1. ECEN 4616/5616 Optoelectronic Design Class website with past lectures, various files, and assignments: http://ecee.colorado.edu/ecen4616/Spring2014/ (The first assignment will be posted here on 1/22) To view video recordings of past lectures, go to: http://cuengineeringonline.colorado.edu and select “course login” from the upper right corner of the page. Lecture #20: 2/28/14

2. Graphical Ray Tracing Review From the equations of Gaussian Optics: Refraction Imaging We can derive the rules of ‘Geometrical Rays’ – rays that can be traced without calculation: 1.A ray parallel to the axis crosses the axis at f’ after refraction. 2.A ray crossing the axis at f will emerge parallel to the axis after refraction. 3.A ray through the middle of the lens (h=0) is undeviated. The imaging equation allows us to deduce a 4 th rule: 4.Any ray from an image point will pass through the corresponding object point after the lens. For these rules to be universal, we must also consider ‘virtual rays’: That is, the extension of rays where they can’t be, such as behind a mirror.

3. f f’ 1.A ray parallel to the axis crosses the axis at f’ after refraction. (We will always assume that rays are traced left to right.) ff’ This is why f’ and f are reversed when considering negative lenses.

4. ff’ 2. A ray crossing the axis at f will emerge parallel to the axis after refraction. f f’ Note that a ray ‘crossing the axis at f here means ‘either the ray or its virtual extension crosses the axis at f.

5. 3. A ray through the middle of the lens (h=0) is undeviated. f f’ f

6. Graphical Ray Tracing Rules for Mirrors First: Mirrors only have one focus, which we will label f’. From the surface power formula, and the sign rule that the index becomes negative when rays reverse direction, we can derive the formula for the power of a mirror (see section 3.4 in the text): Where R is the radius of curvature of the mirror, according to the sign rules. We will always label the center of curvature ‘C’, and hence f’ is halfway between C and the vertex of the mirror. Rules for Graphical Rays in Mirrors: 1.A ray parallel to the axis, crosses the axis at f’ after reflection. 2.A ray crossing the axis at f’ will emerge parallel to the axis after reflection. 3.A ray crossing the axis at the center of curvature (‘C’) will reflect back on itself.

7. 1.A ray parallel to the axis, crosses the axis at f’ after reflection. K > 0 K < 0 f’ C C

8. 2.A ray crossing the axis at f’ will emerge parallel to the axis after reflection. f’ C C

9. 3.A ray crossing the axis at the center of curvature (‘C’) will reflect back on itself. f’ C C

10. C a) Concave Mirror, real object at C Real, inverted image at C m = -1 Graphical Ray Trace Problems on the Homework:

11. C f’ b) Concave Mirror, real object between C and f’ Real, inverted image left of C m < -1

12. c) Convex Mirror, virtual object between f’ and C C f’ Virtual, inverted image right of C m < -1

13. d) Negative lens, real object at f’ f’ Upright, virtual image between f’ and lens. m > 1

14. e) Negative lens, virtual object at f f’ f Virtual, upright image at infinity. m = ∞ (or undefined) To ∞

15. f) Positive lens, virtual object at f’ f f’ Real, upright image at f’ m < 1

16. g) Convex mirror, image object at infinity. ? f’ C From ∞ Upright, virtual image at f’ m = undefined Note that any set of parallel rays is coming from the same object point, when the object is at infinity – we simply pick two whose reflection can be determined graphically – one aimed at the center of curvature, C, and one aimed at the focal point, f’

18. Some (other) Issues with Homework #1 Read the question carefully! OP2: A positive thin lens of 100 mm focal length is followed by a positive thin lens of 50 mm focal length. The distance between the two lenses is 150 mm. L1L1 L2L2 150 mm f1’f1’ f2f2 L1L1 L2L2 100 mm f1’f1’ f2f2 The problem set-up is describing this system: Not this system: A clue is that the second system is not afocal (having K=0)

19. The Whale in the Aquarium In retrospect, this is not a well-defined problem. (But, we can argue that it is representative of problem ‘definitions’ often encountered in the real world.) Whale Image of Whale air water The ‘Correct’ analysis: The whale’s image is closer, but the same size as the whale. What could possibly be wrong with this analysis?

