Optical Instruments (Paraxial Approximation)

Optical Instruments (Paraxial Approximation)
Chapter 2
Optical Instruments (Paraxial
Approximation)
We give only a short description of the most important optical instruments. For
more details see the textbooks, e.g. Hecht [1] or Longhurst [2].
2.1 Camera
The simplest camera has a lens, shutter, an iris diaphragm and a box in which the
back wall contains the film plane. Axial movement of the lens makes focusing on
the object plane possible, see Fig. 2.1.
0
The F-number, defined as F# ¼ 2fa ; where f0 is the focal length of the lens and
pffiffiffi
2a the diameter of the iris, controls exposure time and the depth of focus. A 2
times smaller F-number gives a 2 times smaller exposure time.
The depth of focus Dz is given by
Dz ¼ Dy F#;
where Dy is the acceptable unsharpness, for instance the smallest detail discussed
above.
With a diffraction limited lens the depth of focus is given by 4kðF#Þ2 . This is
called the Rayleigh DOF and constitutes a minimum of Dz.
The distance from lens to film plane s0 is given by the imaging equation
1 1 1
¼ þ
s0 s f
where s is the object distance.
With a height h of the film plane, the field angle w is given by tan w ¼ h=s0 .
The resolution of lenses used in cameras depends on the application of the
camera.
The human eye can resolve details of 0.1 mm at a distance of 250 mm. When a
negative with a format of 24 9 36 mm is enlarged by a factor 4, its smallest
details should have a size of 0.025 mm.
C. Velzel, A Course in Lens Design, Springer Series in Optical Sciences 183,
DOI: 10.1007/978-94-017-8685-0_2,
Springer Science+Business Media Dordrecht 2014
25
26
2 Optical Instruments (Paraxial Approximation)
Fig. 2.1 A simple camera
The lens of a film camera (and also a film projection lens) should have a better
resolution over a smaller field (8 or 16 mm).
With a CCD as target in a camera, the pixel size is related to the required
resolution. With pixels of 5 lm and arguing that 4 pixels per period are necessary
for a good contrast, a resolution of 0.010 mm should be sufficient.
In our review of camera lenses in Sect. 4.1 we will see that a resolution of
30 periods per mm is standard for modern camera lenses.
2.1.1 Camera Obscura
The camera obscura, or pinhole camera, is a predecessor of our modern photocamera. It was used in the renaissance era to record the contours of landscapes and
played an essential role in the discovery of the laws of perspective, see Fig. 2.2.
The image on the backside (film or ground glass) is a projection with the
entrance diaphragm as a center. It is clear that the size of the image is given by ws0 ,
where w is the field angle. We have w ¼ hs ; where h is the object height. With s’
constant we see that a more distant object makes a smaller image. This is called
homocentric perspective.
By imaging the diaphragm with a lens to the left we can make the perspective
so that a more distant object gives a larger image. This is called hypercentric
perspective.
In Sect. 1.5 (Fig. 1.24a) we saw that a telecentric perspective can be obtained
when the diaphragm is in the focal plane of the imaging lens.
To make a sharp image the diaphragm must be made small. Diffraction at a
0
small pinhole gives a blur with the size kds , where k is the wavelength; equating
pffiffiffiffiffiffi
this with the diameter d of the pinhole we get as optimum d ¼ ks0 . With
k = 0.5 lm and s0 = 125 mm we find dopt = 0.25 mm. This is also the optimum
of resolution.
2.2 Human Eye
27
Fig. 2.2 Camera obscura
2.2 Human Eye
In optics the following components of the eye (and their properties) are important.
See Fig. 2.3.
• The cornea, the eye’s first surface, is covered with a tear layer of about 50 lm
thickness. The cornea is transparent and approximately spherical with a radius
of curvature of 7.8 mm (average).
• Between the cornea and the eye lens there is a watery fluid with index
n = 1.336.
• The eye lens is situated about 7 mm behind the cornea and has a refractive index
that varies in the axial direction between 1.386 (outside) and 1.406 (center). Its
axial thickness and curvature are controlled by the ciliary muscle, so that its
power can vary between 17 and 25 dioptres (accommodation) in persons under
the age of 45.
• The iris diaphragm is in front of the eye lens and can vary its diameter from 2 to
8 mm to adapt to the light level.
