1 / 42

Midterm exam

Midterm exam. Review of answers 3 /12/13. Eyes below the surface. #1+2. Aquatic eye in air Too much focusing Terrestrial eye in air Works great. Fig 5.1. Lecture 10. Spherical aberration. #3. Normally, outer rays focus in front of inner rays.

ping
Download Presentation

Midterm exam

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Midterm exam Review of answers 3/12/13

  2. Eyes below the surface #1+2 • Aquatic eye in air • Too much focusing • Terrestrial eye in air • Works great Fig 5.1 Lecture 10

  3. Spherical aberration #3 Normally, outer rays focus in front of inner rays No info on colors so can’t be chromatic aberration Lecture 9 Fig 3.6b

  4. #4 Terrestrial eyes vary with lifestyle Nocturnal eyes Large lens to gather a lot of light Short focal length so good sensitivity but not great resolution Lecture 10 Fig 5.7

  5. #5 Human eye has good resolution and perhaps good sensitivity (a or b) Long focal length Small aperture Good for resolution Lecture 10 Fig 5.7

  6. #6 Lens focal length is determined by curvature of lens surfaces Large radius of curvature = large focal length Small radius of curvature = small focal length a) short r, short f

  7. #7 Why have larger eye? #1 Resolution Fig 5.3 Image size So either small f small I OR large f large I • I / O = f / U • f) a and d Lecture 10

  8. #8 Polarization field • Due to differential scattering (d) • Varies with time of day • Can be detected by verts and inverts

  9. #9 Cephalopod eye vs fish eye: similarities Single chamber, camera style eye Flat cornea Spherical lens with graded index to correct for spherical aberration Retina with photoreceptors Variable iris 3 pairs of muscles to move eye around

  10. #9 Cephalopod eye vs fish eye: differences Microvillarvsciliary receptors One visual pigment (color blind) vs many Receptors point to front vs back Neural processing outside of eye vs processing in the eye Lens has two halves and grows from inside out vs growing from outside in

  11. Attenuation coefficient #10 Trans=e-αl α T UV Blue Grn Red Ocean Trans blue > red or UV Lake Trans blue < red

  12. Attenuation coefficient #10 Trans=e-αl α T UV Blue Grn Red Ocean Trans blue > red or UV Lake Trans blue < red Light reflected in blue ocean > lake red ocean > lake Ocean SWS cone > LWS cone

  13. What you see is projected onto your retina Distance to object Eye focal length Letter size is 2.5 cm or 0.5 cm for width of one line Image size = Object size Focal length object distance Image size = focal length * Object size object distance

  14. Bionic eye: receptors = pins would match projected image size Pin size

  15. Bionic eye #11 a) Each receptor needs to detect 0.5 cm part of E • Can use similar triangles or • in either case

  16. Bionic eye = array #11 b) 2 cm width and 0.016 cm per pin = 125 pins

  17. Bionic eye = array (rotate 90° to show easier) #11 c) What object size does array view? 16mm 50 cm 2 cm

  18. Field of view and # of letters #11 EACH LETTER TAKES 2.5 cm SO 62.5 cm CAN HOLD 25 LETTERS 62.5 cm

  19. Snell’s law #12 a) If go from low to high index - light bends in towards normal θ1 Angles are relative to normal And not relative to surface n1 n2 θ2

  20. #12 Refracted angle for high index plastic • Use Snell’s law • This is better as plastic bends light more than glass (12.74°) so you can get more focusing with less plastic (make lens flatter so less material)

  21. Reflected loss #12 • Use normal reflectance • This is more reflected light than glass (4.6%) so it is worse since more loss

  22. #12 Antireflection (AR) coating with MgF2 • Reflective loss at 1st interface: air / MgF2 • Reflective loss at 2nd interface: MgF2/glass • So this is less loss than 4.6% for air/glass

  23. Dragon vision - baby dragon #13 Lens was oval f=5 mm D=4mm F#=f/D=1.25 Δρ= d/f=3 mm/5mm = 0.6 rad = 34.4° Resolution = 1/2Δρ = Spatial frequency = 0.83 cycles/rad

  24. Resolution #13 Dragon 0.83 cyc/rad 34.4°

  25. Dragon vision - baby dragon #13 Lens was oval f=5 mm D=4mm = 4000 microns (um) S=0.62D2 Δρ2Pabs = 0.62 (4000 um)2(0.6 rad)2 * 0.3 = 1x106um2sr

  26. Sensitivity: all these are given in micron2sr #13 Dragon 1x106 nocturnal

  27. Dragon’s color vision? #13 Like a color blind human = dichromat

  28. Color triangle #14 Monochromatic line establishes a boundary Colors fall within space defined by this line Not possible

  29. Color triangle #14 Impossible C, F, G One cone stim A Two cone stim B American flag A D E Maryland flag A H D or A B D

  30. X1a Sonia • Index of refraction increases with density (as light is slowed down) so nsilurian_air > nearth_air Potential problems? Less refraction at cornea Solutions: Make cornea more rounded or make lens more rounded to decrease focal length and increase focusing

  31. X1b Sonia • If nsilurian_air= 1.1 then angle light bends will be • This is less bending than for earth air / cornea (17.2°)

  32. X2a Sarah – Cerise lake

  33. X2b Sarah • Fish would need to have Long Wavelength Sensitive (LWS) cones to match the downwelling light spectrum • Maybe then at least one offset at green wavelengths • Why no blue cones??

  34. X3 - Jessica • Let’s say in Arthur’s eye his retina is 16 mm behind lens. If his vision were normal his focal length would be

  35. X3 - Jessica • But Arthur’s eyes are only bringing objects 150cm from him into focus so his actual focus is

  36. X3 - Jessica • Then power of Arthur’s eye is • But if his eye were operating “normally” the power would be • So Arthur needs a prescription of 62.5-63.16D =-0.63 D

  37. X3b - Jessica • The prescription is -0.63 which means he needs a convex lens to decrease the focusing power of his eye to move the focusing point back where the retina is

  38. X4a - Brian • Lemurs must be able to change their focal length and bring objects at different distances to focus on their retina. Use the lens equation:

  39. X4b - Brian • Calculate focal length for objects at 20 cm

  40. X4b - Brian • Calculate focal length for objects at 20m • So focal length only has to vary from 14.8 to 15.99 mm

  41. Midterm grade distribution Average 77 A C B

  42. Quiz assignment • Problems similar #11 and #13 • Calculate images projected onto retina • Be able to calculate resolution and sensitivity • Think about units • Compare to other species so can interpret what visual system is for

More Related