complete gradient refractive index lens schematic human ... · complete gradient refractive index...

$: Complete Gradient Refractive Index Lens Schematic Human ... · Complete Gradient Refractive Index Lens Schematic Human Eye Model by Jesús Emmanuel Gómez Correa Adissertationsubmittedinpartialfulﬁllment$
Complete GradientRefractive Index LensSchematic Human Eye

Modelby

Jesús Emmanuel Gómez CorreaA 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 lonecesario, continúa

haciendo lo posible; y derepente te encontraráshaciendo lo imposible".

San Francisco de Asís

Abstract

A new theoretical schematic model of the human eye is introduced. The newmodel considers that the human eye crystalline is a gradient index lens composedby two oblate half spheroids of different heights. By modifying the sphericallysymmetric Luneburg model for a gradient index lens, we created a model for theanterior and posterior half spheroids matching the corresponding geometric andgradient index boundary conditions at the plane of fusion of the spheroids. Wetested the imaging capabilities of our model and found that it is more realisticcompared with those reported in the literature concluding that the gradient indexis dynamic and cannot be modeled by one single explicit equation.

Resumen

Un nuevo modelo teórico del ojo humano es introducido. El nuevo modelo consid-era 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 modi-ficar el modelo de la lente simétrica de Luneburg esférica, creamos un modelo paralas esferoides anterior y posterior, las cuales se ajustan en el eje de fusión con lascondiciones de frontera correspondientes a la geometría y al índice de refracción.Hemos probado las capacidad de formación de imágenes en el ojo completo y en-contramos 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 poruna 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 durantetoda mi vida pero sobretodo por ser el mejor papá de todo el universo yenseñarme a ser el hombre que soy hasta el día de hoy. Porque gracias a susconsejos de todos los días, cariño, dedicación, caricias y sobre todo por suamor he logrado cumplir mis metas y sueños que me he propuesto. Esperoalgún día llegar hacer un papá como tu lo eres conmigo. Papá Te Amo comono tienes idea.

• Victoria Correa: Gracias por buscar siempre que esté bien en todos losaspectos y por ser la mejor mamá de todo el universo y llevarme de la manodía a día sin importar que sucediera y por comprenderme durante toda mivida. Gracias a sus caricias, por cuidarme, por aconsejarme y por buscar mibienestar, pero sobre todo por su gran amor. Espero siempre darle alegríassin importar lo que suceda. Mamá Te Amo como no tienes idea.

• Iker Emmanuel: Por darme la alegría para poder seguir adelante y ser mifuente de inspiración. Te amo demasiado.

• Anel: Gracias por todo lo bueno y lo malo que me has dado a lo largo deestos años, pero sobretodo muchas gracias por darme lo mejor que tengo enmi vida. Te quiero mucho.

• Ian y Lars: Por ser como son conmigo, por quererme y por compartir horasjugando Beisbol y por todas las alegrías y momentos bonitos a mi lado. Losquiero mucho.

• A mis hermanas:

Conchi y Blanca: Por hacerme la vida más feliz y estar conmigo cuandomás 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 mivida sería complicada en todos los aspectos. Gracias por demostrarme suamor día a día. Las Adoro y las Amo hermanitas.

• A mis hermanos:

Sergio y Arcenio: Gracias por ser los hermanos que nunca tuve, sin em-bargo en ustedes los he encontrado. Por considerarme su hermano aunquesea su cuñado, por divertirnos y disfrutar de la vida con muchas sonrisas ysobre todo por quererme tanto. Los quiero demasiado hermanitos.

• A mis sobrinos:

Cheito, Katy, Karol, Arcenito y Ale: Por ser mi alegría de todos losdías y compartir hermosos momentos a mi lado y por demostrarme cuantome quieren cada día que pasa. Los adoro como no tienen idea.

• Dr. Adrián Carbajal: Gracias por sus enseñanzas y por la gran calidadque tienen sus clases, porque gracias a estas clases yo me motivé a seguiradelante en esta área. Muchas Gracias, por todo el conocimiento que me hacompartido a lo largo de estos años de conocernos.

• Dr. Jesús Rogel-Salazar: Muchas gracias por apoyarme en el artículo deguías de ondas y por ayudarme en los detalles de Latex para que esta tesisfuera posible.

• Dr. David Sánchez de la Llave: Muchas gracias por apoyarme durantetodo el doctorado, gracias por los comentarios a mi trabajo y gracias porbuscar lo mejor para mí en cuestiones académicas. Muchas gracias Doctor.

• Barbara Pierscionek: Thank you very much for your help to improve thepapers of this thesis and to believe that this work is important in the visualoptics.

• A los doctores: De la Llave, Cornejo, Arrizón, Balderas y

xi

Malacara: Por tomarse el tiempo de revisar esta tesis, de hacerme co-mentarios para mejorar este trabajo y por aceptar ser mis sinodales.

• A mis amigos:

Julio Ramirez San Juan: Gracias Julio por esas largas horas que nospasamos hablando de box y de fútbol, gracias por tus comentarios tan alen-tadores que me ayudaron ha seguir adelante en gran camino que fue desdela maestría hasta el doctorado. Muchas gracias y sabes que aquí tienes unamigo.