20. Whale air water The whale appears closer to the observer (if the observer is standing close to the aquarium wall), hence looks bigger. The human visual system (HVS), however, is always taking this into account: You don’t think that a car or tree is actually bigger just because it is close to you. That illusion only occurs when you are fooled about the real distance.

21. A student sent me a research paper on this subject: Why do objects appear enlarged under water? Por que objetos parecem maiores debaixo d'agua? Helen E. Ross Shazia Nawaz In the Scottish Journal Arq Bras Oftalmol 2003;66:69-76 “Objects appear enlarged in water by less than the 4/3 angular magnification. They usually appear beyond their optical distance and slightly enlarged in linear size, but not in accordance with size-distance invariance.” The study found that observers saw the objects as both larger than actual, and not as close as actual – Apparently, they split the difference between accurately estimating size and distance: So, if the question had asked “Do people think the whale is larger?”, the answer would have been ‘Usually, yes’.

23. The Coronagraph: A significant problem solved with graphical ray tracing (and ingenuity) Problem: How to image something dim that is close (in angular terms) to a much brighter object? Examples:  The Sun’s corona, when the Sun is not eclipsed by the Moon:  A planet orbiting a star: In both cases, the radiance differs by a factor of millions to one. Optical systems, however, typically have a SNR of ~100:1

24. First Attempt: Image the objects on a plane that contains an occulting disk to block the bright object, but let the dim object pass, then re-image the dim object. Light from bright object Bright Object Image Stop 1 st Image Plane2 nd Image Plane Light from dim object Image of Dim Object The Coronagraph

25. Why it doesn’t work: The system stop and entrance pupil (the first lens, in this example) is brightly and uniformly illuminated by the bright object. Hence, there is significant diffraction from the edges of the entrance pupil, which is spread over both image planes, and not removed by the Image Stop: Diffracted Light from Entrance Pupil Essentially, the edges of the stop have become secondary light sources, due to diffraction. This light might only be a fraction of a percent of the bright object’s light, but that is enough to obscure an image of an object that is thousands (or millions) of times dimmer.

26. Diffracted Light from Entrance Pupil Lyot Stop: (at image of entrance pupil) Light from dim object is passed to image plane. Diffracted light from bright object is stopped. The solution is called the ‘Lyot Stop’: The Coronagraph What is happening here is that the entrance pupil (the first lens, in this case) is re-imaged at the Lyot Stop plane. Hence, the light that diffracts from the edges of the entrance pupil is re-imaged into a bright ring of light at the Lyot plane. The Lyot stop is just an annular opaque ring which blocks the (re- imaged) diffracted light.

27. The Coronagraph This method can be carried out for multiple iterations: 1.The bright object is imaged to a plane, where an opaque stop blocks it. 2.The first brightly illuminated lens (aperture) diffracts light from near its edges, producing a secondary source of interfering light. a.The offending lens is re-imaged and an annular stop blocks the diffracted light. 3.The 2 nd lens (which re-images the first lens) may also be too brightly lit, and diffracts interfering light from its edge. a.The 2 nd lens is re-imaged by a third lens and an annular stop blocks the edge-diffracted light. And etc…. Eventually, scattered light from the surface and/or bulk of the lenses will dominate the stray light. Separating this light from the image light optically requires other techniques, such as spatial filtering.

28. Laying out a Coronagraph K1K1 K2K2 a b d c e 1.The first lens must image the bright object to the first stop: 2.The second lens must image the entrance stop to the Lyot stop: 3.The second lens must also image the first stop (image plane) to the second image plane: These equations constrain, but do not determine the design: The designer must find reasonable values to use.

29. Laying out a Coronagraph Another serious design constraint is to determine the rate at which the PSF from the first lens decays, and whether that is enough to mask the image from the dim object(s). The larger the first lens, the smaller the angular extent of the PSF, so the PSF dropoff and angular separation between dim and bright objects constrain the minimum diameter of the first lens. The difficultly of building low F/# lenses means the the focal length will also increase with the diameter. Since extra-Solar planets are so close (angularly) from the associated star, another line of research is to find aperture shapes which have extremely small amounts of diffracted light: An example is the “Starshade” designs being created at CU: