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Complete Gradient
Refractive Index Lens
Schematic Human Eye
Model
by
Jesús Emmanuel Gómez Correa
A dissertation submitted in partial fulfillment
of the requirements for the
PhD Degree in Optics
at the
Instituto Nacional de Astrofísica,
Óptica y Electrónica
Thesis Advisor:
Dr. Sabino Chavez Cerda
Researcher at INAOE
Mayo 2015
Tonantzintla, Puebla
c
INAOE
2015
"Empieza haciendo lo
necesario, continúa
haciendo lo posible; y de
repente te encontrarás
haciendo lo imposible".
San Francisco de Asís
Abstract
A new theoretical schematic model of the human eye is introduced. The new
model considers that the human eye crystalline is a gradient index lens composed
by two oblate half spheroids of different heights. By modifying the spherically
symmetric Luneburg model for a gradient index lens, we created a model for the
anterior and posterior half spheroids matching the corresponding geometric and
gradient index boundary conditions at the plane of fusion of the spheroids. We
tested the imaging capabilities of our model and found that it is more realistic
compared with those reported in the literature concluding that the gradient index
is dynamic and cannot be modeled by one single explicit equation.
Resumen
Un nuevo modelo teórico del ojo humano es introducido. El nuevo modelo considera que el cristalino del ojo humano es una lente con índice de refracción gradiente,
la cual está compuesta por dos esferoides achatadas con diferente radio. Al modificar el modelo de la lente simétrica de Luneburg esférica, creamos un modelo para
las esferoides anterior y posterior, las cuales se ajustan en el eje de fusión con las
condiciones de frontera correspondientes a la geometría y al índice de refracción.
Hemos probado las capacidad de formación de imágenes en el ojo completo y encontramos que es más realista en comparación con los reportados en la literatura,
en conclusión el índice del gradiente es dinámico y no puede ser modelado por
una sola ecuación explícita.
To my dad, my mom, and my son.
Para mi papá, mi mamá y mi hijo.
Acknowledgments/ Agradecimientos
• Macario Gómez: Gracias por todo el apoyo que me has dado durante
toda mi vida pero sobretodo por ser el mejor papá de todo el universo y
enseñarme a ser el hombre que soy hasta el día de hoy. Porque gracias a sus
consejos de todos los días, cariño, dedicación, caricias y sobre todo por su
amor he logrado cumplir mis metas y sueños que me he propuesto. Espero
algún día llegar hacer un papá como tu lo eres conmigo. Papá Te Amo como
no tienes idea.
• Victoria Correa: Gracias por buscar siempre que esté bien en todos los
aspectos y por ser la mejor mamá de todo el universo y llevarme de la mano
día a día sin importar que sucediera y por comprenderme durante toda mi
vida. Gracias a sus caricias, por cuidarme, por aconsejarme y por buscar mi
bienestar, pero sobre todo por su gran amor. Espero siempre darle alegrías
sin importar lo que suceda. Mamá Te Amo como no tienes idea.
• Iker Emmanuel: Por darme la alegría para poder seguir adelante y ser mi
fuente de inspiración. Te amo demasiado.
• Anel: Gracias por todo lo bueno y lo malo que me has dado a lo largo de
estos años, pero sobretodo muchas gracias por darme lo mejor que tengo en
mi vida. Te quiero mucho.
• Ian y Lars: Por ser como son conmigo, por quererme y por compartir horas
jugando Beisbol y por todas las alegrías y momentos bonitos a mi lado. Los
quiero mucho.
• A mis hermanas:
Conchi y Blanca: Por hacerme la vida más feliz y estar conmigo cuando
más las he necesitado. Por apoyarme cada día que pasa y darme ánimos
x
Chapter 0. Acknowledgments/ Agradecimientos
para seguir adelante. Por quererme y amarme tanto. Porque sin ustedes mi
vida sería complicada en todos los aspectos. Gracias por demostrarme su
amor día a día. Las Adoro y las Amo hermanitas.
• A mis hermanos:
Sergio y Arcenio: Gracias por ser los hermanos que nunca tuve, sin embargo en ustedes los he encontrado. Por considerarme su hermano aunque
sea su cuñado, por divertirnos y disfrutar de la vida con muchas sonrisas y
sobre todo por quererme tanto. Los quiero demasiado hermanitos.
• A mis sobrinos:
Cheito, Katy, Karol, Arcenito y Ale: Por ser mi alegría de todos los
días y compartir hermosos momentos a mi lado y por demostrarme cuanto
me quieren cada día que pasa. Los adoro como no tienen idea.
• Dr. Adrián Carbajal: Gracias por sus enseñanzas y por la gran calidad
que tienen sus clases, porque gracias a estas clases yo me motivé a seguir
adelante en esta área. Muchas Gracias, por todo el conocimiento que me ha
compartido a lo largo de estos años de conocernos.
• Dr. Jesús Rogel-Salazar: Muchas gracias por apoyarme en el artículo de
guías de ondas y por ayudarme en los detalles de Latex para que esta tesis
fuera posible.
• Dr. David Sánchez de la Llave: Muchas gracias por apoyarme durante
todo el doctorado, gracias por los comentarios a mi trabajo y gracias por
buscar lo mejor para mí en cuestiones académicas. Muchas gracias Doctor.
• Barbara Pierscionek: Thank you very much for your help to improve the
papers of this thesis and to believe that this work is important in the visual
optics.
• A los doctores:
De la Llave, Cornejo, Arrizón, Balderas y
xi
Malacara: Por tomarse el tiempo de revisar esta tesis, de hacerme comentarios para mejorar este trabajo y por aceptar ser mis sinodales.
• A mis amigos:
Julio Ramirez San Juan: Gracias Julio por esas largas horas que nos
pasamos hablando de box y de fútbol, gracias por tus comentarios tan alentadores que me ayudaron ha seguir adelante en gran camino que fue desde
la maestría hasta el doctorado. Muchas gracias y sabes que aquí tienes un
amigo.
Sandra: Por reírnos de cualquier cosa, por ayudarme, aconsejarme y por
ser una de mis mejores amigas de una buena parte de mi vida. Gracias
Sandrita y aquí sabes muy bien que tienes a un muy buen amigo.
José Adán: Por tu amistad dentro y fuera del INAOE, la cual, día a día
se va haciendo más Grande y por la gran recomendación que me diste para
salir a tiempo del doctorado.
Julio García y Karla Sánchez: Gracias por dejarme compartir mis ocurrencias, alegrías y tristeza con ustedes. Muchas Gracias a los dos y por
tantas salidas a comer, cenar y al cine. Los quiero mucho.
Marco Canchola: Por darme tu apoyo cada día y por compartir muy
buenos momentos conmigo y hacerme sentir que tengo un muy buen amigo
de verdad. Gracias viejo sabes que te quiero mucho.
Juan Pablo (JP): Gracias por tantas horas compartidas en el cubículo,
porque sin tu amistad hubiera sido complicado estar tantas horas en el
INAOE. Gracias mi buen JP.
Jorge Ugalde: Gracias por tu amistad y por compartir tantas horas en el
mismo cubo, un gran futuro te espera.
Juan Carlos: Gracias por las largas horas que hemos pasado riéndonos y
disfrutando de la vida. Gracias mi estimado Juan Carlos.
Sergio Mejia: Por divertirnos tanto y pasar varias horas riéndonos, por Tu
xii
Chapter 0. Acknowledgments/ Agradecimientos
apoyo y sinceridad, por la gran compañía que me has dado y por tu gran
amistad
Eicela: Por ser mi amiga y por todos los consejos que me ha dado en estos
6 años de conocernos y por facilitarme la vida en el INAOE.
Paty: Por el gran apoyo que me dio a lo largo de estos 6 años en y por
facilitarme la vida en el INAOE.
• CONACYT: Por otorgarme la beca de doctorado 235164, para que fuera
posible obtener el grado.
• INAOE: Por darme las comodidades para poder trabajar a gusto y darme
el privilegio de estudiar un doctorado en esta reconocida institución.
Special Acknowledgment
Dr. Sabino Chávez-Cerda: Le quiero agradecer por las tres tesis que me ha
dirigido en estos 8 años. Pero sobretodo le quiero agradecer por todos los consejos
que me ha dado tanto personales como académicos. Gracias por enseñarme que el
conocimiento no se mide con una regla de 30 centímetros si no que el conocimiento
se mide con la regla más grande que se tiene.
Después de estos 8 años de conocernos yo no lo considero mi asesor, yo lo
considero un muy buen amigo en mi vida. Muchas Gracias por todo, espero que
sigamos colaborando como hasta el día de hoy lo hemos hecho. Muchas Gracias
Dr. Sabino.
Contents
i
Abstract
iii
Resumen
v
vii
Acknowledgments/ Agradecimientos
ix
Special Acknowledgment
xiii
1 Preface
1
2 Gradient Index Media
5
2.1
The Ray Equation . . . . . . . . . . . . . . . . . . . . . . . . . .
5
2.2
The Linear Gradient Index Medium . . . . . . . . . . . . . . . . .
10
2.3
The Radial Cylindrical Gradient Index Medium . . . . . . . . . .
14
2.4
Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
3 The Spherical Luneburg Lens
19
3.1
The Spherical Gradient Index Medium . . . . . . . . . . . . . . .
19
3.2
The Ray Integral Equation . . . . . . . . . . . . . . . . . . . . . .
22
3.2.1
26
3.3
Generalized Snell Law for Inhomogeneous Media with
Spherical Symmetry . .q. . . . . . . . . . . . . . . . . . .
Rays in a medium with n (r) =
. . . . . . . . . . . . . .
29
3.4
Maxwell’s Fisheye . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
3.5
The Luneburg Lens . . . . . . . . . . . . . . . . . . . . . . . . . .
40
3.6
The Generalized Luneburg Lens . . . . . . . . . . . . . . . . . . .
49
3.7
The Elliptical Luneburg Lens . . . . . . . . . . . . . . . . . . . .
52
C+
1
r
xvi
Contents
.
.
.
.
53
55
57
60
4 The Human Eye As An Optical System
4.1 The Human Eye . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Refracting components of the human eye: Cornea and Lens
4.1.2 Pupil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.3 Retina . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Schematics Eye . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
62
63
67
68
70
73
5 Schematic Eye with Composite Luneburg Crystalline
5.1 Composite Modified Luneburg Lens . . . . . . . . . . . . . . . . .
5.2 Schematic Luneburg Eye . . . . . . . . . . . . . . . . . . . . . . .
5.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
76
88
94
6 Conclusions and Future Work
95
List of Figures
99
3.8
3.9
3.7.1 Linear Transformation of The Spherical
3.7.2 GRIN of The Elliptical Luneburg Lens
Ray Tracing . . . . . . . . . . . . . . . . . . .
Conclusions . . . . . . . . . . . . . . . . . . .
Bibliography
Luneburg
. . . . . .
. . . . . .
. . . . . .
Lens
. . .
. . .
. . .
.
.
.
.
103
Chapter 1
Preface
Some laws of optics have been known for the centuries, for example, the Greeks
were aware with some of the properties of light, they understood the law of reflection. However, they were unable to understand the nature of the eye because the
eye has an extraordinary complexity. They believed that light was emitted by the
eye and only produced a visual response when the emitted rays struck an object.
Many centuries passed before it was realized that light passes from the object to
the eye and not from the eye to the object [1].
Nowadays, the human eye is considered an optical system and its properties
are studied by the Visual Optics. Many studies from the Visual Optics establish
that the eye is very complex due to the fact that its refractive surfaces are not
strictly spherical and its lens has a gradient refractive index (GRIN).
At the present time, one of the challenges in Visual Optics is creating a
schematic model of the human eye. Since the GRIN profile may change from
one individual to another, the human lens has been the most challenging task in
order to have a complete and realistic schematic model.
Many generic expressions for the refractive index based in biometric data of
animal and human lens have been proposed over the years that provide a good
estimation of the actual GRIN distribution [2, 3, 4]. In some models the anterior
and posterior faces are considered to be symmetric [2] while in more recent models
a realistic asymmetry of the faces is taken into account [3, 4]. In the latter, the
GRIN is described by two different equations with respect to a plane or a curved
surface that intersects the human lens at its equator [3, 4, 5]. A drawback is that
a ray (or its derivative) travelling in the proposed GRIN distribution may undergo
a discontinuity at any of such surfaces.
2
Chapter 1. Preface
The most famous lens with a gradient refractive index is the Spherical Luneburg Lens. This lens was introduced in 1944 by Rudolf K. Luneburg in the book
"Mathematical Theory of Optics" [6]. A Luneburg lens is a GRIN lens with
spherical geometry with normalized unitary radius and stigmatic property that
focuses a sphere into a sphere [7]. As a particular example, Luneburg solved the
problem for incident rays at the anterior surface coming from infinity (infinite
sphere) focused at the opposite side surface of the spherical lens (sphere with
radius r = 1).
The importance of this lens in visual optics is due that four decades ago it
was proposed that the human lens can be studied as a Luneburg lens [8]. It is
very important to say that the Maxwell’s Fisheye Lens also has been proposed
for represent the GRIN of the human lens [9]. However, the Maxwell’s Fisheye
has the drawback that only points within or on the surface of the lens are sharply
imaged while that the lens proposed by Luneburg images sharply every parallel
bundle of rays incident on the outer surface of the lens [10].
In this work we propose a more realistic theoretical schematic eye using the
idea that the human lens can be represented as a composite asymmetric Luneburg
lens calculating a continuos GRIN function from the anterior to the posterior
conicoid faces. Our model takes into account the obliquity of the rays coming
from the cornea illuminated by a source at infinity. We have used the biometric
parameters provided by A. M. Rosen, et al. [11] but our method allows obtaining
custom made GRIN distributions given the corresponding biometric parameters.
This work is divided in six chapters: The Chapter 1 is this introduction. In the
Chapter 2 we make a analysis for the gradient mediums with planar and cylindrical
geometry. Right after, the spherical gradient index medium is explained and we
study fully the classical Luneburg lens and the Elliptical Luneburg Lens in the
Chapter 3. The Chapter 4 explains the human eye as an optical system, where
the optical components are described. Also, in this Chapter the most importants
schematic human eye models are presented and its properties and its drawbacks
are studied. In the Chapter 5 we introduce a new model of the human lens
and a new schematic human eye model using this new lens. Finally, in the last
3
Chapter we sum up the material and project future work and perspectives of our
investigations.