• Between the lens and the retina there is a watery gel with index 1.336.
• The retina is situated at an axial distance of about 25 mm behind the cornea; it is
curved with a radius of about 12 mm.
At the backside of the retinal tissue we find the sensitive elements, ‘‘cones’’ and
‘‘rods’’. The cones are concentrated in the macula, the cone density is greatest in
the fovea with distances between their centres from 2.5 to 5 lm. The cones are
used for seeing with high resolution (about 0.5 mrad) in daylight.
The rods are distributed over the peripheral retina; they are connected by nerve
patterns and serve for pattern recognition and movement detection. The resolution
in the periphery is considerably lower than in the macula; the sensitivity of the
rods is highest at low light levels.
In Fig. 2.4 we give a model of the eye based on work of Gullstrand [2, p. 425].
In this figure, 1 is the cornea, 2 and 3 are surfaces of the eye lens, 4 is the retina.
The iris (pupil) is just in front of the eye lens. The refractive indices behind cornea
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2 Optical Instruments (Paraxial Approximation)
Fig. 2.3 The human eye, horizontal cross-section of the eye. Nose is below, so it is the right eye
Fig. 2.4 Gullstrand’s eye
model
and lens are those of aqueous and vitreous bodies. The refractive index of the lens
in this model is an average value.
The data of this eye model are
OBJ
IMA
#
Radius
Distance
Index
0
1
2
3
4
–
7.8
10.0
-6.0
-12.0
?
7
3.6
13.4
–
1
1.336
1.413
1.336
–
2.2 Human Eye
29
The stop (iris) is in front of the first lens surface. The principal planes lie at
distances VH = 1.47 mm and VH0 = 1.50 mm from the cornea. The distance VF0
is 24 mm. The radius of curvature of the image surface is 12 mm. The power is 60
diopters. See [2, Sect. 17.2].
2.2.1 Nodal Points
The nodal points of an optical system are defined as axial points with an angular
magnification of +1, so that u0 = u. In Fig. 2.4 N and N0 are the nodal points and H
and H0 the principal points.
With n0 u0 = nu-Kh (1.19) and u0 = u we have
HN ¼ H0 N0 ¼ h= u ¼ ðn0 nÞ=K ¼ f 0 þ f
Note that HN = H0 N0 = 0 when n0 = n.
2.2.2 Exercise 4, Nodal Points of the Eye
When the eye is rotated around the second nodal point (N0 in Fig. 2.4) the image
on the retina of a far object will not move (check this with your own eye).
Calculate the position of the nodal points in Gullstrand’s eye model.
Tip: first find the position of the principal planes.
A few words about eye corrections:
•
•
•
•
cornea too steep gives nearsightedness, to be corrected by a negative glass,
cornea too shallow farsightedness, to be corrected by a positive glass,
at old age accommodation fails, one needs reading spectacles then,
when the cornea has different curvature in two perpendicular directions (with
arbitrary orientation) we have astigmatism, to be corrected by ‘‘cylinder power’’,
• the eye lens can become opaque by cataract; then a plastic ‘‘intra-ocular lens’’
can be inserted.
In this course we do not consider
• color vision
• binocular vision
• eye movements.
Cones can discriminate between red, green and blue accurately, so that many
different shades of color can be perceived. The cones are more sensitive for green
than for red and blue, with maximum sensitivity of 680 lumen/watt at 555 nm.
By comparing the images of both eyes, the brain can see depth.
The eye can be rotated in two dimensions by two sets of muscles (see Fig. 2.3)
to direct the eye axis to objects of interest.
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2 Optical Instruments (Paraxial Approximation)
2.3 Magnifier and Microscope
With the unaided eye we can resolve details of the order of 0.1 mm at a distance of
250 mm.
When we use a magnifier lens in front of the eye, with focal length f0 , we have
an angular magnification
M¼
250
f0
ð1Þ
compared to the original situation, when we put the object in the focal plane. See
Fig. 2.5.
One could expect that with a magnification M, details of the order of magnitude
of 100/M lm could be resolved. This was confirmed in the work of Anthoni van
Leeuwenhoek.
Anthoni van Leeuwenhoek (1632–1723) used lenses in the form of small glass
spheres, where
K¼
2ð n 1Þ
nr
With r = 4 mm and n = 1.6 we have f0 = 5.3 mm and M = 48. It turned out
that the observable details were of the order of 2 lm.
The compound microscope consists of an objective lens that makes a magnified
image of the object, and an ocular that projects an image at infinity, so that the eye
has a focused image on its retina (see Fig. 2.6).
The angular magnification of the compound microscope is given by
M ¼ Mob
250
;
f oc
ð2Þ
where Mob is the (linear) magnification of the objective and foc the focal length of
the ocular. With Mob = 20 and foc = 25 we have a magnification of 2009 and we
expect to see details of 0.5 lm.
This is of the order of the wavelength of light, so we should consider the effect
of diffraction.
According to the theory of diffraction the smallest detail that can be resolved by
an objective with aperture angle u is given by [see Sect. 4.7, (4.53)]
Dy ¼ 0:6
k
;
n sin u
where n is the index in object space, k is the wavelength in vacuo; n sin u is called
the numerical aperture. With k = 0.5 lm (green light) and sin u = 0.5 (quite
normal for a 209 objective), n = 1, we have Dy = 0.6 lm. With such an
objective details of 0.5 lm cannot be observed.
2.3 Magnifier and Microscope
31
Fig. 2.5 Magnifier
Fig. 2.6 Compound microscope
By making the refractive index higher than 1 in object space we can improve
the resolution. Microscopy in a transparent medium, for which water and other
fluids are being used in practice, is called immersion microscopy.
With a water immersion we have n = 1.33, and the diffraction limit becomes,
with the same values of u and k as before, equal to 0.45 lm; smaller than 0.5 lm.
2.3.1 Maximum Magnification
We calculate the maximum magnification as function of the diameter of the eye
pupil. This is important for visual microscopy (and also for observation with a
telescope).
According to Abbe’s sine rule we have, for a well corrected objective (see
Fig. 2.6)
n0 sin u0 ¼
n sin u
Mob
Abbe’s sine rule and its consequences are explained in Sect. 14-3 of Longhurst [2].
Usually n = 1, Mob 1, so that n0 sin u0 is small and
n0 sin u0 ¼
U
;
2f oc
where U is the diameter of the eye pupil, is a good approximation.
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2 Optical Instruments (Paraxial Approximation)
The ocular magnification is Moc ¼ 250
f oc , so that we find for the total
magnification
M ¼ Mob Moc ¼ 500
n sin u
:
U
ð3Þ
Because the smallest pupil diameter is U = 2, the maximum magnification is
Mmax ¼ 250 n sin u
ð4Þ
when the eye pupil is the limiting diameter.
When we make the exit pupil of the ocular narrower, we can take a somewhat
higher value for M; but beware of diffraction (a Uoc of 1 mm gives a spot of 5 lm
on the retina). Van Heel recommends to take the maximum magnification as
1,000 n sin u [3].
2.4 Telescopes
Kepler’s (astronomical) telescope consist of an objective and a positive ocular, see
Fig. 2.7.
It gives an inverted image; the inversion can be corrected by inserting a prism
arrangement (for instance that of Porro) or an inverting system between objective
and ocular. Here we will not consider this any further, but in the design of telescopes of this type it is an important issue (see Sect. 6.2).
In this arrangement, with a single lens ocular, we take the aperture stop at the
objective. Then the exit pupil is at a distance behind the ocular given by
1
1
1
¼
þ
s02 f 1 þ f 2 f 2
Because f 1 þ f 2 f 2 , we have s02 ffi f 2 ; s02 is called the eye relief. The linear
magnification, defined as the ratio between the ray heights in the exit and entrance
pupils, is ML ¼ ff 2 .
1
0
The angular magnification, defined as the ratio w
w (see Fig. 2.8) is given by
MA ¼ ff 1 :
2
(ML ¼ M1 is an example of a more general relationship, known as Lagrange’s
A
invariant, see Sect. 2.5.)
Usually we designate a hand-held telescope by a product like 8 9 30, where the
first figure denotes MA and the second the diameter of the entrance pupil.
The diameter of the exit pupil is now 30
8 ¼ 3:75 mm.
