Transcript Cardinal planes/points in paraxial optics

Optical systems:
Cameras and the eye
Hecht 5.7
Friday October 4, 2002
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Optical devices: Camera
Multi-element lens
AS=Iris Diaphragm
Film: edges
constitute field stop
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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
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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
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Camera
AS=EnP=ExP Why?
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Camera: Light Gathering Power
D = diameter of entrance pupil
L = object distance (L>> d)
D
l
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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’
dA D 2 D 2
d  2 

2
r
4s '
4f 2
Irradiance at any point
on film is proportional
to (D/f)2
s’ ≈ f
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f-number of a lens
Define f-number,
f
A
D
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
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I  2
A
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Standard settings on camera lenses
f# = f/D
1.2
1.8
2.8
4.0
5.6
8
11
16
22
Good lenses, f# =
(f#)2
1.5
3.2
7.8
16
31.5
64
121
256
484
1.2 or 1.8 (very fast) Difficult to get f/1
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Total exposure on Film
 watts 
E  I  2   t (exposuretim e)
 m 
J
 2
m
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
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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,

2
5 10
M 
103
  5 10
5
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
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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.
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Telephoto Lens, Example
Suppose d = 9.0 cm, f2=-1.25 cm f1 = 10 cm
Then for this telephoto lens
P  P1  P2  dP1P2
f  50cm
Choose f = |h’| + bfl
Now the principal planes are located at
f'
h'  H 2 ' H '   d
 45cm
f1 '
f
h  H1 H  d
 360cm
f2
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Telephoto Lens, Example
H’
h’ = - 45 cm
9 cm
5 cm
f’= s’TP = 50 cm
s'TP
50
mTP
5  5  10

sc
5
M
5
5
Airplane now 1 cm long
instead of 1 mm !!!!
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Depth of Field
s2’
s2
s1
s1’
d
x
so
x
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
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Depth of Field (DOF)
ds o '
x
D
so f  f  Ad 
s1 
f 2  Adso
so f  f  Ad 
s2 
f 2  Adso
α
α
x
x
d
D
so’
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Depth of field
2 Adso (so  f ) f
DOF  s2  s1 
2
f 4  A2 d 2 so
2
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
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Depth of field
Strongly dependent on the f# of the lens
Suppose, so = 4m, f = 5 cm, d = 40 µm
s1 
so f  f  Ad 
10,000

f 2  Adso
25  1.6 A
1200
1000
s2
DOF = s2 – s1
s1,s2 (cm)
800
s f  f  Ad 
10,000
s2  o 2

f  Adso
25  1.6 A
600
Depth of field (focus)
400
s1
200
0
0
2
4
6
8
f#
10
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Human Eye, Relaxed
20 mm
15 mm
n’ = 1.33
H H’
F
F’
3.6 mm
7.2 mm
P = 66.7 D
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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

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Myopia: Near Sightedness
Eyeball too large ( or power of lens too large)
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Myopia – Near Sightedness
Far point of the eye is much less than ∞, e.g. lf
Must move object closer to eye to obtain a clear image
Normal N.P.
Myopic
Myopic
F.P.
N.P.
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Myopia
e.g. lf = 2m
1 n' 1
 
lf
s'
f
How will the
near point be
affected?
0.5 + 66.7 = 67.2 D
is relaxed power of eye – too large!
To move far point to ∞, must decrease power to 66.7
Use negative lens with P = -0.5 D
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Laser Eye surgery
Radial Keratotomy – Introduce radial cuts to the
cornea of the elongated, myopic eyeball
Usually use the 10.6 µm line of a CO2 laser for
almost 100% absorption by the corneal tissue
Front view
Blurred
vision
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Laser Eye surgery
Radial Keratotomy – Introduce radial cuts to the
cornea of the elongated, myopic eyeball
Usually use the 10.6 µm line of a CO2 laser for
almost 100% absorption by the corneal tissue
Distinct
vision
Front view
Flattening
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Hyperopia – Far Sightedness
Eyeball too small – or lens of eye can’t fully accommodate
Image of close objects formed behind retina
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Hyperopia – Far Sightedness
Suppose near point = 1m
1 n'
  1  66 .7  67 .7 D
1 s'
Recall that for a near point of 25 cm, we need 70.7D
Use a positive lens with 3 D power to correct this
person’s vision (e.g. to enable them to read)
Usually means they can no longer see distant
objects - Need bifocals
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Correction lenses for myopia and hyperopia
http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/V/Vision.html
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