Chapter 2
Gradient Index Media
Gradient index media have many applications in telecommunications to model
slim antennas and in Visual Optics to model the human eye lens. Because of
this, it is necessary to start by defining: what is a gradient index medium?. The
gradient index medium is an inhomogeneous medium in which the refractive index
varies from point to point [12].
The gradient index type depends on the geometry of the medium is the gradient
index type. The three most important types of gradient index are: The linear
gradient, The radial gradient and The spherical gradient.
We know that rays propagate as straight lines in a homogeneous medium, i e,
in mediums that have a constant refractive index. For a medium with Gradient
Refractive index, the rays are represented as curves. The shape of these curves
are given by the solution of the Ray Equation.
For this reason, we will begin this section by describing Fermat’s Principle in
order to define the ray equation so that we will be able to analyze the different
types GRIN types.
2.1
The Ray Equation
If we take any two points P1 and P2 , these points can be connected with an
infinite number of curves. If each curve travels a distance ds with a velocity
v = c/n (x, y, z), where n (x, y, z) is the space dependent refractive index in a
given point and c represents the speed of light in free space. The time taken to
transverse the geometrical path ds in a medium of refractive index n (x, y, z) is
6
Chapter 2. Gradient Index Media
given by
n (x, y, z) ds
(2.1)
c
where the distance ds is a small part of a curve between P1 and P2 , as show in
the Fig. 2.1.
dt =
Now, if t represents the total time taken by the ray to transverse the path
from P1 to P2 along the curve, we have
t=
1X
ni (x, y, z) dsi
c i
(2.2)
dsi represents the ith ds along of a curve, to each dsi corresponds only one
ni (x, y, z). We can observe that this equation represents the time for a discrete
refractive index, i.e., we have the sum for all segments dsi .
Figure 2.1: Fermat’s Principle.
If we have a continuously refractive index, where the refractive index is changing by each point, the equation can be rewritten as
1
t=
c
Z
P2
n (x, y, z) ds
P1
(2.3)
2.1. The Ray Equation
7
from this equation is very easy see that the path length of ray is
P2
Z
n (x, y, z) ds.
L=
(2.4)
P1
Once obtained this result we can define Fermat’s Principle: ”The actual ray
path between two points is the one for which the optical path length is stationary
with respect to variations of the path". This represents that
Z
P2
n (x, y, z) ds = 0.
δ
(2.5)
P1
where δ variation of the integral means that it is a variation of the path of the
integral such that the endpoints P1 and P2 are fixed.
The equation (2.5) represents that the ray path is an extremum and it may be
maxima, minimum or stationary. For example, if n (x, y, z) is a constant at each
point, i e, a homogenous medium, the rays are straight lines that correspond to a
minimum value of the optical path connecting two points in the medium.
The trajectory of rays can be compared with the trajectory of the particles in
classical mechanics. The Hamilton’s principle in classical mechanics establishes
that the trajectory of a particle between times t1 and t2 is such that
Z
t2
Ldt = 0.
δ
(2.6)
t1
where L is called the Lagrangian. The difference between Eq. (2.5) and Eq. (2.6)
is that the integration is over time in the first equation and in the second one is
over the space. This difference can be eliminated if the infinitesimal arclenght ds
is rewritten as
s 2
q
2
dx
dy
2
2
2
ds = (dx) + (dy) + (dz) = dz
+
+1
(2.7)
dz
dz
8
Chapter 2. Gradient Index Media
if ẋ = dx/dz and ẏ = dy/dz, then
ds = dz
p
ẋ2 + ẏ 2 + 1
(2.8)
where dz is equivalent to dt in the Eq. (2.6) and in this case, Eq. (2.5)
Z
P2
n (x, y, z)
δ
p
1 + ẋ2 + ẏ 2 dz = 0.
(2.9)
P1
From this equation we can define an optical Lagrangian as
L (x, y, ẋ, ẏ, z) = n (x, y, z)
p
1 + ẋ2 + ẏ 2 .
(2.10)
As in classical mechanics, the solution of this problem must satisfy the EulerLagrange equations, in this case these equations are given by
and
d
dz
d
dz
∂L
∂ ẋ
∂L
∂ ẏ
=
∂L
∂x
(2.11)
=
∂L
.
∂y
(2.12)
If we substitute Eq. (2.10) into Eq (2.11), we obtain
"
d
n (x, y, z) ẋ
p
dz
1 + ẋ2 + ẏ 2
#
=
p
∂n (x, y, z)
1 + ẋ2 + ẏ 2
∂x
(2.13)
from Eq. (2.7) is easy to see that Eq. (2.13) can be reduced to
because
d
dx
∂n (x, y, z)
n (x, y, z)
=
ds
ds
∂x
(2.14)
d
1
d
=p
.
ds
1 + ẋ2 + ẏ 2 dz
(2.15)
2.1. The Ray Equation
9
Similarly for y and z components, we have
and
d
dy
∂n (x, y, z)
n (x, y, z)
=
ds
ds
∂y
(2.16)
dz
∂n (x, y, z)
d
n (x, y, z)
=
ds
ds
∂z
(2.17)
If we define ~r as a position vector, we can obtain a vector equation which contains
equations (2.14), (2.16) and (2.17), i e,
d
d~r
n (x, y, z)
= ∇n (x, y, z)
ds
ds
(2.18)
this equation is known as the Ray Equation. It is very important to say that Eq.
(2.18) can also be obtained from the Eikonal equation.
The Eikonal equation is important in geometrical optics because it describes
the phasefront of a wave. In order to obtain the Ray Equation from the Eikonal
equation, we start with the latter one as [14]
[∇S (x, y, z)]2 = n2 (x, y, z)
(2.19)
where S (x, y, z) = Constant and S (x, y, z) can be interpreted as the function
describing the phasefront of the wave. Now, if we defined a unit vector normal to
the phase fronts and tangent to the light ray ŝ, i.e., along the ray, then
∇S (x, y, z) = n (x, y, z) ŝ
(2.20)
d
d
d~r
[∇S (x, y, z)] =
n (x, y, z)
ds
ds
ds
(2.21)
where ŝ = d~r/ds, hence
10
Chapter 2. Gradient Index Media
since as d/ds = ŝ · ∇, this equation can be represented as
d
d~r
n (x, y, z)
= (ŝ · ∇) ∇S
ds
ds
(2.22)
We want to know the direction of the rays, hence
∇ [∇S (x, y, z)]2 = ∇n2 (x, y, z)
(2.23)
∇ [∇S (x, y, z)]2 = 2n (x, y, z) (ŝ · ∇) ∇S
(2.24)
∇n2 (x, y, z) = 2n (x, y, z) ∇n (x, y, z)
(2.25)
(ŝ · ∇) ∇S = ∇n (x, y, z)
(2.26)
d~r
d
n (x, y, z)
= ∇n (x, y, z)
ds
ds
(2.27)
using
and
we obtain
from the Eq. (2.22)
we end up with the Ray equation, which was obtained from the Eikonal equation.
Using the Eq. (2.27), we will study the axial and the radial gradients index
medium in sections 2.2 and 2.43, respectively.
2.2
The Linear Gradient Index Medium
In the linear gradient index medium, the refractive index varies in a continuous
way along the optical axis of the inhomogeneous medium. The isoindicial surfaces
(surfaces of constant index) are planes that are parallel to the optical axis, as is
show in Fig. 2.2.
2.2. The Linear Gradient Index Medium
11
Figure 2.2: The linear gradient index medium.
A linear refractive index in the x-axis is given by
(
2
n (x) =
n20 − αx x > 0
n20 x < 0
(2.28)
and in the y-axis by
(
2
n (y) =
n20 − αy y > 0
n20 y < 0
(2.29)
where α is any small number and n0 is the refractive index in x = 0 or y = 0.
These refractive indixes are represented in the Fig. 2.2.
To learn how rays propagate within a gradient index medium is necessary to
solve the ray equation, which is
d~r
d
n (x, y, z)
= ∇n (x, y, z)
ds
ds
(2.30)
this equation can be rewritten as a set of three equations, shown in equation
12
Chapter 2. Gradient Index Media
(2.31).
∂n(x,y,z)
d
n (x, y, z) dx
= ∂x
ds
ds
dy
d
n (x, y, z) ds = ∂n(x,y,z)
ds
∂y
∂n(x,y,z)
d
dz
n
(x,
y,
z)
=
.
ds
ds
∂z
(2.31)
If the refractive index does not depend on z, we have
d
dz
∂n (x, y)
n (x, y)
=
= 0.
ds
ds
∂z
(2.32)
implying that
dz
=β
(2.33)
ds
where β is an invariant of the ray path. From Fig. 2.3 we can observe that
n (x, y)
Figure 2.3: An arc lenght along the ray path.
dz
= cos θ (x, y)
ds
(2.34)
⇒ β (x, y) = n (x, y) cos θ (x, y) .
(2.35)
where θ (x, y) is the angle make the optical axis (z-axis) and the tangent line to
the curve at one point (x, y). Using Eqs. (2.8) and (2.35), we can rewrite (2.31)
2.2. The Linear Gradient Index Medium
as
13
∂n2 (x, y)
d2 x
1
= 2
dz 2
2β (x, y)
∂x
(2.36)
d2 y
∂n2 (x, y)
1
=
dz 2
2β 2 (x, y)
∂y
(2.37)
dβ (x, y)
1
∂n2 (x, y)
=
= 0.
dz
2β (x, y)
∂z
(2.38)
If the linear refractive index is given by Eq. (2.28), then Eq. (2.36), for x > 0,
becomes
α
d2 x
=− 2
(2.39)
2
dz
2β
and the general solution is
x (z) = −
α 2
z +C
4β 2
(2.40)
where C = Constant, β = n0 cos θ0 and θ0 = θ (0, 0). In this case, θ0 is the
incident angle of the ray at the point x = 0 and y = 0.
Thus for a ray that passes through the points x = 0 and z = 0, the constant
C, takes the value of
C = z tan θ0
(2.41)
therefore
x (z) = −
α 2
z + z tan θ0 .
4β 2
(2.42)
We can observe in the Fig. 2.4 that if x < 0 the trajectory of the ray is a
straight line and if x > 0 the trajectory is a parabola.
The solution for a linear refractive index in the y-axis is the same that for the
x-axis, for this reason, it is not necessary solve it.
14
Chapter 2. Gradient Index Media
Figure 2.4: Solution for the linear gradient index medium.
2.3
The
Radial
Cylindrical
Gradient
Index
Medium
In the radial cylindrical gradient index medium, the index profile has a maximum
at the cylinder axis and decreases continuously from the axis to the periphery along
the transverse direction in such a way that the isoindicial surfaces are concentric
cylinders about the optical axis [12], as is show in Fig. 2.5.
Figure 2.5: The radial gradient index medium.
The radial cylindrical gradient index is represented by
2.3. The Radial Cylindrical Gradient Index Medium
 2
2
2
r
x
+
y

2
2
 n1 1 − 2∆
= n1 1 − 2∆
0<r<a
a2
a2
n2 (r) =


n1 [1 − 2∆]
r>a
n22 =
15
(2.43)
where ∆ is any small number, n1 is the refractive index in r = 0, n2 is a constant,
a is the maximum radius of the refractive gradient index (r = a). This equation
is known as parabolic refractive index and it has no dependency on z.
To find the solution of the rays in radial cylindrical gradient index medium,
we substitute Eq. (2.43) into Eqs. (2.36) and (2.37), then
d2 x
dz 2
d2 y
dz 2
where Γ =
√
n1 2∆
aβ
+ Γ2 x = 0
+ Γ2 y = 0
(2.44)
and the solutions are given by
x = A sin Γz + B cos Γz
y = C sin Γz + D cos Γz.
(2.45)
The easiest solution is when we only work in a sagittal plane, therefore, it is
necessary to impose initial launching conditions [13], like these
x (z = 0) = 0
y (z = 0) = 0
dx
|
=0
dz z=0
dy
|
= 0.
dz z=0
(2.46)
These launching conditions make that the meridional rays are confined in the
sagittal plane; we may mention that the meridional rays are defined such that they
are confined in a plane and intersect the sagittal plane. Using these conditions
16
Chapter 2. Gradient Index Media
the solution is of the form
x=
aβ
√
n1 2∆
tan θ0 sin
h
√
2∆
z
a cos θ0
i
=
a√
sin θ0
2∆
sin
h
√
2∆
z
a cos θ0
i
(2.47)
y=0
where θ0 is the incident angle of the ray at the point x = 0 and y = 0. The
solution for two different values of β are shown in the Fig. 2.6.
Figure 2.6: Solution for the Sagittal plane.
The difference between the gradient in the sagittal plane of the radial cylindrical gradient index and the linear gradient in the same plane is that in the first
the gradient is given in an interval of [−a, a] while in the second one the gradient
is in an interval of [0, a], where a is any number, for this reason the solutions are
different.
The most complete solution is when we impose the initial launching conditions
as
x (z = 0) = a0
y (z = 0) = 0
(2.48)
dx
|
=
0
z=0
dz
dy
|
= tan θ0
dz z=0
where a0 is the distance on the x axis, which was launched the ray in the y−z plane
and θ0 is the angle that the ray makes with the z axis. With these conditions, one
obtains
x = a0 cos Γz
(2.49)
y = a0 sin Γz
2.4. Conclusions
17
Equations (2.49) are represented in the Fig. 2.7.
Figure 2.7: Solution for the radial gradient index medium.
These solutions describe which are know as skew rays which do not remain
confined to a plane and represent the solution for the radial cylindrical gradient
index medium, as show in Fig. 2.7.
2.4
Conclusions
The importance of this Chapter is that we have reviewed the theory of propagation
of the rays in gradient media and with this we have presented the ray equation
which we will be used from now on.
In the next Chapter we will present the extension to a spherical gradient index.
This study must be carried out carefully, due to the fact that the crystalline is a
inhomogenous lens with a gradient index that can be represented as a spherical
gradient index.
Chapter 3
The Spherical Luneburg Lens
This chapter provides a description of light propagation through the spherical
gradient medium. We will analyse the spherical Luneburg lens and Maxwell’s
Fish Eye who are the most famous examples of lenses with spherical gradient
index. The difference between the spherical Luneburg lens and the Maxwell’s
Fisheye is the representation of the mathematical approach of the refractive index
and that the Maxwell fish eye lens has a drawback; only points within or on the
surface of the lens are sharply imaged. The lens proposed by Luneburg images
sharply every parallel bundle of rays incident on the outer surface of the lens [10].
This analysis will be done with the integral ray equation because we want to
know the form of the ray in an interval, i.e., from point P0 to point P1 .