The system of Fig. 2.9 suffers from vignetting: a chief ray from the centre of the
entrance stop (at the objective) misses the ocular when its angle w with the axis
(the field angle) is larger than
2.4 Telescopes
33
Fig. 2.7 Kepler’s telescope
Fig. 2.8 Magnification
w¼
1 Uoc
2 ðf 1 þ f 2 Þ
where Uoc is the diameter of the ocular. With our 8 9 30 telescope and
f1 = 200 mm we have f1 ? f2 = 225 mm. With Uoc = 4.5 mm we could have
w = 0.01. The full field would be 2w = 0.02 or 20 mrad.
Van Heel [3] points out that with the exit pupil XP rather far from the eye lens
and partial illumination of this pupil due to vignetting, the observer has to move
his head to see objects at the edge of the field. This is shown in Fig. 2.9.
An improvement of this situation can be obtained by the use of a field lens, see
Fig. 2.10.
When we put a lens in (or near) the point F where intermediate image is, and
take the power of this lens so that the objective is imaged by it on the eye lens, we
can take the diameter of the field lens Fig. 2.16, so that the field angle U0F is what
f1
we want.
In our example UF = 4 mm would give 2w = 0.02. Because, as we will see
later, a lens near the image plane does not contribute much to the aberrations, we
can make its diameter easily larger to obtain a larger field angle.
With telescope and microscope oculars UF = 20 mm is frequently used.
Now the eye lens acts as aperture stop, with a minimum diameter of 3.75 mm,
and the field lens acts as a field stop. Over the whole field we have (nearly) no
vignetting.
34
2 Optical Instruments (Paraxial Approximation)
Fig. 2.9 Vignetting in a simple telescope
Fig. 2.10 Telescope with
field lens
In this set-up the eye of the observer should be immediately behind the eye lens;
in many situations this is not practical, an eye relief of a few centimeters is usually
required. This leads to a compromise, by taking a slightly larger Uoc and increasing
the focal length of the field lens, we obtain the final lay-out. See Fig. 2.11.
The parameters of this lay-out are given in the following table:
Objective
Field lens
Eye lens
Stop
1
2
3
4
Focal length
Distance
Diameter
300
81.8
37.5
300
37.5
25
30
20
17.2
3.75
Galilei’s telescope consists of a positive objective and a negative eye lens. With
the stop at the objective its exit pupil lies between the lenses, and therefore far
from the eye, so that vignetting cannot be avoided.
This telescope has no inversion, so that it can be used directly for terrestrial
applications; it is applied (at low magnifications) in the theatre and in sports. See
Fig. 2.12.
A modern application of the telescope, in reverted order, as a beam expander
for laser beams. Between the lenses (in F) a pinhole filter takes away stray light.
See Fig. 2.13.
2.5 Illumination
35
Fig. 2.11 Lay-out of a telescope, scale 2:1
Fig. 2.12 Galilei’s
telescope, MA = 2
Fig. 2.13 Beam expander
2.5 Illumination
Some instruments, such as the microscope and the projector, need artificial illumination. Others, like the camera and the telescope (and also the human eye) have
objects that send out light themselves, or light reflected by light sources already
present.
A simplified illumination system for a microscope contains a light source, a
condenser lens, a transparent object, the objective and the stop. See Fig. 2.14.
In the figure the object size is greatly exaggerated; the full field of a 109
microscope objective has a diameter of about 2 mm.
The condenser and the objective image the source in the aperture stop. It can be
shown that with a flat source, the light flux (Watt/m2) through the object is given by
U ¼ pB Ssin2 uc
ð5Þ
where S is the area of the source and uc is the aperture angle of the condenser.
36
2 Optical Instruments (Paraxial Approximation)
Fig. 2.14 Illumination
system
Fig. 2.15 Lagrange invariant
The constant B, called radiance, is a property of the source. With a thermal
light source it is a function of the temperature of the source.
In paraxial optics we have the invariant
S0 u02 ¼ Su2 :
ð6Þ
This equation is a special case of the optical invariant.
H ¼ nðhu huÞ;
where (h, u) and ðh; uÞ are two paraxial rays. See Sect. 1.4, (1.22).
H is equal in all planes of the system. Here we consider object and image plane.
See Fig. 2.15.
In the above object plane, when we take h = 0, we have H ¼ nhu. This is
equal to H0 ¼ n0 h0 u0 in the image plane, so that, with n0 = n, we have
Su2 = S0 u0 2. This is called Lagrange’s invariant or Helmholtz-Lagrange invariant
in the literature.
