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Optical systems

Optical systems. Hecht 5.7 Wednesday October 2, 2002. Field Stop. The aperture that controls the field of view by limiting the solid angle formed by chief rays. As seen from the centre of the entrance pupil (E n P), the field stop (or its image) subtends the smallest angle.

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Optical systems

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  1. Optical systems Hecht 5.7 Wednesday October 2, 2002

  2. Field Stop The aperture that controls the field of view by limiting the solid angle formed by chief rays As seen from the centre of the entrance pupil (EnP), the field stop (or its image) subtends the smallest angle. In previous example, the lens is the field stop

  3. Entrance Window (EnW) The image of the field stop in all elements preceding it Defines the lateral dimension of the object that will be viewed Example: Camera Where is the entrance window? AS FS

  4. Exit Window (ExW) The image of the field stop in all elements following it Defines the lateral dimension of the image that will be viewed Example: Camera Where is the exit window? AS FS

  5. Entrance window and Angular field of view Angular field of view in object plane angle subtended by entrance window (EnW) at centre of entrance pupil (EnP) Exit window and Angular field of view Angular field of view in image plane angle subtended by exit window (ExW) at centre of exit pupil (ExP) Field Stop That component whose entrance window(EnW) subtends the smallest angle at the centre of the entrance pupil (EnP)

  6. Field of a positive thin lens FS=EnW P Image field Eye pupil Q’ Object field α α’ F Q P’ Object point must be within cone (to left of lens) to be seen α = field of view in object space α’ = field of view in image space AS=ExP Entrance pupil (small)

  7. Field of a positive thin lens FS=EnW P Q’ F Q P’ Full cone of rays received from all points on object AS=ExP Entrance pupil (small)

  8. Stops, pupils and windows in an optical system α α’ EnP EnW AS FS ExP ExW

  9. Optical devices: Camera Multi-element lens Film: edges constitute field stop AS=Iris Diaphragm

  10. Camera Most common camera is the so-called 35 mm camera ( refers to the film size) 27 mm 34 mm Multi element lens usually has a focal length of f =50 mm

  11. Camera Object s = 1 m Image s’ ≈ 5.25 cm Object s = ∞ Image s’ = 5.0 cm Thus to focus object between s = 1 m and infinity, we only have to move the lens about 0.25 cm = 2.5mm For most cameras, this is about the limit and it is difficult to focus on objects with s < 1 m

  12. Camera AS=EnP=ExP Why?

  13. Camera: Light Gathering Power D = diameter of entrance pupil L = object distance (L>> d) D l

  14. Camera: Brightness of image Brightness of image is determined by the amount of light falling on the film. Each point on the film subtends a solid angle D D’ Irradiance at any point on film is proportional to (D/f)2 s’ ≈ f

  15. f-number of a lens Define f-number, This is a measure of the speed of the lens Small f# (big aperture) I large , t short Large f# (small aperture) I small, t long

  16. Standard settings on camera lenses Good lenses, f# = 1.2 or 1.8 (very fast) Difficult to get f/1

  17. Total exposure on Film Exposure time is varied by the shutter which has settings, 1/1000, 1/500, 1/250, 1/100, 1/50 Again in steps of factor of 2

  18. Photo imaging with a camera lens In ordinary 35 mm camera, the image is very small (i.e. reduced many times compared with the object An airplane 1000 m in the air will be imaged with a magnification, Thus a 30 m airplane will be a 2 mm speck on film (same as a 2 m woman, 50 m) Also, the lens is limited in the distance it can move relative to the film

  19. Telephoto lens L1 L2 50 mm d A larger image can be achieved with a telephoto lens Choose back focal length (bfl ≈ 50 mm) Then lenses can be interchanged (easier to design) The idea is to increase the effective focal length (and hence image distance) of the camera lens.

  20. Telephoto Lens, Example Suppose d = 9.0 cm, f2=-1.25 cm f1 = 10 cm Then for this telephoto lens Choose f = |h’| + bfl Now the principal planes are located at

  21. Telephoto Lens, Example H’ h’ = - 45 cm 9 cm 5 cm f’= s’TP = 50 cm Airplane now 1 cm long instead of 1 mm !!!!

  22. Depth of Field s2 s2’ s1 s1’ d x x so so’ If d is small enough (e.g. less than grain size of film emulsion ~ 1 µm) then the image of these points will be acceptable

  23. Depth of Field (DOF) α α d D x x so’

  24. Depth of field E.g. d = 1 µm, f# = A = 4, f = 5 cm, so = 6 m DOF = 0.114 m i.e. so = 6 ± 0. 06 m

  25. Depth of field Strongly dependent on the f# of the lens Suppose, so = 4m, f = 5 cm, d = 40 µm DOF = s2 – s1

  26. Human Eye, Relaxed 20 mm 15 mm n’ = 1.33 F H H’ F’ 3.6 mm P = 66.7 D 7.2 mm

  27. Accommodation • Refers to changes undergone by lens to enable imaging of closer objects • Power of lens must increase • There is a limit to such accommodation however and objects inside one’s “near point” cannot be imaged clearly • Near point of normal eye = 25 cm • Fully accommodated eye P = 70.7 for s = 25 cm, s’ = 2 cm

  28. Myopia: Near Sightedness Eyeball too large ( or power of lens too large)

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