The importance of this analysis is due to the fact that both lenses have been
proposed to represent the refractive gradient index of the crystalline. It is very
important to mention that it has been proposed, but for the case of the Luneburg
lens does not exist a complete analysis.
For this reason, in this chapter we will focus our attention to study the spherical Luneburg lens.
3.1
The Spherical Gradient Index Medium
Lets suppose that we have a refractive index that is constant on concentric spheres,
i.e., each sphere has a thickness of dr and in this thickness, the refractive index
is constant. The smaller sphere has the highest refractive index and it decreases
from the inner sphere to a sphere of radius a, then we can say that we have a
spherical gradient index medium, see Fig. 3.1.
20
Chapter 3. The Spherical Luneburg Lens
Figure 3.1: The spherical gradient.
The spherical gradient index can be represented by different mathematical
expressions. An interesting case is when we define the spherical gradient index as

C + 1 0 < r < a
2
r
n (r) =

n22
r>a
(3.1)
p
where r = x2 + y 2 + z 2 , C is a constant, n2 = C − a1 , where a is the maximum
radius of the refractive gradient index.
This medium is interesting because the light rays in this medium are identical
with the paths of particles which move in a Coulomb field of potencial φ = −1/2r,
and with the energy C/2.
The variation of the spherical gradient index given by Eq. (3.1) is represented
in the Fig. 3.2 for different values of C. Notice that the spherical gradient index
has a singularity at r = 0, for this reason, the radius is r > 0 in Eq. (3.1). The
singularity at r = 0 is not a problem for the ray tracing in this medium, because
we will find a ray integral equation that depend on a new function ρ (r). This
function will be define in the section 3.3.
At this point, it is important to answer: which is the the ray integral equation?,
so we can solve any gradient index with spherical symmetry. In the next section
we will answer this question and later we will solve the refractive gradient index
3.1. The Spherical Gradient Index Medium
Figure 3.2: The Index Variation.
21
22
Chapter 3. The Spherical Luneburg Lens
given by the Eq. (3.1).
3.2
The Ray Integral Equation
We know from the literature about media with radial symmetry that the rays
are confined to a single plane [13]. In this work we are interested in knowing the
solutions the a x − y plane, for this reason it is possible to reduce this problem of
three variables into a two variables problem.
Now, if we consider a continuous medium with a refractive index that has
radial symmetry n (r), as shown in the Fig. 3.3, we can easily know the ray
integral equation.
Figure 3.3: Medium with radial symmetry
The ray integral equation is obtained from the Eikonal equation, then we can
rewrite the Eq. (2.19) as
∂ψ
∂x
2
+
∂ψ
∂y
2
= n2 (x, y)
(3.2)
as we are working in a medium with radial symmetry, we make a change of
3.2. The Ray Integral Equation
23
variables of the form
x = r cos θ
y = r sin θ
(3.3)
and
p
r=
we obtained
∂ψ
∂r
2
1
+ 2
r
x2 + y 2
∂ψ
∂θ
2
(3.4)
= n2 (r)
(3.5)
This equation can be solved with the method of Separation of Variables, then the
Eq. (3.5) can be rewritten as
∂ψ
=r
∂θ
s
n2 (r)
−
∂ψ
∂r
2
(3.6)
we can observe that the left side of the equation does not depend on r, then
dψ(θ)
dθ
=K
dψ(θ) = Kdθ
R
R
dψ(θ) = Kdθ
R
R
dψ(θ) = K dθ
ψ(θ) = Kθ
(3.7)
24
Chapter 3. The Spherical Luneburg Lens
and the right of the equation does not depend on θ, we have
r
n2 (r) −
K=r
2
K =r
2
2
n (r) −
dψ(r)
dr
dψ(r)
dr
2
2 2
− n2 (r) = − dψ(r)
dr
q
2
dψ(r)
= n2 (r) − Kr2
dr
q
2
dψ(r) = n2 (r) − Kr2 dr
q
R
Rr
2
dψ(r) = r0 n2 (r) − Kr2 dr
Rr q
2
ψ(r) = r0 n2 (r) − Kr2 dr
K2
r2
(3.8)
if the solution is given by
therefore
ψ = ψ(θ) ± ψ(r)
(3.9)
Z rr
K2
n2 (r) − 2 dr
ψ = Kθ ±
r
r0
(3.10)
where K is for the moment an arbitrary constant.
We know that the Eikonal equation describes the phase front of a wave, and
the rays are lines that are normal to these wavefronts, showing the direction of
energy flow at one particular point, for this reason it is necessary to apply the
Jacobi Theorem to obtain the integral ray equation. This theorem is represented
by
∂ψ
=α
(3.11)
∂K
3.2. The Ray Integral Equation
25
applying the Jacobi theorem, we have
∂ψ
∂K
Rr q
K2
2
=
Kθ ± r0 n (r) − r2 dr
Rr q
2
∂ψ
∂
∂
= ∂K [Kθ] ± ∂K r0 n2 (r) − Kr2 dr
∂K
Rr ∂ q
2
∂ψ
∂K
=
θ
±
n2 (r) − Kr2 dr
∂K
∂K
∂K
r
0
1/2 h
Rr ∂ 2
K2
∂
= θ ± r0 ∂K n (r) − r2
n2 (r) −
∂K
−1/2 Rr 2
∂ψ
K2
2K
n
(r)
−
=
θ
±
dr
∂K
r2
r2
r0
2K
R
r
∂ψ
= θ ± r0 q 2 r2 K 2 dr
∂K
n (r)− 2
r
Rr
∂ψ
q 2K
=
θ
±
dr
∂K
r0 r2 n2 (r)− K 2
r2
Rr
∂ψ
= θ ± 2K r0 r2 √ 2dr 2 2
∂K
n (r)r −K
Rr r
∂ψ
= θ ± 2K r √ 2 dr 2 2
∂K
∂ψ
∂K
∂
∂K
0
α = θ ± 2K
−
θ
2
= ±K
i
dr
(3.12)
n (r)r −K
Rr
√ dr
r0 r n2 (r)r2 −K 2
Rr
√ dr
r0 r n2 (r)r2 −K 2
α − θ = ±2K
α
2
r
K2
r2
Rr
r0 r
√
(3.13)
dr
n2 (r)r2 −K 2
we defined
α
=θ
2
θ
= θ0
2
(3.14)
the rays of light in the x − y plane are obtained by the following equation
Z
r
θ − θ0 = ±K
r0
dr
p
r n2 (r)r2 − K 2
(3.15)
The latter equation is known as the integral ray equation. The integration constants θ0 and r0 are the initial coordinates of the ray on the spherical refractive
gradient index.
In this analysis, the constant K is very important because it represents the
ray direction at the point (r0 , θ0 ), then we need to know the value of K. In the
26
Chapter 3. The Spherical Luneburg Lens
next section, we will found this value.
3.2.1
Generalized Snell Law for Inhomogeneous Media with
Spherical Symmetry
From the literature [14], it is known that the path of rays in a medium with
gradient index n = n (r) are planar curves in a plane containing the origin. We
−
can consider a position vector →
r along the path of the curve that represents a light
−
ray inside the spherical refractive gradient index, and a unit vector →
s tangent to
the ray, as shown in the Fig. 3.4.
Figure 3.4: The path of rays in a medium with gradient index
−
−
If we consider the vector variation →
r × [n (r) →
s ] along the ray (curve) and we
derive with respect to s, we obtain
−
d →
d→
r
d
−
−
−
−
[−
r × [n (r) →
s ]] =
× [n (r) →
s]+→
r ×
[n (r) →
s]
ds
ds
ds
(3.16)
−
d→
r
−
=→
s
ds
(3.17)
but
and from Eq. (2.27)
d
ds
→
→
−−
d−
r
n
= 5→
n
ds
(3.18)
3.2. The Ray Integral Equation
27
therefore
→
− −−→
d →
−
−
−
−
[−
r × [n (r) →
s ]] = →
s × [n (r) →
s]+→
r × 5 n (r)
ds
it is very easy to see that the first term is 0 and the equation reduces to
→
− −−→
d →
−
−
[−
r × [n (r) →
s ]] = →
r × 5 n (r)
ds
(3.19)
(3.20)
as n (r) is a function that only depend on r
−
→
− −−→ dn (r) →
r
5 n (r) =
dr r
(3.21)
from this equation it is clear to see
d →
−
[−
r × [n (r) →
s ]] = 0
ds
(3.22)
→
−
−
r × [n (r) →
s ] = cte
(3.23)
this leads us to
a vector product that can be expressed as
→
−
−
−
−
r × [n (r) →
s ] =k →
r kk n (r) →
s k sin ϕ
(3.24)
−
−
−
where ϕ is the angle between the vectors →
r and →
s , and as the vector →
s is a unit
vector, we have
−
k→
r k= r
(3.25)
−
s k= n (r)
k n (r) →
if we substitute these equations into Eq. (3.25)
→
−
−
r × [n (r) →
s ] = rn (r) sin ϕ
(3.26)
rn (r) sin ϕ = cte
(3.27)
hence
28
Chapter 3. The Spherical Luneburg Lens
and this constant is K, i.e.,
K = rn (r) sin ϕ
(3.28)
The Eq. (3.28) is known as the generalized Snell law for inhomogeneous media
whit spherical symmetry. This demonstrates that the path of rays in a medium
with gradient index n = n (r) are planar curves because K is always a constant
with positive or negative sign and there is not sign change along the ray.
Figure 3.5: Rays on Horizontal axis
It is easy to calculate the variation range of K, due the fact that the maximun
point of incidence can be obtained when r = 1, n = 1 and ϕi = π2 . From the
generalized Snell’s law, we have that K = 1 and the minimum point is when we
are on the axis of propagation, which leads us to have ϕi = 0 and for consequence
K = 0, then the variation range of K is
0 ≤ K ≤ 1.
(3.29)
If we considered an angle α0 , which is measured relative to the normal and it is
on the negative side on the same axis. We observe that the incident angle relative
to the normal is given by θ − π, i.e., θ0 = π, ϕ = α0 , and K = n0 r0 sin α0 , by
3.3. Rays in a medium with n (r) =
q
C+
1
r
29
substituting these values in Eq. (3.15), we obtain
r
Z
θ − π = ±n0 r0 sin α0
r0
dr
p
r n2 (r)r2 − n20 r02 sin2 α0
(3.30)
dr
p
r n2 (r)r2 − n20 r02 sin2 α0
(3.31)
dr
p
r n2 (r)r2 − n20 r02 sin2 α0
(3.32)
for α0 < π2 , we have
r
Z
θ − π = +n0 r0 sin α0
r0
and for α0 > π2 , we have
r
Z
θ − π = −n0 r0 sin α0
r0
If we considered that α0 is given from 0 to π2 , we obtain that
0 ≤ K ≤ n0 r0
3.3
(3.33)
q
Rays in a medium with n (r) = C + 1r
We have that the integral ray equation is
Z
r
θ − θ0 = ±K
r0
dr
p
r n2 (r)r2 − K 2
(3.34)
we can define a function ρ (r), that has the form
ρ(r) = n(r)r.
(3.35)
1
r
(3.36)
if
n2 (r) = C +
then
ρ2 (r) = Cr2 + r
(3.37)
30
Chapter 3. The Spherical Luneburg Lens
by substituting Eq. (3.37) into Eq. (3.34) and we obtain
Z
r
θ − π = −K
r0
dr
√
2
r Cr + r − K 2
(3.38)
in this case θ0 = π, because we are considering that origin of ray is on the horizontal axis. We can find the solution of the integral when we rewrite the lower
term as
v"
r
#
u
K
1 2
u
√
−
1
2K
t 1− r
(3.39)
Cr2 + r − K 2 = r C +
4K 2
C + 4K1 2
and by making a change of variable
K
r
z=q
r2
1
2K
−
C+
q
C+
dr = −
z
θ−π =−
z0
1
4K 2
K
we observe that
Z
(3.40)
1
4K 2
Kr2
q
C+
dz
1
4K 2
dz
K
q
√
r2 C + 4K1 2 1 − z 2
therefore
Z
z
θ−π =−
z0
(3.41)
dz
√
1 − z2
(3.42)
(3.43)
the solution is very easy if we assume
then
Z
dz
√
=
1 − z2
Z
z = sin u
(3.44)
dz = cos udu
(3.45)
cos udu
p
=
1 − sin2 u
Z
du = u = arcsin z
(3.46)
3.3. Rays in a medium with n (r) =
hence
Z
z
√
θ−π =−
z0
Now, if arcsin z0 = β −
π
2
q
C+
1
r
dz
= arcsin z0 − arcsin z
1 − z2
31
(3.47)
is constant, we have
z = sin
π
2
− [θ − β] = cos (θ − β)
(3.48)
we substitute Eq. (3.48) into Eq. (3.40) and obtain
r=
1+
√
2K 2
4CK 2 + 1 cos (θ − β)
(3.49)
the latter equation determines the type of curves and these curves can be easily
seen if us assume that
β=0
x = r cos (θ)
(3.50)
y = r sin (θ)
p
r = x2 + y 2
with equations (3.50) we find from the Eq. (3.49)
2
C
1 √
1 + 4CK 2 − 2 y 2 = 1
4C x −
2C
K
(3.51)
h
i
√
y 2 = K 2 4Cx2 − 4x 1 + 4K 2 C + 4K 2 .
(3.52)
2
or
Equation 3.51 tell us that the curves are conic sections and that a given C all
these conics have the same principal axes A = 1/2C. From this equation we can
observe that the eccentricity is [6]
1 √
e=
1 + 4K 2 C =
2C
r
1
K2
+
.
4C 2
C
The type of these conics is determined by the value of C, i.e.:
(3.53)
32
Chapter 3. The Spherical Luneburg Lens
If C > 0: Equation 3.51 represents hyperbolas with the same principal axis
A = 12 C, with the point x = y = 0 as common focal point, as shown in Fig. 3.6.
Figure 3.6: The rays are hyperbolas when C > 0.
If C = 0: Equation 3.51 represents parabolas with x = y = 0 as common focal
point, as shown in Fig. 3.7.
If C < 0: Equation 3.51 represents ellipses with the same principal axis A =
1
and x = y = 0 as common focal point, as shown in Fig. 3.8. In this case, the
2|C|
1
.
light rays cannot penetrate into the region r > − 2C
In this section our aim was to find the shape of the rays, but it is important to
see that the refractive index given by Eq. (3.36) can be transformed into another
vital refractive index. By recalling the Eikonal equation
ψx2 + ψy2 + ψz2 = n2 (r)
(3.54)
and substituting the Eq. (3.36) in cartesian coordinates into the Eq. (3.54), we
obtain
1
ψx2 + ψy2 + ψz2 = C + p
(3.55)
x2 + y 2 + z 2
3.3. Rays in a medium with n (r) =
q
C+
1
r
Figure 3.7: The rays are parabolas when C = 0.
Figure 3.8: The rays are ellipses when C < 0.
33
34
Chapter 3. The Spherical Luneburg Lens
using Legendre transformations
ψ = ψ (x, y, z)
ω = ω (ξ, η, ς)
ψ + ω = xξ + yη + zς
(3.56)
ξ = ψx
η = ψy
ς = ψz
x = ωξ
y = ωη
z = ως
(3.57)
where
it is very easy to see that the Eq. (3.55) is transformed into
1
ξ 2 + η2 + ς 2 = C + q
ωξ2 + ωη2 + ως2
(3.58)
and this equation can be rewritten as
ωξ2
+
ωη2
+
ως2
=
1
2
−C + ξ + η 2 + ς 2
2
(3.59)
if C = −1, so this equation is reduced to the Eikonal equation with a refractive
index given by
1
n (ξ, η, ς) =
(3.60)
2
1 + ξ + η2 + ς 2
or
1
n (r) =
.
(3.61)
1 + r2
This optical medium is know as Maxwell’s Fisheye. A medium with this
refractive index will be studied in next section.
3.4. Maxwell’s Fisheye
3.4
35
Maxwell’s Fisheye
The Maxwell’s Fisheye has been proposed to represent the refractive gradient
index of the crystalline [9], for this reason is important to study it. The gradient
refractive index of Maxwell’s fisheye is generally represented as
n=
(b2
a
+ r2 )
(3.62)
if a and b are constants, and a = b = 1, we have
n=
1
.
(1 + r2 )
(3.63)
We want to know the form of the rays, in order to do so we use the integral
ray equation
Z
dr
√
(3.64)
θ−π =K
2
r n r2 − K 2
substituting the gradient refractive index into Eq. (3.64) and by doing some
algebra, we obtain
Z
K (1 + r2 )
q
θ−π =
dr
(3.65)
r r2 − K 2 (1 + r2 )2
this integral can be solve by using
f (r) = arcsin
where
r2 − 1
K
√
r
1 − 4K 2
(3.66)
K (1 + r2 )
df (r) = q
dr
2
2
2
2
r r − K (r + 1)
(3.67)
we can observe that
K (1 + r2 )
Z
θ−π =
r
q
r2
−
K2
(1 +
Z
dr =
r2 )2
df (r)
(3.68)
36
Chapter 3. The Spherical Luneburg Lens
and the solution is
θ − π = arcsin
K
r2 − 1
√
r
1 − 4K 2
+C
(3.69)
where C is the integration constant. Equation (3.69) is rewritten as
√
1 − 4K 2
r2 − r
sin (θ − π − C) − 1 = 0
K
C can be defined as
C=β−
3π
2
substituting the value of C and using the trigonometric identities sin
cos θ and, cos θ = cos (−θ) into Eq. (3.70), we have
√
2
r −r
1 − 4K 2
cos (θ − β) − 1 = 0
K
if β = 0, then
1 − 4K 2
cos θ − 1 = 0
K
if we introduce cartesian coordinates into this equation, we have
√
x−
1
,
2K
π
2
(3.71)
−θ =
(3.72)
√
r2 − r
and if R =
(3.70)
then
1 − 4K 2
2K
2
+ y2 =
1
4K 2
2
√
2
x − R − 1 + y 2 = R2 .
(3.73)
(3.74)
(3.75)
The form of the rays is given by Eq. (3.75). We can observe that this equation
√
represents a set of displaced circles on the x axis by a factor of R2 − 1 and all
the circles intersect at y = ±1. We can say that all rays have a circular path, as
shown in the Fig 3.9.
So far, we have not said where the rays were originated, we only know that the
rays have a circular path. Now, we will consider that all rays are originated in a
3.4. Maxwell’s Fisheye
37
Figure 3.9: Maxwell Solution
point P0 (x, y) on the x-axis, i.e., x = x0 and y = 0. From Eq. (3.72) in cartesian
coordinates, we have
where R =
1
2K
x−
√
R2 − 1 cos β
2
2
√
+ y − R2 − 1 sin β = R2
(3.76)
and this equation is a generalization of Eq. (3.75).
If y = 0 in Eq. (3.76) represents the interception points on the x axis, and
this leads us to the equation
√
x2 − 2x R2 − 1cosβ − 1 = 0.
It can be solved if we use
x0,1 =
b±
(3.77)
√
b2 − 4ac
2a
(3.78)
p
(R2 − 1) cos2 +1
(3.79)
√
if a = 1, b = −2 R2 − 1 cos β and c = −1, then
x0,1 =
√
R2 − 1 cos β ±
38
Chapter 3. The Spherical Luneburg Lens
Figure 3.10: Maxwell Solution
and the solutions are
p
√
x0 = R2 − 1 cos β + (R2 − 1) cos2 +1
p
√
x1 = R2 − 1 cos β − (R2 − 1) cos2 +1
(3.80)
the relationship between x0 and x1 is given by
x0 x1 = −1
(3.81)
we can observe that this relationship is independent of the parameters R and β
and this tells us that all rays that originate at x0 are intersected at a point x1 ,
where
1
x1 = − .
(3.82)
x0
the solution is sketched in Fig. 3.10.
This result can be generalized if we have a source that is outside the x axis,
i.e., if all rays are originated in a point P0 = (x0 , y0 ), there is an intersected point
3.4. Maxwell’s Fisheye
39
Figure 3.11: Maxwell Solution Off-Axis
P1 = (x1 , y1 ), where
x1 = − xr20
0
y1 = − yr20
(3.83)
0
the solution is sketched in Fig. 3.11.
From the latter solution we can observe that all the rays are focusing in one
only point, i.e., it is an optical instrument free of the aberrations. For this reason,
the Maxwell fish eye is considered a perfect optical Instrument in the x − y plane,
the image produced is inverted and it has a magnification of M = − rr10 .
These solutions are interesting, because the curves that represent the rays are
inside the medium, i.e., the Maxwell’s Fisheye is not a lens, for this reason, it
cannot be proposed to represent the refractive gradient index of the human lens.
The well known perfect optical Instrument is the Luneburg lens which is our
main interest in this work and it will study in the next section.
40
Chapter 3. The Spherical Luneburg Lens
3.5
The Luneburg Lens
The Luneburg lens is the answer to the problem posed by R. K. Luneburg in
1944, this problem is stated as follows: If we have a refractive index with radial
symmetry, where the maximum radius is 1 and it is submerged in a refractive
index n = 1 (air), What should be the refractive index of the medium where all
incident rays focus at a single point?.
In order to solve this problem, we begin by considering a sphere with maximum
radius equal to 1 and where the refractive index is
n = n (r) , r < 1
n = 1,
r≥1
n (1) = 1, r = 1.
(3.84)
At this point, we do not know as is the refractive index n (r), but we are
considering some boundary conditions, i.e., the maximum radius (rm ) is 1 and also
we have n (rm ) = 1. The first boundary condition (rm = 1) does not generate any
loss of generality of this problem solution, because r is converted into a normalized
variable. The second boundary condition (n (rm ) = 1) does not mean that this
medium is air or vacuum, this means that the refractive index of the element
considered is normalized with respect to the refractive index where the sphere is
submerged [15].
Figure 3.12 has the parameters that we use for solving the Luneburg problem.
From this figure we can observe that the incident ray is originated at the point
r00 and the point r10 is the point where all rays are focused after passing through
the sphere. The point P0 indicates the point where the incident ray enters at the
sphere and the point P1 is the point at which the ray exits the same sphere. The
radius r∗ = r (θ) is the minimum radius at the angle θ∗ .
The rays outside the sphere propagates in straight lines, i.e., the ray from r00
to P0 and the ray P1 to r10 , because they are in a medium with constant refractive
index.
Now, our interest is to know the propagation of the rays inside of the sphere.
3.5. The Luneburg Lens
41
Figure 3.12: Luneburg Sphere Parameters
For this reason, it is necessary to know the variation from the angle θi to the angle
θs . In the point P0 we have that
π = α0 + Ψi + π − θi
θi = α0 + Ψi + π − π
θi = α0 + Ψi
(3.85)
and K in this point is
K = n (r0 ) r0 sin Ψ
(3.86)
from the boundary conditions is very easy to see that n (r0 ) = 1 and r0 = 1, then
sin Ψ = K
Ψ = arcsin K
(3.87)
Ψi + Ψ = π
Ψi + arcsin K = π
Ψi = π − arcsin K
(3.88)
we observe that
42
Chapter 3. The Spherical Luneburg Lens
substituting the value of Ψi into Eq. (3.85)
Ψi + α0 + π − θi = π
Ψi + α0 − θi = 0
π − arcsin K + α0 − θi = 0
(3.89)
θi = π − arcsin K + α0 .
(3.90)
therefore
In the outter point P1 , we have
θs + α1 + π − Ψs = π
θs = Ψs − α1 + π − π
θs = Ψs − α1
(3.91)
but from the Generalized Snell’s Law in this point
Ψs = arcsin K
(3.92)
and substituting the value of Ψs into Eq. (3.91)
θs = arcsin K − α1 .
(3.93)
The angular variation inside the sphere is given by
θs − θi = arcsin K − α1 − [π − arcsin K + α0 ]
θs − θi = arcsin K − α1 − π + arcsin K − α0
θs − θi = 2 arcsin K − π − [α0 + α1 ]
(3.94)
using the trigonometric identity
arcsin k + arccos k =
π
2
(3.95)
3.5. The Luneburg Lens
43
we have
θs − θi = − [2 arccos K − (α0 + α1 )] .
(3.96)
The sum of the angles α0 + α1 can be defined as the deflexion function given by
αT (K) = α0 (K) + α1 (K)
(3.97)
this function represents the direction change of the ray passing through the
medium. With this equation, we can rewrite Eq. (3.96) as
θs − θi = − [2 arccos K + αT (K)] .
(3.98)
From Eq. (3.33) is easy to find the K value range, because we know the value of
n0 and r0 , then
0 ≤ K ≤ 1.
(3.99)
It is possible to obtain from the ray integral equation, the variation from θi to
θs , i.e.,
Z r1
dr
p
(3.100)
θs − θi = K
ρ2 − K 2
r0 r
to evaluate the integral is necessary to divide the variation of the angle into two
variations. The first variation is from r0 to r∗ and the second is r∗ to r1 , this leads
to
Z r∗
Z r1
dr
dr
p
p
θs − θi = K
−K
(3.101)
ρ2 − K 2
ρ2 − K 2
r0 r
r∗ r
but r0 and r1 are equal to 1, then
Z
1
θs − θi = −2K
r∗
dr
p
.
r ρ2 − K 2
(3.102)
Since Eq. (3.96) is equal to Eq. (3.102) we have
Z
1
K
r∗
dr
1
p
= arccos K + αT (K)
2
r ρ2 − K 2
(3.103)
44
Chapter 3. The Spherical Luneburg Lens
where 0 ≤ K ≤ 1.
Since the Luneburg lens is a stigmatic system is possible to find the deflexion
function [7]. From the triangle given by the points r00 , P0 , and the origin, we can
obtain the value α0
r0
1
= sin0Ψi
sin α0
sin α0 = sinr0Ψi
h0 i
(3.104)
α0 = arcsin sinr0Ψi
h 0i
α0 = arcsin rK0
0
and from the triangle given by the points r10 , P1 , and the origin; we have that the
α1 value is
1
r10
(3.105)
=
sin α1
sin (π − Ψs )
but
sin (π ± x) = ∓ sin x
(3.106)
sin (π − Ψs ) = sin Ψs
(3.107)
then
we obtain
r0
1
sin α1
= sin 1Ψs
sin α1 = sinr0Ψs
h1 i
α1 = arcsin sinr0Ψs
h 1i
α1 = arcsin rK0
(3.108)
1
α0 and α1 only depend on K, then
α0 (K) = arcsin
h i
K
;
r0
0
α1 (K) = arcsin
h i
K
r10
(3.109)
therefore, the deflexion function is given by
K
K
αT (K) = arcsin 0 + arcsin 0 .
r0
r1
(3.110)
3.5. The Luneburg Lens
45
Substituting this equation into Eq. (3.103),
Z
1
K
r∗
if
K
1
K
p
+ arcsin
= arccos K +
arcsin
0
2
r0
r10
r ρ2 − K 2
(3.111)
1
K
K
f (K) = arccos K +
arcsin
+ arcsin
0
2
r0
r10
(3.112)
dr
we have
Z
1
K
r∗
dr
p
= f (K) .
r ρ2 − K 2
(3.113)
Since we want to know the refractive index of the medium, we will consider a
change of variable of the form
τ = log r
(3.114)
where
dr
(3.115)
r
we know that ρ is a function of r, but with the change of variable, ρ becomes a
function of τ , i.e., ρ = ρ (τ ). With this change of variable, we have
dτ =
Z
0
K
−∞
dτ
p
= f (K) .
ρ2 (τ ) − K 2
(3.116)
the minimum value that r∗ can have is 0, then, the lower limit of integration is
−∞, and when ρ is a function of r the upper limit of integration is 1, and with
the change of variable it is 0.
Now, we assume that ρ (r) and ρ (τ ) are invertible functions, i.e.,
r = r (ρ)
τ = τ (ρ)
(3.117)
τ (ρ) = log r (ρ)
(3.118)
this leads us to
46
Chapter 3. The Spherical Luneburg Lens
if
and
Ω (ρ) = τ (ρ) = log r (ρ)
(3.119)
dτ (ρ)
1 dr (ρ)
dΩ (ρ)
=
=
dρ
dρ
r (ρ) dρ
(3.120)
we can rewrite this derivative as
Ω0 (ρ) dρ =
where
Ω0 (ρ) =
dr (ρ)
r (ρ)
dΩ (ρ)
dρ
(3.121)
(3.122)
Equation (3.116) is transformed with these equations in
Z
1
K
K
Ω0 (ρ) dρ
p
= f (K)
ρ2 − K 2
(3.123)
from the denominator we can see that the highest value of ρ is 1 and the smallest
value is K, because 0 ≤ K ≤ 1. This is an of Abel’s type integral where the
solution is known [16], but we will find its solution using the Luneburg theorem
which tells us [15]:
If the function f (K) is defined by the integral
Z
λ
K
K
Ω0 (ρ) dρ
p
= f (K)
ρ2 − K 2
(3.124)
in the interval 0 ≤ K ≤ λ, then Ω (ρ) is determined by the integral
2
Ω (λ) − Ω (ρ) =
π
in the interval 0 ≤ ρ ≤ λ.
Z
ρ
1
f (K)
p
dK
K 2 − ρ2
(3.125)
3.5. The Luneburg Lens
47
Using the Luneburg theorem we can solve the integral of Eq. (3.123), then
log (1)−log (r) =
2
π
1
Z
ρ
arccos K
1
p
dK +
π
K 2 − ρ2
Z
Z
1
1
ρ
h
i
arcsin rK0 + arcsin rK1
p
dK (3.126)
K 2 − ρ2
but as log (1) = 0
2
π
− log (r) ==
Z
ρ
1
arccos K
1
p
dK +
π
K 2 − ρ2
ρ
h
i
arcsin rK0 + arcsin rK1
p
dK.
K 2 − ρ2
(3.127)
The first integral is easy to solve when we define
f (K) = arccos K
(3.128)
where
Z
1
f (K) = K
K
Z
dρ
p
=K
ρ K 2 − ρ2
d
dρ
1
K
p
log ρdρ
K 2 − ρ2
Z
1
=K
K
d log ρ
p
K 2 − ρ2
(3.129)
and applying the Luneburg theorem, we have
2
− log ρ =
π
1
Z
ρ
arccos K
p
K 2 − ρ2
(3.130)
therefore
− log (r) = − log ρ +
1
π
Z
ρ
1
h
i
arcsin rK0 + arcsin rK0
1
p0
dK.
2
2
K −ρ
(3.131)
We introduce a new function given by
1
ω (ρ, a) =
π
Z
ρ
1
arcsin at
p
dt
t2 − ρ2
(3.132)
48
Chapter 3. The Spherical Luneburg Lens
we obtain
ρ
= ω (ρ, r00 ) + ω (ρ, r10 )
r
but knowing that ρ = nr, then we will have that
log
(3.133)
0
0
n = e[ω(ρ,r0 )+ω(ρ,r1 )]
(3.134)
r = ρe−[ω(ρ,r0 )+ω(ρ,r1 )]
(3.135)
and
0
0
Finally, if the solution for the ω (ρ, a) function is known, it is possible to solve
the system of equations given by Eqs. (3.134) and (3.135), and we will be able to
know the refractive index n (r).
For example, if we have a system where r00 = ∞ and r10 = 1, the solutions for
the ω (ρ, a) function when a = r00 and a = r10 are
arcsin ∞t
p
dt = 0
t2 − ρ2
(3.136)
h
i
p
arcsin t
1
p
dt = log 1 + 1 − ρ2 .
2
t2 − ρ2
(3.137)
1
ω (ρ, ∞) =
π
and
1
ω (ρ, 1) =
π
Z
ρ
1
Z
1
ρ
Substituting these expressions into the Eqs. (3.134) and (3.135), we have that
q
p
n = 1 + 1 − ρ2
r=
ρ
n
(3.138)
If we solve this system of equations, we obtain the refractive index n (r) given by
n (r) =
where r =
√
2 − r2
(3.139)
p
x2 + y 2 .
A sphere with this refractive index makes that all rays from infinity focused
on the point r10 = 1, i.e., all the rays focused on the horizontal axis and on the
3.6. The Generalized Luneburg Lens
49
surface of the sphere, as shown in the Fig. 3.13. This sphere is known as The
Luneburg Lens.
Figure 3.13: The Luneburg Lens
The Luneburg lens can be generalized to focus the rays outside or inside the
lens because Luneburg’s original lens is a limiting case in which r00 is at infinity
and r10 is on the surface of the lens. We are interested in the first case because in
the eye all rays focus on the retina rather than inside the crystalline. Recalling
that our goal is to build a model of the crystalline based on the Luneburg lens.
3.6
The Generalized Luneburg Lens
In 1958, Morgan extended Luneburg’s analysis in his paper entitled "General
Solutions of the Luneburg Lens Problem", where he demonstrates the possibility
of a spherical gradient lens that images one finite sphere sharply onto another
[17]. In some cases both; the object and image surfaces lie outside the lens itself
[10]. The principle of the generalized Luneburg lens is shown in Fig. 3.14.
But Morgan was not the only person who studied this problem. In 1957,
Toraldo di Francia published his paper entitled "Il Problema matematico del sis-
50
Chapter 3. The Spherical Luneburg Lens
Figure 3.14: The Generalized Luneburg Lens
tema ottico concentrico stigmatico", where he does the analysis of Luneburg lens,
but his method can be applied easily to the most general case [18].
In this section, we rely on the Morgan paper because this work provides a
more complete and explicit analysis of the generalized Luneburg’s problem [17].
The initial conditions to solve this problem are: a refractive index n (r) must
exist in the interval 0 ≤ r ≤ 1, and the external foci are given by r00 > 1 and
r10 > 1.
To find the refractive index n (r) we start from Eq. (3.113),
Z
1
K
r∗
dr
p
= f (K) .
r ρ2 − K 2
(3.140)
Before solving this equation we can note that, if the index of the lens at r = a,
where a ≤ 1, is less than unity, some rays from r00 will be totally reflected. Then,
the full aperture of the lens will not be used [10]. To avoid this problem, let us
introduce a new refractive index in the interval a ≤ r ≤ 1, which is given by
n (r) =
P (r)
1
≥
r
r
(3.141)
if P (r) is known and a ≤ 1, the solution of the Eq. (3.140) is
Z
a
K
r∗
dr
p
= f (K) − F (K)
r ρ2 − K 2
(3.142)
3.6. The Generalized Luneburg Lens
where
1
Z
51
dr
p
r P 2 (r) − K 2
F (K) = K
a
(3.143)
and f (K) is given by Eq. (3.112).
If we consider that ρ (r) is a invertible function then we can introduce the
Ω (ρ) function (Eq. (3.119)) to solve Eq. (3.142), i.e., the integral reduces to
Z
1
K
K
Ω0 (ρ) dρ
p
= f (K) − F (K)
ρ2 − K 2
(3.144)
using the Luneburg theorem, the solution of this integral is
2
a
− log =
r
π
1
Z
ρ
f (K) − F (K)
p
K 2 − ρ2
(3.145)
it can rewrite as
2
a
− log =
r
π
Z
ρ
1
2
p
−
K 2 − ρ2 π
f (K)
Z
1
ρ
F (K)
p
K 2 − ρ2
(3.146)
from Eqs. (3.130) and (3.132) is very easy to see that
2
π
Z
ρ
1
f (K)
p
= ω (ρ, r00 ) + ω (ρ, r10 ) − log ρ
K 2 − ρ2
(3.147)
and we define a new Υ (ρ) function given by
2
Υ (ρ) =
π
Z
ρ
1
F (K)
p
K 2 − ρ2
(3.148)
where Υ (ρ) can be known if P (r) is known.
Substituting Eqs. (3.147) and (3.148) into Eq. (3.146), we have
0
0
n = e[ω(ρ,r0 )+ω(ρ,r1 )−Υ(ρ)]
(3.149)
52
Chapter 3. The Spherical Luneburg Lens
and
ρ
(3.150)
n
with these two equations we can find a refractive index for any value of r00 and r10 .
For example, if a = 1 is very easy to see that Υ (ρ) = 0 and leading to Luneburg
solution, i.e., when r00 = ∞ and r10 = 1.
r=
Morgan established a condition for the existence of a solution, that is, in order
for a solution to exist for any value of r00 and r10 it is necessary that the following
condition is satisfied
Z 1
1
1
dr
p
arcsin 0 + arcsin 0 ≥ 2
.
(3.151)
r0
r1
P 2 (r) − 1
a r
Nowadays, Luneburg lens and the generalized Luneburg lens have great interest in various areas. For example; in telecommunications, the Luneburg lens
is studied because it eliminates the aberrations caused by the antennas and the
scanners [19]. Also, this lens provides excellent steering capabilities for incident
wide angles [20, 21].
The Luneburg lens have a spherical geometry, this geometry can be changed
to a ellipsoidal geometry when we make a linear transformation. The new lens
is named as the elliptical Luneburg lens and it has the same property of the
Luneburg lens, i.e., all rays are focused on one point after the lens.
In the next section, a detailed analysis of the Luneburg lens with an elliptical
geometry will be made.
3.7
The Elliptical Luneburg Lens
We can define the Luneburg lens as a marvellous optical lens because it is a
aberrations free lens. In telecommunications is extremely difficult to be applied
in any practical antenna system due to its large spherical shape.
Currently, it has been proposed a transformation that reduces the profile of
the original Luneburg lens without affecting its unique properties. The new trans-
3.7. The Elliptical Luneburg Lens
53
formed slim lens is then discretized and simplified for a practical antenna application [21].
The Luneburg lens and The ellipsoidal Luneburg lens have been designed experimentally using Polymeric nanolayered. The first lens is presented as a developing application of the nanolayered polymer technology and the second lens is
used to model a human crystalline lens using the anterior and posterior shape of
the crystalline for a age=5 years. The importance of this work is that the anterior
and posterior GRIN lenses were assembled into a bio-inspired GRIN human eye
lens through which a clear imaging was possible [22, 23].
Due of these experimental advances is necessary to analyze the Elliptical
Luneburg lens.
3.7.1
Linear Transformation of The Spherical Luneburg
Lens
We will demonstrate that from a linear transformation of a circle we can generate
an ellipse. If we have a vector of the form
"
C=
xc
yc
#
where this vector represents the components of a unit circle and it has the property
of
C t C = x2c + yc2 = 1
(3.152)
and also has a imaging vector given by
"
C0 =
xc0
yc0
#
where
C 0 = AC
(3.153)
54
Chapter 3. The Spherical Luneburg Lens
then
C = A−1 C 0
t
C t = (A−1 C 0 )
(3.154)
from Eq. (3.152)
t
C t C = (A−1 C 0 ) (A−1 C 0 )
t
(A−1 C 0 ) (A−1 C 0 ) = 1
if
"
A=
a b
c d
(3.155)
#
"
A−1
#
a
c
At =
b d
"
# "
#
d −b
d0 −b0
1
=
=
det A −c a
−c0 a0
#
"
0
0
d
−c
t
A−1 =
−b0 a0
now, we have
"
A−1 C
0
=
d0 −c0
−b0 a0
#"
xc0
yc0
#
"
=
d0 xc0 − c0 yc0
−b0 xc0 + a0 yc0
#
i
d0 xc0 − c0 yc0 −b0 xc0 + a0 yc0
"
#
i
0 0
0 0
h 0
d
x
−
c
y
c
c
−1 0 t
−1 0
A C
A C = d xc0 − c0 yc0 −b0 xc0 + a0 yc0
−b0 xc0 + a0 yc0
i
t
h
A−1 C 0 A−1 C 0 = (d0 xc0 − c0 yc0 )2 + (−b0 xc0 + a0 yc0 )2
A−1 C 0
t
=
h
2
2
(d0 xc0 − c0 yc0 ) + (−b0 xc0 + a0 yc0 ) = 1.
(3.156)
Taking the appropriate values of the matrix A, Eq. (3.156) becomes to the
3.7. The Elliptical Luneburg Lens
55
equation of an ellipse, given by
"
A=
then
(A)−1
1
=
α
"
α 0
0 1
#
#
1 0
0 α
"
=
1
α
0
0 1
#
is clear to see that if a0 = 1, b0 = 0, c0 = 0 y d0 = α1 , then Eq. (3.156) becomes
1
xc0
α
2
and if
"
A=
+ yc20 = 1
1 0
0 α
(3.157)
#
Equation (3.156) becomes
x2c0
+
1
y c0
α
2
= 1.
(3.158)
The parameter α is very important because it represents a compression or a
expansion of the x-axis or y-axis, as it is shown in the Eqs. (3.157) and (3.158).
With these two equations it is possible to generate the elliptical Luneburg lens
and we can to know the refractive index shape.
3.7.2
GRIN of The Elliptical Luneburg Lens
We know that the Luneburg lens has a refractive index given by
n (r) =
where r =
√
2 − r2
(3.159)
p
x2 + y 2 is the the radius on one point inside the lens and 0 ≤ r ≤ 1.
56
Chapter 3. The Spherical Luneburg Lens
Now, if we defined a new radius given by
s
re =
1
x
α
2
+ y2
(3.160)
and we substitute it into Eq. (3.159), we have
v
#
" u
2
u
1
x + y2 .
n (re ) = t2 −
α
(3.161)
This equation represents the refractive index of the elliptical Luneburg lens. If
we consider that the α parameter is α < 1, we have an elliptical refractive index
with semi-major axis in the vertical axis, as shown in the Fig. 3.15.
Figure 3.15: The Elliptical Luneburg Lens
We have been designed theoretically the ellipsoidal Luneburg lens using a linear
transformation. Now, we need to know how the rays propagate inside this lens.
3.8. Ray Tracing
3.8
57
Ray Tracing
If we know the refractive index n (r), it is possible to obtain the rays inside of
this refractive index from Eq. (3.15). But in this section, we will use the theory
of J. A. Grzesik published on the paper entitled "Focusing properties of a threeparameter class of oblate, Luneburg-like inhomogeneous lenses", because it is a
lot easier to understand and also because this theory can apply to both spherical
and elliptical Luneburg Lenses [24].
The refractive index of spherical or elliptical Luneburg lenses can be represented as
p
(3.162)
n (x, y) = nc − (nc − ns ) (µ2 x2 + y 2 )
√
√
where nc and ns is the central and surface refractive index of the lens, respectively, and µ is a compression factor along x-axis. If µ−1 < 1 we obtain an oblate
ellipsoidal lens with |x| < µ−1 , and if µ = 1 we have a sphere of unit radius.
We can observe that when we have nc = 2 and ns = 1, Eq. (3.163) generates
the same refractive index that in Eq. (3.161) if α = µ1 . At this point, this is all
what we need to consider.
Figure 3.16: Parameters
As the refractive index depends only on x and y then is possible to use the
ray equation given by Eq. (2.14) in order to see how the rays propagate in the
58
Chapter 3. The Spherical Luneburg Lens
medium. Equation (2.14) with some minor manipulation, adopts the form
∂n2 (x, y, z)
d n2 (x, y)
=
dx 1 + ẏ 2
∂x
(3.163)
free from any square root entanglement. To solve this equation, we will consider
only two parameters α0 and P0 . The parameter α0 is the angle that the ray makes
with respect to the direction of compression and P0 = (x0 , y0 ) is the point where
the ray enters the lens, as shown in Fig. 3.16.
