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 (E n P), the field stop (or its image) subtends the smallest angle. In previous example, the lens is the field stop

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

4 Exit Window (E x W) The image of the field stop in all elements following it Defines the lateral dimension of the image that will be viewed Example: Camera ASFS Where is the exit window?

5 Entrance window and Angular field of view Angular field of view in object plane angle subtended by entrance window (E n W) at centre of entrance pupil (E n P) Angular field of view in image plane angle subtended by exit window (E x W) at centre of exit pupil (E x P) Exit window and Angular field of view That component whose entrance window (E n W) subtends the smallest angle at the centre of the entrance pupil (E n P) Field Stop

6 Field of a positive thin lens Eye pupil AS=E x P P Q P’ Q’ Entrance pupil (small) Object field Image field F 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 FS=E n W α α’α’α’α’

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

8 Stops, pupils and windows in an optical system ASFS ExPExPExPExP ExWExWExWExW EnWEnWEnWEnW EnPEnPEnPEnP α’α’α’α’ α

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

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

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

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’ s’ ≈ f D Irradiance at any point on film is proportional to (D/f) 2

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

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 L1L1L1L1 L2L2L2L2 d 50 mm 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, f 2 =-1.25 cm f 1 = 10 cm Then for this telephoto lens Now the principal planes are located at Choose f = |h’| + bfl

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

22 Depth of Field s2s2s2s2 s2’s2’s2’s2’ s1s1s1s1 s1’s1’s1’s1’ sosososo so’so’so’so’ xx d 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) xx dαα D so’so’so’so’

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

25 Depth of field Strongly dependent on the f# of the lens Suppose, s o = 4m, f = 5 cm, d = 40 µm DOF = s 2 – s 1

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27 Human Eye, Relaxed 3.6 mm 7.2 mm 20 mm n’ = mm F F’ HH’ P = 66.7 D

28 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

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