Ernst Abbe showed that for well corrected optical systems
S0 sin2 u0 ¼ S sin2 u
ð7Þ
Without loss of light (by absorption or scattering) in the system therefore the
radiance of the source is equal to that of its image B = B0 .
This is in agreement with the second law of thermodynamics (Clausius): the
temperature in the image cannot be higher than in the source itself.
The quantity S sin2u is called the throughput of the system. It is proportional to
2
H ; we have seen that the throughput determines the transport of energy through
the system.
2.5 Illumination
37
Fig. 2.16 Slide projector
Throughput is also important in the description of the channel capacity;
according to Shannon [4] this is the number of bits that can be transported per
second through the system.
The channel capacity is proportional to the number of degrees of freedom N,
defined as
S0
N¼
SD
where S0 is the area of the object and SD is the area of the object-side diffraction
spot,
SD ¼
k2 p
n2 sin2 u 4
so that we have
N ¼ S0 sin2 u
4n2
pk2
Because throughput is an invariant (*H2) also N is invariant (the ‘‘pixel
theorem’’).
k
With a field radius of 1 mm and nsinu
¼ 1 lm we have N = 4.106. This is
typical for a 209 microscope objective. With a stepper lens N can be of the order
of 5 9 1010.
Example
As an example of an illumination system we discuss a simple projection device:
the slide projector. Slide projectors are not in frequent use anymore, but their
modern successors operate according to the same principle. See Fig. 2.16 for a
scheme of a slide projector.
A light source S is imaged by a condenser lens C in the pupil of the projection
lens L. Between the source and the condenser there is a filter F that transmits only
the visible part of the source spectrum, so as to prevent heating of the transparency
(slide) T that follows the condenser. The slide is imaged on a screen (not in the
figure) that is far from L, typically a few meters (at home) or some tens of meters
(in a lecture room).
Let us assume that the projector source is a 100 W halogen lamp, that will
produce about 3,000 lumen.
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2 Optical Instruments (Paraxial Approximation)
How much of this light will reach the screen depends on the aperture of the
condenser and the transmission of the components between the source and screen.
Assuming that the lamp sends an equal amount of light in all directions (it is a
uniform source, not Lambertian) the condenser will receive a portion u2/4, where u
is the condenser aperture. With u = p/6, or 30 this factor is equal to 0.0685. Let
the filter F have a transmission coefficient of 0.8 and let us assume that we have
reflection losses of 2 % for each optical surface and there are 10 of these; the
transmission coefficient without transparency then becomes 0.64.
A blank slide does not contain useful information; let us assume that the
average transparency transmits 50 % of the light. Then the luminous flux that hits
the screen is given approximately by
U ¼ 3000 0:0685 0:64 0:5 66 lm
When we assume that the screen is a diffuse scatterer that absorbs 30 % of the
light impending on it and the spectator is at a distance of 4 m from the screen in an
otherwise dark room, his two eyes will receive a portion of u2e of the light reflected
from the screen, where ue is the aperture angle of the eye measured from the
screen. With a pupil diameter of the eye of 8 mm we have ue = 0.001.
The luminous flux received by the eye pupil is therefore given by
U ¼ 66 0:7 ð0:001Þ2 ¼ 0:46 104 lm
When we estimate that the screen image on the retina occupies an area of about
100 mm2, the illuminance on the retina will be of the order of 0.5 lm/
m2 = 0.5 lux.
This is enough for a well-adapted eye to perceive a clear screen image.
More on the subject of illumination is found in Mouroulis and Macdonald [5].
References
1.
2.
3.
4.
E. Hecht, Optics, 4th ed. (Addison-Wesley, Boston, 2002)
R.S. Longhurst, Geometrical and Physical Optics, 2nd ed. (Longman, London, 1973)
A.C.S. van Heel, Inleiding in de Optica (M. Nijhoff, Boston, 1950)
C. Shannon, W. Weaver, The Mathematical Theory of Communication (University of Illinois
Press, Urbana, 1964)
5. P. Mouroulis, J. Mcdonald, Geometrical Optics and Optical Design (Oxford University Press,
New York, 1997)
http://www.springer.com/978-94-017-8684-3
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