Now, if the rays are given by y (x) function, the solution of this equation is
y (x) = Λ2 sin sin−1 y0 Λ−1
± µ−1 sin−1 xΛ−1
− sin−1 x0 Λ−1
2
1
1
where
Λ21 =
and
Λ22
ns cos2 (α0 ) + µ2 (nc − ns ) x20
µ2 (nc − ns )
ns sin2 (α0 ) + (nc − ns ) y02
.
=
(nc − ns )
(3.164)
(3.165)
(3.166)
The ± sign in Eq. (3.164) provides the possibility to have an ascendent or a
descent ray, as dictated by the sign of injection angle α0 .
The ray tracing with this solution is very easy to obtain. For example, if we
have nc = 2, ns = 1, µ = 1 and the rays come from infinity, i.e. α0 = 0, the Λ1
and Λ2 parameters are
q
Λ1 = 1 + x20
(3.167)
and
Λ2 = y0 .
(3.168)
With these parameters and if x goes from x0 to 1 for each ray, we obtain the ray
tracing of the Luneburg lens when we substituted these values into Eq. (3.164),
as shown in the Fig (3.17).
For the elliptical Luneburg lens, we need nc = 2.5, ns = 1, µ > 1 (in this case
µ = 2), α0 = 0 and x goes from x0 to x1 for each ray. With these values we have
3.8. Ray Tracing
59
Figure 3.17: Ray tracing of The Luneburg Lens
that
r
Λ1 =
1
+ x20
6
(3.169)
and
Λ2 = y 0 .
(3.170)
then we can find the ray tracing of the elliptical Luneburg lens, as shown in Fig.
3.18.
Figure 3.18: Ray tracing of The Elliptical Luneburg Lens
60
Chapter 3. The Spherical Luneburg Lens
Figure 3.19: Ray tracing of The Elliptical Luneburg Lens with nc = 2, ns = 1,
µ = 2 and α0 = 0
We can note that the central refractive index is different for both examples.
This should not worry us, because the ray propagation depends on three parameters (nc , ns and µ) and we must choose very carefully the values of these
parameters so that the rays are focused.
For example, if nc = 2, ns = 1, µ = 2 and α0 = 0, we obtain a lens which only
focuses rays entering in the center of the lens, as shown in Fig. 3.19.
3.9
Conclusions
Throughout this chapter we have studied the properties of a spherical gradient
index medium; its refractive index and how the rays propagate inside that medium.
We have given several examples using the ray equation, the ray integral equation
and the Grzesik’s theory to have a better understanding of the ray propagation
in a spherical gradient index medium.
We have mainly studied the classical Luneburg lens (Spherical Luneburg lens)
and the elliptical Luneburg lens. With the latter, we finally have the fundament
to build our proposed crystalline. But, before we move on, we need to know the
anatomy of the eye and the geometrical parameters (measurements) in order to
use them in our model.
Chapter 4
The Human Eye As An Optical
System
The human visual system is composed, in one hand, by the optics of the eye, and
in the other one, by a signal processing system wired to the brain. In this work we
are interested in the optics of the eye, for this reason, the signal processing system
will not be treated here. However, there is a considerable number of information
about this area in the literature that can be consulted [25, 26, 27].
The eye is an optical system with an extraordinary complexity. The description
or modeling of the eye is done by organizing average measurements and properties
into simplified models called schematic eyes [25, 27]. These models are useful to
systematically study the properties and performance of individual components of
the eye.
The complexity of the eye is due to the fact that its refractive surfaces are
not strictly spherical and its lens has a gradient refractive index. Also, the eye is
complex because it can be seen as an optical imaging and detection instrument.
The complex structure of its components make the eye to be difficult to study.
The most complex component is the lens, for this reason, almost all schematic
models do not include the lens or the gradient refractive index of it, and it is
changed by a constant refractive index.
In this chapter, the components of the human eye are studied and also we
describe some interesting schematic models are described.
62
Chapter 4. The Human Eye As An Optical System
4.1
The Human Eye
Physiologically the eye has many elements, as shown in Fig. 4.1. The inside of
the eye is divided into three compartments [28, 29]:
• The anterior chamber, between the cornea and the iris, which contains the
aqueous fluid.
• The posterior chamber, between the iris, the ciliary body and the lens, which
contains the aqueous fluid.
• The vitreous chamber, between the lens and the retina, which contains a
transparent colourless and gelatinous mass called the vitreous humour or
vitreous body.
Figure 4.1: Human eye and its optical elements [28].
The number of elements can be minimized if we consider the eye as an optical
system, i.e., the elements to be considered in this section are cornea, pupil, lens,
retina, aqueous humor and vitreous humor.
4.1. The Human Eye
63
The elements of the eye not considered in this chapter can be consulted with
more detailed in an anatomical description sense in many specialized books in this
area [30, 29].
4.1.1
Refracting components of the human eye: Cornea and
Lens
In the human eye, there are two refracting optical elements; the cornea and the
lens. These elements are very different in their geometrical shape and in their refractive index, but in order to provide a good quality retinal image, these elements
must be transparent and have appropriate curvatures and refractive indices [28].
Given the importance of these two elements in the human eye, the cornea and
the lens are described in this section.
4.1.1.1
Cornea
The cornea is the curved transparent front surface of the eye. It has approximately
spherical shape with a radius of curvature of about 8 mm.
The anterior surface of the cornea is protected by the sclera which is the
outermost structure of the human eye. It is a dense, white, opaque, fibrous tissue
that is mainly protective in function and is approximately spherical with a radius
of curvature of about 12 mm. The centres of curvature of the sclera and cornea
are separated by about 5 mm [28, 29].
The posterior surface of the cornea is in contact with the aqueous humour that
is a colorless liquid and is composed by a 98% of water. The aqueous humour is
a liquid with constant refractive index. The value of its refractive index is well
defined and is of 1.336.
The cornea in optical terms is one of the two refracting elements of the eye.
This element contributes with one third of the optical power of the human vision
system.
The refractive index of the cornea is usually taken as 1.376 and this value is
considered constant over the entire cornea. However, the anterior corneal surface
64
Chapter 4. The Human Eye As An Optical System
is not a smooth optical surface due to its cellular structure, although an optically
smooth surface is provided by the very thin tear film which covers it. While this
has an index less than 1.376, for most optical calculations this tear film can be
regarded as a very thin optical element consisting of two concentric surfaces of
almost equal radii of curvature and therefore has negligible power [1].
In some works, the refractive index of the cornea is considered a gradient
refractive index [31, 32], because the cornea is composed by many layers, as shown
in Fig. 4.2, and each corneal layer has its own refractive index.
Figure 4.2: The structure of the cornea [28].
From Fig. 4.2, we can observe that the stroma is by far the thickest layer, i.e.,
its refractive index dominates. For this reason, the value of refractive index of the
cornea is usually taken as 1.376.
Since the cornea is the first and most accessible optical element in the human
eye and by its important optical power makes it the object of several investigations
and treatments [33]. And it is probably also the most measured.
4.1. The Human Eye
4.1.1.2
65
Lens
The other refracting element is the crystalline lens and it is commonly called lens.
The anterior surface of the lens is in contact with the posterior surface of the iris
and the aqueous humour. The posterior surface is in contac with the vitreous
humour that is a clear gel that occupies the posterior segment of the eye and its
refractive index can be considered equal to aqueous humour, 1.336.
The geometric shape of the lens and its refractive index are very difficult to
measure in vivo, because the lens is a dynamic lens responsible for adjusting the
focusing distance. This changed of focusing distances is called accommodation.
Also, as the crystalline lens ages, its geometric shape changes, and it becomes less
flexible, and consequently a person’s ability to accommodate is lost slowly with
time [25]. It is very important to say that also its refractive index changes with
age.
The lens is supported by tiny muscular fibers, called ciliary muscles that pull
the lens. During accommodation, when the eye needs to change focus from distant
to closer objects, the ciliary muscle contracts and causes the suspensory ligaments,
which support the lens, to relax. This allows the lens to become more rounded,
thickening at the centre and increasing the surface curvatures. The front surface
moves slightly forward. These changes result in an increase in the equivalent
power of the eye. When the eye has to focus from close to more distant objects,
the reverse process occurs [25, 28, 29].
However, measures of the geometric shape and refractive index of the lens in
vitro, exist [11]. These show that the lens is not axially symmetric, as can be seen
in Fig. 4.3, and the lens due to its biological nature and protein distributions is
an optically inhomogeneous lens with a gradient refractive index.
Since the lens does not has an axial symmetry, it is represented by two lenses:
one anterior and one posterior, which intersect at the equator.
The most important characteristic of the lens is that the refractive index is
represented by a gradient refractive index where the iso-indicial surfaces are ellipsoidal curves.
Throught the years, the gradient index of the lens has been represented by
66
Chapter 4. The Human Eye As An Optical System
Figure 4.3: The geometric shape of the lens. za and zp are differents.
two mathematical equations, one equation for anterior lens and one equation
for the posterior lens [3, 4]. These models present drawbacks because they are
discontinuous at the equator. These discontinuities can be classified into two main
groups:
• In the first group the gradient index variation in the equator is discontinuaous between the anterior lens and posterior lens [3].
• In the second group the discontinuities are between the derivatives of first
and second order with respect to the iso-indical surface geometry [34, 4].
These problems are generated by the accommodation of the lens, because it
reshapes is required to change its internal gradient index distribution imposing a
constraint to stablish a definitive expression for modeling its gradient refractive
index [35]. In other words, there cannot exists an unique mathematical expression
to simulate a living biological crystalline.
The paper of A. M. Rosen, et al. entitled "In vitro dimensions and curvatures
4.1. The Human Eye
67
of human lenses" provides excellent experimental data for the geometric shape
and refractive index of the lens [11]. In this paper, the refractive index ranges
from nc = 1.4181 ± 0.075 in the core to ns = 1.3709 ± 0.0039 at the boundary
surface of the lens. And also, it provides the equations for the parameters of the
lens with age dependence given by
R = [0.0138 (±0.002) ∗ Age + 8.7] /2
za = 0.0049 (±0.001) ∗ Age + 1.65
zp = 0.0074 (±0.002) ∗ Age + 2.33
(4.1)
where the ratio of anterior thickness (za ) to posterior thickness (zp ) is constant at
0.70. Measurements were made on 37 human lenses ranging in age from 20 to 99
years.
Numerous studies and measurements of the defects of the cornea and crystalline have given the basis to the hypothesis that the crystalline compensates the
defects of the cornea by minimizing or balancing them. A discussion on this can
be found in a review by Artal [33].
These and other important characteristics of the lens will be studied with more
detail in the Chapter 5. Also, we will studied the light propagating inside lens.
4.1.2
Pupil
The iris forms the aperture stop of the eye. Its aperture or opening is known as
the pupil. The pupil size is determined by two antagonistic muscles, which are
under autonomic (reflex) control [28]:
• The sphincter pupillae, which is a smooth muscle forming a ring around the
pupillary margin of the iris. When it contracts, the pupil constricts. It is
innervated by the parasympathetic fibres from the oculomotor (3rd cranial)
nerve by the way of the ciliary ganglion and the short ciliary nerves.
• The dilator pupillae, which is more primitive and consists of myo-epithelial
cells that extend radially from the sphincter into the ciliary body. It dilates
68
Chapter 4. The Human Eye As An Optical System
the pupil and is innervated by sympathetic nerve fibres, which synapse in
the superior cervical ganglion and enter the eye by way of the short and
long ciliary.
The diameter of the pupil may vary from about 2-3 mm at high illumination
to about 8 mm in darkness.
If we considered identical illumination conditions is possible to observe that
the diameter is different for each person. This is due to pupil size decreases with
increasing in age and pupils react less to changes to light levels.
For example, for a 10 years old person can consider a typical diameters of 4.8
mm, while at 45 will be of 4.0 mm, and 3.4 mm at 80 years. For an eye in total
darkness the most common diameters are 7.6 mm at 10 years, 6.2 mm at 45, and
5.2 mm at 80 years [29].
4.1.3
Retina
The retina is the ocular surface where images are projected and is the lightsensitive tissue of the eye. It is composed of a number of cellular and pigmented
layers, and a nerve fibre layer, i.e., the retina consists of ten layers, as shown in
Fig. 4.4.
The study by direct observation of this membrane is known as Retinoscopy.
This procedure has undergone significant evolution since the adaptive optics systems were first implemented [36].
The retina is especially interesting since it is directly connected to the central
nervous system and some experts even consider it to be part of the brain since
neural cells can be found in it [37]. Here is where specialized cells that can convert
light into nervous impulses can be found. There are two main types of corneal
cells which are responsible for light conversion. The cones are the ones responsible
for color discrimination, they come in three types, each one capable of detecting a
specific frequency band associated to color blue, green, and red, respectively. The
rods are responsible for night vision, since they are very sensitive to light. This
high sensitivity is somehow related to the wiring between rods which combines
4.1. The Human Eye
Figure 4.4: The layers at the back of the human eye [28].
69
70
Chapter 4. The Human Eye As An Optical System
the signals into the brain, giving them as a consequence a poor spatial resolution
[33, 37]. There are around seven millions cones, a hundred and twenty five millions
of rods and as much as a million nervous fibers. The fovea is the region of the
retina where a higher cone density is found.
4.2
Schematics Eye
During the last decades different schematic eye models have been proposed. The
paraxial schematic models are the most relevant [28, 1]. However, more complicated schematic eye models containing a Gradient Refractive Index lens have been
published in recent years [3, 38].
The paraxial schematic eye models have a problem; they use a homogeneous
index lens and a limited numbers of surfaces. In the Emsley reduced model, the
eye is represented only by a single refractive surface; being the anterior surface of
the cornea [28, 1, 39], as shown in Fig. 4.5.
Figure 4.5: The Emsley reduced schematic eye.
The Gullstrand-Emsley model has three refracting surfaces, as shown in Fig.
4.6; one for the cornea and two for the lens, moreover the aqueous and vitreous
refractive index were modified up to 1.416 when it is well known that the real
values are 1.336 [28].
