginal ray height zero keeps the rod exit surface in focus. The distance from the intermediate pupil to the second lens is optimized for correct image height on the integrator exit according to the choice of the first focal length. We choose f’ = 30 mm for the first lens at a distance of 30 mm from the display (to allow some space for thick lenses) and optimize for a rod exit size of 15 mm.

Now we can turn the paraxial system into one with real lenses. We replace each paraxial lens with two face-to-face plano-convex lenses. Start with 30 mm radius for the first group and

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23 mm for the second group. Also we add a condition for telecentric image space to the merit function and add the standard RMS spot size operands. The lenses within the two groups should be identically shaped plano-convex lenses. Variables are the curvatures, the intermediate distance and the image distance.

This is the system before optimization:

After optimization the system looks like this:

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c) Choose File Convert To NSC Group:

De-select “Add Sources & Detectors” and click OK. Zemax converts the system and switches to non-sequential mode:

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We now add a Detector Color for the projection screen, the rod integrator and the light source. Remember that non-sequential mode is object oriented and all coordinates refer to the reference object’s coordinate system (i.e. global when zero) and not to intermediate distances! Also, the order in which the objects are defined in the Non-Sequential Component Editor does not matter (with the exception of coinciding objects). For diagnostics we also add a Detector Rectangle at the integrator exit and at the display plane.

Since we are building the system in the reverse direction and want to have the light propagating in -Z direction, make sure to reverse the ray direction from the source:

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With this model we can run a first ray trace:

Empty image on screen:

Display plane (left: position space, right: angle space):

Integrator exit (left: position space, right: angle space):

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Note that unless we check “Use Polarization” in the ray trace control we will not see losses due to reflection and transmission on optical surfaces or due to absorption in optical media. We started with a flux of 1W from the LED and see 0.87W on the screen. The remaining flux was lost due to the size mismatch of rod exit and illumination optics (we designed for the width of the field, not the diagonal), and due to aberrations. Looking at the real-lens illumination system above, we see that the F-number for the field points on the integrator side of the illumination system is smaller than on axis. Thus, we collect less light in the field. A better corrected design using aspheres could overcome this due to the better correction. To also include Fresnel losses and bulk absorption in the calculation we check the option “Use Polariziation” when raytracing:

Now the flux on screen is down to 0.45W. 23% alone are lost in the integrator due to reflection losses. Another 26% percent disappear in the illumination optics. So it is good idea to add some anti-reflection coatings to the illumination system:

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Now the total loss in the illumination system is reduced to 12%. This could be further improved by choosing a better anti-reflection coating. The reflection losses in the integrator could be reduced by shortening the integrator or by using an integrator slab made from glass utilizing total internal reflection. We now simulate the projection of an image. We reduce the Y size of the integrator exit to 5.625 mm to match the aspect ratio of the display. The angular extent of the light exiting the integrator will thus increase a little in the Y direction. We now place a Slide object at the position of the display. Make sure to displace it from the detector object by a distance larger than the “Glue Distance” specified in the Non-Sequential settings of the System Explorer. Objects closer than this distance will be considered coinciding with the object specified last in the NSC Editor taking precedence! By appending the Slide object we would effectively “cut” a rectangular hole in the detector.

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At the moment we are just tracing three wavelengths so the simulated RGB image looks like this:

We can define a spectrum for our source and have the wavelengths randomly chosen from the whole range, thus improving the color rendering of the image. For the LED we will choose the spectrum of a black body with a temperature of 6000K:

Now the simulated image looks much better:

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