The Le Grand full theoretical model is represented by four surfaces; two for
the cornea and two for the lens, as shown in Fig. 4.7. Both models provide the
4.2. Schematics Eye
71
Figure 4.6: The Gullstrand-Emsley schematic eye.
accommodation of the lens.
Figure 4.7: The Le Grand full theoretical schematic eye.
Gullstrand was the first that proposed a model with an inhomogeneous lens,
but the lens only have four refracting surfaces [40]. This lens has a core and a
cladding, the core has a high refractive index limited by the interior surfaces and
the cladding has lower refractive index limited by the exterior surfaces. The lower
refractive index is surrounding the core, i. e., Gullstrand represented the lens as
a lens inside other lens, but it is very important to say that it does not represent
a GRIN lens, because both lenses, independently, have a homogeneous refractive
index.
In 1974, R. G. Zainullin and et al. in the paper entitled "The crystalline lens
as a Luneburg lens" proposed to use the Luneburg lens to represent the crystalline
72
Chapter 4. The Human Eye As An Optical System
lens of the eye of vertebrates [8]. We say that they proposed, because they do not
make a schematic eye in that paper.
Using the same idea of R. G. Zainullin and et al., Campbell and Hughes
published a paper entitled "An analytic, gradient index schematic lens and eye for
the rat which predicts aberrations for finite pupils", where the lens is represented
as a Luneburg lens [2]. But, there is a problem in this paper, it can not be used
to represent a schematic human eye since the anterior and posterior faces of the
lens are considered to be symmetric, as shown in Fig. 4.8.
Figure 4.8: The schematic eye is analytically derived from the refractive index
profile of the crystalline lens and anatomical measurements of a rat eye. [2].
Also, the Maxwell fish eye has been considered to build a schematic human
eye, however, its design is a spherically symmetric lens, and the index of refraction
varies symmetrically about a point [9]. For this reason, the schematic eye have
the same problem that the schematic eye of Campbell and it is not realistic.
This Schematic eye has many conceptual mistakes, because it does not exist a
generalized Maxwell’s Fisheye, and the Maxwell’s Fisheye is not a lens (see Section
3.4), then It is not possible to make a ray tracing outside the environment of the
4.3. Conclusions
73
Maxwell’s Fisheye. This design was proposed by Yun Wu and et al. in 2010 and
its schematic human eye is shown in Fig. 4.9.
Figure 4.9: Schematic eye model with Maxwell fish-eye spherical lens [9].
In this thesis we propose a more realistic theoretical schematic eye using the
idea that the lens can be represented as a composite modified Luneburg lens.
4.3
Conclusions
In this chapter we have analysed the anatomy of the human eye and we have
presented the optical characteristics of each element of the human eye.
Also, we have studied the different schematic eyes, in which we have observed
their problems to represent a realistic schematic eye model.
With the characteristics of the human eye given in this chapter, we may be able
to build a schematic human eye using a new lens that we will call the composite
modified Luneburg lens, which will be discussed in the next chapter. This lens
is based on the elliptical Luneburg lens and the new model can be modified for
74
Chapter 4. The Human Eye As An Optical System
different eye conditions, as we will show in Chapter 5.
Chapter 5
Schematic Eye with Composite
Luneburg Crystalline
The original Luneburg lens is a sphere with gradient index that has the property
of being spherical aberrations free [6, 17]. For every pair of object and image
conjugate points the internal refractive index distribution is different presenting
a similitude with the human lens [8].
Now, it has been demonstrated that it is possible to perform a geometric
transformation in such a way that the original spherical Luneburg lens can be
transformed into an oblate spheroidal shape maintaining its aberration free property [24, 21]. This can be used as a first approximation to the human crystalline.
However, the human crystalline is not symmetrical; to take this into account an
alternative model will be proposed having an asymmetric bi-spheroidal shape but
assuming that the gradient refractive index can be represented with a given analytical expression [41, 42]. As discussed above, the crystalline shape and refractive
index are being modified continuously as the vision point changes, so a realistic
model of the crystalline must consider the dynamics of both of them.
In this chapter we propose a dynamic model of the crystalline constructed with
two separate spheroidal hemispheres with variable curvatures for the anterior and
posterior sections and considering a varying Gradient Refractive Index (GRIN).
Its imaging properties are investigated based on the Luneburg lens theory modified
accordingly to the proposed composite lens. We impose the condition in our model
that the isoindical lines be continuous as well as their first order derivatives at the
equatorial plane. This condition creates a perfect match of the gradient refractive
index of both hemispheres guaranteeing the smooth continuity of the rays. This
76
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
model will be referred to as the composite modified Luneburg lens.
Also, in this chapter we will propose a more realistic theoretical schematic
eye using the composite modified Luneburg lens. This schematic eye is based in
biometric parameters reported in Ref. [11].
5.1
Composite Modified Luneburg Lens
As mentioned above, the problem of the spherical Luneburg lens after a simple
transformation can be reformulated into that of a spheroid maintaining its aberration free property [17, 21]. The index in the spherical Luneburg lens is described
p
by a function n (r) where r = x2 + y 2 + z 2 and x, y and z are the Cartesian
p
coordinates. We set ρ = x2 + y 2 and by symmetry around the z-axis we will
work only with y. As r defines a spherical shape, the transformation to obtain
the desired spheroidal shape can be either
1 0
y
s
2
+ z 02 = r2
or
02
y +
1 0
z
s
2
= r2 .
(5.1)
(5.2)
In these equations s is a constant parameter. By properly choosing the value of
s we can compress or expand the sphere along the y or z axis as needed. The model
of crystalline will be approximated by two spheroidal hemispheres [21, 24, 41, 42].
We now proceed to construct the model by imposing the condition that the
gradient indices and the axial derivatives of the anterior and posterior hemispheres
match at every point on the equatorial plane. For this purpose the refractive
indices, na (r) for the anterior and np (r) for the posterior hemisphere, must be of
the form
5.1. Composite Modified Luneburg Lens

s


0
0

n2c −

 na (y , z ) =
s
n(y 0 , z 0 ) =



 np (y 0 , z 0 ) =
n2c −

y0 2
R
y0 2
R
77
+
1
z0
Rsa
+
1
z0
Rsp
2 2 z0 ≤ 0
(5.3)
z0 ≥ 0
where nc is the refractive index at the center of the lens, R is the radius of the lens
p
measured on the equatorial plane. The scaling factor is s(a,p) = z(a,p) /R n2c − n2s
where za is the anterior vertex and zp the posterior vertex. For na (r) the variable
z 0 is negative and for np (r) the variable z 0 is positive. Figure 5.1 displays the
projection on a meridional plane of the constructed crystalline lens with the bielliptical iso-indical lines showing continuity at the equatorial plane.
Figure 5.1: Geometry of the bi-spherical model showing the continuity of the
isoindical lines at the equator plane.
78
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
For the numerical example under study we used as a reference the biometric
parameters reported in Ref. [11]. The refractive index ranges from nc = 1.4181 ±
0.075 at the core to ns = 1.3709 ± 0.0039 at the boundary surface of the lens.
The age dependence of geometric parameters of their lens are estimated by the
equations given by Eq. (4.1).
For a lens aged 35 years the corresponding parameters are
R = 4.4005mm
za = 1.8215mm
zp = 2.5890mm.
(5.4)
The curvatures of the anterior and posterior hemispheres are such that the ratio
between za and zp is 0.7035. The lens aged 35 years is shown in Fig. 5.1.
Now, to investigate the imaging properties of our model we follow a procedure
similar to that described in Ref. [24]. this is, to modified it accordingly to fit the
new geometry and the imposed continuity conditions. We assumed that the lens
is embedded in a medium with refractive index equals to 1.336 and by keeping
constant the surface refractive index ns .
The first case that will be studied simulates a relaxed crystalline assuming an
object at infinity so that the rays impigne parallel to its anterior surface. The
chosen geometrical parameters and refractive indices produce practically a perfect focus at the image plane placed at a distance of 63.05mm from the posterior
surface of the lens as shown in Fig. 5.2 a). For the next two cases we modify
the vertex ratio za /zp and calculate the refractive index that is necessary to form
a point image at the same plane of a point object placed at 250 mm from the
anterior surface. In Fig. 5.2 b) we see that for a reduced ratio the Composite
Modified Luneburg (CML) lens is thinner resulting in a GRIN distribution such
that the central refractive index has been increased. The opposite case of making thicker the CML lens with a larger ratio za /zp = 0.7185 results in a GRIN
redistribution with a reduced refractive index at its center, as is shown in Fig 5.2
c). The results of our simulations confirm the prediction discussed in Ref. [35].
By changing the shape of the crystalline implies a gradient refractive index dis-
5.1. Composite Modified Luneburg Lens
79
Figure 5.2: Ray tracing through the proposed CML lens embedded in a medium
with refractive index of 1.336. a) Rays incident from an infinite distance. Rays
incidents from a finite distance of 250 mm with b) za /zp = 0.6872 and c) za /zp =
0.7185.
80
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
tribution change. Although in the present model we have used perfect geometric
surfaces, implementing minor modifications to our model, by using real biometric
data it is possible to simulate imaging of a real human lens imaging [38].
The GRIN profiles and the GRIN distribution for each lens are shown in Fig.
5.3. From this figure, we can observe the refractive index changes when we change
the ratio za /zp .
Figure 5.3: The GRIN distribution and the GRIN profiles for each lens with
different ratio za /zp .
We have said the central refractive index increases when za /zp decreases and
the central refractive index decreases when za /zp decreases, this is much easier to
see when all the profiles are plotted in the same figure, as is shown in Fig 5.4.
If we observe the rays inside the CML lens, we can say that its bending due
to the GRIN is almost imperceptible appearing as if the refractive index was
homogeneous. However, from Fig. 5.5 we observe that the rays are planar curves
5.1. Composite Modified Luneburg Lens
81
Figure 5.4: The GRIN profiles for each lens with different ratio za /zp .
as is predicted from the Luneburg theory.
These rays are different for each CML lens; in its length and its trajectory,
because the propagation distance is different for each ratio za /zp , and the change
on the refractive index causes a different trajectory for the rays are different, i.e.,
a higher refractive index causes the ray to have a greater curvature, while a lower
refractive index causes the ray to have a smaller curvature, as shown in Fig 5.5.
The second case that is investigated simulates different CML lenses keeping
the surface refractive index ns constant and the ratio za /zp of a lens aged 35 years.
Also, we assumed that the lens is embedded in a medium with a refractive index
equals to 1.336.
In these computational simulations, the only parameter of the lens that can be
modified is the central refractive index nc , because we decided that the parameters
ns and za /zp to be constants. A parameter that also can be modified, but this
is not a parameter of the lens, is the plane of a point object, these planes were
placed at different distances from the anterior surface, i.e., the distances were:
infinity, 2000 mm, 1000 mm, 500 mm, and 250 mm.
We calculate the central refractive index nc that is necessary to produce prac-
82
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
Figure 5.5: Trajectory of the inner rays for each CML lens with different ratio
za /zp .
tically a perfect focus at the image plane placed at a distance of 63.05 mm from
the posterior surface of the lens.
From Fig. 5.6, we can observe that the central refractive index decreases when
we placed the point object plane near the anterior surface of the lens. This means
that the refractive power of the lens is lower when the objects are closer to the
lens.
These results give us a better idea of the changes in the refractive index of the
lens, because we are not altering the geometry of the lens and due to this fact, we
do not make assumptions of changes in the ratio za /zp for different object point
distances. The change of the refractive index is much easier to see when all GRIN
profiles are plotted in the same figure, as is shown in Fig 5.7.
If we knew experimentally, how the ratio za /zp changes for specific distances,
we could find how the refractive index varies in the real life. Unfortunately, it
is very difficult to measure this change in vivo, because the lens is dynamic and
5.1. Composite Modified Luneburg Lens
83
Figure 5.6: The GRIN distribution and the GRIN profiles for each lens with the
same ratio za /zp and different central refractive index.
84
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
Figure 5.7: The GRIN profiles for each lens with the same ratio za /zp and different
central refractive index.
there are no measurements of this change.
The propagation of the rays are shown in Fig. 5.8, in which can be observed
that the rays are focused on an image plane placed at 63.05 mm from the posterior
surface of the lens. The incident rays have been cut, because the distances of where
the rays come from are large compared to the size of the CML lens.
From Fig. 5.8 a) to Fig. 5.8 e), the rays and the gradient refractive index
appear to be equal to each other, however, the rays and the gradient refractive
index are different in each case, as shown in Fig. 5.9 and Fig. 5.6, respectively.
From Fig. 5.9, the rays inside the lens appear to have larger curvature when
we have a lower gradient refractive index, but this is not true, because we must
remember that we are changing the incidence point on the lens. From Fig. 3.16
is easy to see that the angle α0 is larger when the point of incidence approaches
to the lens. This implies that the constant K given by generalized Snell law for
inhomogeneous media with spherical symmetry is changing when the angle α0 is
changing, i.e., the K values depende on the values of α0 , as we see from Eq. 3.86.
The two cases have been studied with a human lens of 35 years old. However,
5.1. Composite Modified Luneburg Lens
85
Figure 5.8: Ray tracing through the proposed CML lens embedded in a medium
with a refractive index of 1.336 and a constant ratio za /zp = 0.7036. Rays incidents from a finite distance of a) Infinity, b) 2000 mm, c)1000 mm, d) 500 mm,
and e)250 mm.
86
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
Figure 5.9: Trajectory of the inner rays for each CMLL with constant ratio za /zp =
0.7036.
we have equations where the geometrical parameters of the human lens depend
on age. Then, it is very interesting to make an analysis of second case when we
have a human lens of a different age.
For example, the Fig. 5.10 represents the GRIN profiles of a human lens that
has 20 years old for different object distances. The different with a human lens of
35 years old is the increment of 0.0002 in the central refractive index and a image
distance of 62.35 mm. The increment in the central refractive index is due to the
lens dimensions of 20 years old is lower compared a lens of 35 years old.
Now, the question is: Why the image distance changed?. The response is: It
is always possible to find a distance where practically all the rays are focused, but
this distance depend on the geometrical parameters and the refractive index of
the lens. In the case where the lens has 20 years old, it is impossible to find a
refractive index where all the rays are focused in a distance of 63.05 mm.
Rosen in his paper say: "The ratio of anterior thickness to posterior thickness
is constant at 0.70 for all age". It is very interesting because we have the same
5.1. Composite Modified Luneburg Lens
87
Figure 5.10: The GRIN profiles for each lens with the same ratio za /zp and
different central refractive index. The human lens is 20 years old.
88
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
ratio for all age and different Za and Zp , for this reason, we have different image
distances for all age.
The ratio Za /Zp changes in the accommodation process of the human lens.
This change allows that we can focus all rays at retina for different object distances.
We can observe the Fig. 5.2 and the question arises: why in the Fig. 5.2 was
possible to find the image distance of 63.05 mm when we have different Za /Zp ?.
In this case, we have fixed the image distance and the central refractive index (nc)
and we have found a ratio Za /Zp where all rays are focused at 63.05 mm. Observe
that the ratio Za /Zp is different to 0.70 as we said in previous paragraphs. To
find this ratio is very difficult because we have a big number of possibilities.
We must understand that we have two cases very different when we fixed
the geometrical parameters or when we fixed the image distance and the central
refractive index.
With the carried out analysis of the CML lens is possible to construct a new
Schematic Model of the Human Eye using the CML lens as a human lens. This
model is studied in next section.
5.2
Schematic Luneburg Eye
In the previous section, we constructed a CML lens where its geometrical parameters and refractive indices were chosen to produce practically a perfect focus. It is
woeth to note that, the human lens does not have those central refractive indices
used in Section 5.1.
In this section, we are interested in understanding the behavior of the light
inside the complete eye, but first we want to know about the behavior of the light
when passing through the lens inside the eye.
The lens to be used is 35 years old with a ratio of za /zp = 0.7036, R = 4.7,
nc = 1.4181, and ns = 1.3709, as is shown in Fig. 5.1. With these parameters
and refractive indices of the eye, we can generate the ray tracing along of this new
CML lens. It is very important to say that the refractive indices and parameters
5.2. Schematic Luneburg Eye
89
will be modified along the simulations to study the function of accommodation in
the eye.
We can see in Fig. 5.11 that rays inside the lens are planar curves and also
we can see that the lens produce a negative spherical aberration when rays are
impigne from an infinite distance.
Figure 5.11: Negative spherical aberration from the human lens.
This aberration is important when we considered the cornea as it is perfectly
spherical, because the cornea has a positive spherical aberration as it is shown
in Fig. 5.12. The positive aberration from the cornea can be suppressed by
the negative aberration from the crystalline [43] when we choose an appropriate
refractive index.
For example, if we have nc = 1.41262, ns = 1.3709, za /zp = 0.7036, and
rays that impigne from infinity distance, then the aberration from the cornea is
suppressed by the lens aberration as is shown in Fig. 5.13. In this figure, we can
see that the rays focused on the retina.
With this model, by knowing the curvature it is possible to know the refractive
index or if knowing the refractive index it is possible to know the curvatures of
the lens. Also, we can change the distance from which the rays are coming from
and we can construct a new lens that can suppress the cornea positive spherical
90
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
Figure 5.12: Positive spherical aberration due to the cornea. Rays incident from
infinity distance, the anterior and posterior surfaces of the cornea are immersed
in a refractive index of 1 and 1.336, respectively.
Figure 5.13: Schematic eye model with CML lens.
5.2. Schematic Luneburg Eye
91
aberration. The parameters of the human lens for different distances are given in
Table 5.1 and the propagations are presented in Fig. 5.14.
Infinity
Ratio za /zp
nc
0.7036
1.41262
250 mm
Ratio za /zp
nc
0.7185
1.41477
0.7323
1.41249
0.7450
1.41088
0.7568
1.40897
0.7677
1.40803
500 mm
Ratio za /zp
nc
0.7185
1.41255
0.7323
1.41092
0.7450
1.40939
0.7568
1.40801
0.7677
1.40669
ns
1.3709
ns
1.3709
1.3709
1.3709
1.3709
1.3709
ns
1.3709
1.3709
1.3709
1.3709
1.3709
Table 5.1: Parameters of the crystalline for different rays impigned distances.
Simulations in Fig. 5.14 are made having the refractive index of the surface
fixed, then the ratio between vertex, za /zp , is change, and a suitable refractive
index is search for the centre so the light rays focused on the retina.
We can observe from Table 5.1 that the accommodation of the surface does
affect the refractive index of the crystalline. When the vertex ratio is increased,
the refractive index decreases.
In particular, when the crystalline has a vertex ratio of 0.7185 and rays coming
from a distance of 250 mm, we can observe that the refractive index is bigger than
when the ratio is of 0.7036 and rays come from infinity. This means that for closer
distances, the accommodation must be greater than for larger distances.
It is clear, from Fig. 5.13 that the marginal rays do not focused on the retina.
This can be seen as a problem, but due to the fact that these rays never enter
92
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
Figure 5.14: Examples of Schematic model eyes with CML lens: a) Incident rays
coming from a 250 mm distance and b) Incident rays coming from a 500 mm
distance. Both Schematic eyes have a relation of 0.7450.
5.2. Schematic Luneburg Eye
93
into the eye, because they are blocked by the eye’s pupil.
Figure 5.15: Schematic eye when rays are off-axis.
This analysis is completed when an off-axis analysis is made. When rays are
off-axis, the rays reach the retina with a coma aberration. This aberration is
larger when the rays are farther from the axis as shown in Fig. 5.15.
This schematic model can be modified for different object distances, different
refractive indices and different curvatures of the crystalline; this is to make it as
versatile as possible. Also, it is possible to change the corneal topography to get a
more realistic scenario, since every single eye is different. The lens accommodation
is obtained by reshaping the lens, resulting in the necessity of a modification in
its gradient refractive index.
As can be seen from the simulations, the rays inside the crystalline are planar
curves.
The analysis of the off-axis case shows that when the rays enter with a certain
angle with respect to the optical- or z-axis into the eye, a coma aberration is
presented. As this angle increase the resultant coma aberration increase.
94
5.3
Chapter 5. Schematic Eye with Composite Luneburg Crystalline
Conclusions
In conclusion, a schematic model eye based on a composite modified Luneburg
lens acting as the eye lens has been presented.
Chapter 6
Conclusions and Future Work
Along this work, we have presented a new model of the human lens and a new
schematic human eye using a composite modified Luneburg lens.
This lens has been proposed as a bi-spheroidal model of the human lens based
on the gradient index Luneburg lens. Using biometrical data reported in the
literature our model accurately predicts the imaging properties of a human lens,
where we have observed that the composite modified Luneburg lens has the same
physiological properties of the human lens. It was also obtained that is necessary
a modification in its gradient refractive index when simulating accommodation by
reshaping the lens it is necessary a modification in its gradient refractive index
implying that it is not possible to establish a definitive mathematical expression
for the gradient refractive index of a biological crystalline.
Also, with this lens model it was possible to create a schematic GRIN eye
that emulates the human lens behavior. With this model, we did an analysis that
had never been done, i.e., this model allows the evaluation of different imaging
situations; i.e., the effects of the accommodation in the human lens due to the
changed in distance from which rays impigne.
The analysis predicts that the accommodation of the human lens is accompanied by a change in the gradient refractive index of this lens, which depends on
the ratio za /zp .
This analysis was completed when an off-axis analysis was made. This demonstrated that the rays off-axis reach the retina with a coma aberration, due to
the inclination of the rays. In the specialized literature, the coma aberration
is obtained when the lens has a inclination on the optical axis, however, in our
schematic model, the rays off-axis act in a similar manner as the lens inclination,
96
Chapter 6. Conclusions and Future Work
for this reason, the coma aberration is present.
It is very important to say that our schematic model can be modified for
different object distances, different refractive indices and different curvatures of
the human lens; also, our model can be modified for different topographies of the
cornea, i.e., it can be modified fully to emulate any experimental data.
The work presented in this thesis has opened various research topics in the
optical area. For example, in Visual Optics , it will be of interest to study the ray
propagation in a three-dimensional elliptical Luneburg lens, the lens is shown in
Fig. 6.1.
Figure 6.1: Three-dimensional elliptical Luneburg lens.
The importance of considering the study of the elliptical Luneburg lens is
because we can do a study of a three-dimensional human lens, as shown in Fig.
6.2, with which we can generate a three-dimensional schematic human eye, in
which we will be able study the effect that causes the fovea to be on the visual axis.
Also, this study is important in the communications area because the elliptical
Luneburg lens can be used to replace conventional antenna systems.
In the fiber integrated optics, it is possible to produce a loss-free waveguide
using the Luneburg lens [44]. The problem with this waveguide will be that the
97
Figure 6.2: Three-dimensional human lens.
incident rays must always be parallels to its optical axis. However, we can do
an analysis using the off-axis rays to obtain a loss-free waveguide that does not
depend on the angle of incidence of the rays. Note that this idea generates a
critical angle of π.
In general, we have introduced a model of the human lens built as a composite
Luneburg lens whose gradient index function and its derivatives are continuous at
the equatorial plane. This allowes the construction of a more realistic schematic
model of the eye. We demonstrated its imaging capabilities of the whole schematic
eye for sources at infinity. In our model we have used conicoid surfaces but it can
be generalized to any surface with the condition of having continuos derivatives
at the equator plane. Our method allows us to obtain custom made GRIN distributions once the biometric parameters are given.
List of Figures
2.1
Fermat’s Principle. . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2.2
The linear gradient index medium. . . . . . . . . . . . . . . . . .
11
2.3
An arc lenght along the ray path. . . . . . . . . . . . . . . . . . .
12
2.4
Solution for the linear gradient index medium. . . . . . . . . . . .
14
2.5
The radial gradient index medium. . . . . . . . . . . . . . . . . .
14
2.6
Solution for the Sagittal plane. . . . . . . . . . . . . . . . . . . . .
16
2.7
Solution for the radial gradient index medium. . . . . . . . . . . .
17
3.1
The spherical gradient. . . . . . . . . . . . . . . . . . . . . . . . .
20
3.2
The Index Variation. . . . . . . . . . . . . . . . . . . . . . . . . .
21
3.3
Medium with radial symmetry . . . . . . . . . . . . . . . . . . . .
22
3.4
The path of rays in a medium with gradient index . . . . . . . . .
26
3.5
Rays on Horizontal axis . . . . . . . . . . . . . . . . . . . . . . .
28
3.6
The rays are hyperbolas when C > 0. . . . . . . . . . . . . . . . .
32
3.7
The rays are parabolas when C = 0. . . . . . . . . . . . . . . . . .
33
3.8
The rays are ellipses when C < 0. . . . . . . . . . . . . . . . . . .
33
3.9
Maxwell Solution . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
3.10 Maxwell Solution . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
3.11 Maxwell Solution Off-Axis . . . . . . . . . . . . . . . . . . . . . .
39
3.12 Luneburg Sphere Parameters . . . . . . . . . . . . . . . . . . . . .
41
3.13 The Luneburg Lens . . . . . . . . . . . . . . . . . . . . . . . . . .
49
3.14 The Generalized Luneburg Lens . . . . . . . . . . . . . . . . . . .
50
3.15 The Elliptical Luneburg Lens . . . . . . . . . . . . . . . . . . . .
56
3.16 Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
3.17 Ray tracing of The Luneburg Lens . . . . . . . . . . . . . . . . .
59
3.18 Ray tracing of The Elliptical Luneburg Lens . . . . . . . . . . . .
59
3.19 Ray tracing of The Elliptical Luneburg Lens with nc = 2, ns = 1,
µ = 2 and α0 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
100
List of Figures
4.1
Human eye and its optical elements [28]. . . . . . . . . . . . . . .
62
4.2
The structure of the cornea [28]. . . . . . . . . . . . . . . . . . . .
64
4.3
The geometric shape of the lens. za and zp are differents. . . . . .
66
4.4
The layers at the back of the human eye [28].
. . . . . . . . . . .
69
4.5
The Emsley reduced schematic eye. . . . . . . . . . . . . . . . . .
70
4.6
The Gullstrand-Emsley schematic eye. . . . . . . . . . . . . . . .
71
4.7
The Le Grand full theoretical schematic eye. . . . . . . . . . . . .
71
4.8
The schematic eye is analytically derived from the refractive index
profile of the crystalline lens and anatomical measurements of a rat
eye. [2]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
72
4.9
Schematic eye model with Maxwell fish-eye spherical lens [9]. . . .
73
5.1
Geometry of the bi-spherical model showing the continuity of the
isoindical lines at the equator plane. . . . . . . . . . . . . . . . . .
77
Ray tracing through the proposed CML lens embedded in a medium
with refractive index of 1.336. a) Rays incident from an infinite
distance. Rays incidents from a finite distance of 250 mm with b)
za /zp = 0.6872 and c) za /zp = 0.7185. . . . . . . . . . . . . . . . .
79
The GRIN distribution and the GRIN profiles for each lens with
different ratio za /zp . . . . . . . . . . . . . . . . . . . . . . . . . .
80
5.4
The GRIN profiles for each lens with different ratio za /zp . . . . .
81
5.5
Trajectory of the inner rays for each CML lens with different ratio
za /zp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
The GRIN distribution and the GRIN profiles for each lens with
the same ratio za /zp and different central refractive index. . . . .
83
The GRIN profiles for each lens with the same ratio za /zp and
different central refractive index. . . . . . . . . . . . . . . . . . . .
84
Ray tracing through the proposed CML lens embedded in a medium
with a refractive index of 1.336 and a constant ratio za /zp = 0.7036.
Rays incidents from a finite distance of a) Infinity, b) 2000 mm,
c)1000 mm, d) 500 mm, and e)250 mm. . . . . . . . . . . . . . . .
85
5.2
5.3
5.6
5.7
5.8
List of Figures
5.9
5.15
Trajectory of the inner rays for each CMLL with constant ratio
za /zp = 0.7036. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The GRIN profiles for each lens with the same ratio za /zp and
different central refractive index. The human lens is 20 years old.
Negative spherical aberration from the human lens. . . . . . . . .
Positive spherical aberration due to the cornea. Rays incident from
infinity distance, the anterior and posterior surfaces of the cornea
are immersed in a refractive index of 1 and 1.336, respectively. . .
Schematic eye model with CML lens. . . . . . . . . . . . . . . . .
Examples of Schematic model eyes with CML lens: a) Incident rays
coming from a 250 mm distance and b) Incident rays coming from
a 500 mm distance. Both Schematic eyes have a relation of 0.7450.
Schematic eye when rays are off-axis. . . . . . . . . . . . . . . . .
6.1
6.2
Three-dimensional elliptical Luneburg lens. . . . . . . . . . . . . .
Three-dimensional human lens. . . . . . . . . . . . . . . . . . . .
5.10
5.11
5.12
5.13
5.14
101
86
87
89
90
90
92
93
96
97
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