Sandra: Por reírnos de cualquier cosa, por ayudarme, aconsejarme y porser una de mis mejores amigas de una buena parte de mi vida. GraciasSandrita 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 díase va haciendo más Grande y por la gran recomendación que me diste parasalir a tiempo del doctorado.

Julio García y Karla Sánchez: Gracias por dejarme compartir mis ocur-rencias, alegrías y tristeza con ustedes. Muchas Gracias a los dos y portantas salidas a comer, cenar y al cine. Los quiero mucho.

Marco Canchola: Por darme tu apoyo cada día y por compartir muybuenos momentos conmigo y hacerme sentir que tengo un muy buen amigode 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 elINAOE. Gracias mi buen JP.

Jorge Ugalde: Gracias por tu amistad y por compartir tantas horas en elmismo cubo, un gran futuro te espera.

Juan Carlos: Gracias por las largas horas que hemos pasado riéndonos ydisfrutando 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 granamistad

Eicela: Por ser mi amiga y por todos los consejos que me ha dado en estos6 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 porfacilitarme la vida en el INAOE.

• CONACYT: Por otorgarme la beca de doctorado 235164, para que fueraposible obtener el grado.

• INAOE: Por darme las comodidades para poder trabajar a gusto y darmeel privilegio de estudiar un doctorado en esta reconocida institución.

Special Acknowledgment

Dr. Sabino Chávez-Cerda: Le quiero agradecer por las tres tesis que me hadirigido en estos 8 años. Pero sobretodo le quiero agradecer por todos los consejosque me ha dado tanto personales como académicos. Gracias por enseñarme que elconocimiento no se mide con una regla de 30 centímetros si no que el conocimientose mide con la regla más grande que se tiene.

Después de estos 8 años de conocernos yo no lo considero mi asesor, yo loconsidero un muy buen amigo en mi vida. Muchas Gracias por todo, espero quesigamos colaborando como hasta el día de hoy lo hemos hecho. Muchas GraciasDr. Sabino.

Contents

i

Abstract iii

Resumen v

vii

Acknowledgments/ Agradecimientos ix

Special Acknowledgment xiii

1 Preface 1

2 Gradient Index Media 52.1 The Ray Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 The Linear Gradient Index Medium . . . . . . . . . . . . . . . . . 102.3 The Radial Cylindrical Gradient Index Medium . . . . . . . . . . 142.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 The Spherical Luneburg Lens 193.1 The Spherical Gradient Index Medium . . . . . . . . . . . . . . . 193.2 The Ray Integral Equation . . . . . . . . . . . . . . . . . . . . . . 22

3.2.1 Generalized Snell Law for Inhomogeneous Media withSpherical Symmetry . . . . . . . . . . . . . . . . . . . . . 26

3.3 Rays in a medium with n (r) =√C + 1

r. . . . . . . . . . . . . . 29

3.4 Maxwell’s Fisheye . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.5 The Luneburg Lens . . . . . . . . . . . . . . . . . . . . . . . . . . 403.6 The Generalized Luneburg Lens . . . . . . . . . . . . . . . . . . . 493.7 The Elliptical Luneburg Lens . . . . . . . . . . . . . . . . . . . . 52

xvi Contents

3.7.1 Linear Transformation of The Spherical Luneburg Lens . . 533.7.2 GRIN of The Elliptical Luneburg Lens . . . . . . . . . . . 55

3.8 Ray Tracing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.9 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4 The Human Eye As An Optical System 614.1 The Human Eye . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4.1.1 Refracting components of the human eye: Cornea and Lens 634.1.2 Pupil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.1.3 Retina . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

4.2 Schematics Eye . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5 Schematic Eye with Composite Luneburg Crystalline 755.1 Composite Modified Luneburg Lens . . . . . . . . . . . . . . . . . 765.2 Schematic Luneburg Eye . . . . . . . . . . . . . . . . . . . . . . . 885.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

6 Conclusions and Future Work 95

List of Figures 99

Bibliography 103

Chapter 1

Preface

Some laws of optics have been known for the centuries, for example, the Greekswere aware with some of the properties of light, they understood the law of reflec-tion. However, they were unable to understand the nature of the eye because theeye has an extraordinary complexity. They believed that light was emitted by theeye 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 tothe eye and not from the eye to the object [1].

Nowadays, the human eye is considered an optical system and its propertiesare studied by the Visual Optics. Many studies from the Visual Optics establishthat the eye is very complex due to the fact that its refractive surfaces are notstrictly spherical and its lens has a gradient refractive index (GRIN).

At the present time, one of the challenges in Visual Optics is creating aschematic model of the human eye. Since the GRIN profile may change fromone individual to another, the human lens has been the most challenging task inorder to have a complete and realistic schematic model.

Many generic expressions for the refractive index based in biometric data ofanimal and human lens have been proposed over the years that provide a goodestimation of the actual GRIN distribution [2, 3, 4]. In some models the anteriorand posterior faces are considered to be symmetric [2] while in more recent modelsa realistic asymmetry of the faces is taken into account [3, 4]. In the latter, theGRIN is described by two different equations with respect to a plane or a curvedsurface that intersects the human lens at its equator [3, 4, 5]. A drawback is thata ray (or its derivative) travelling in the proposed GRIN distribution may undergoa discontinuity at any of such surfaces.

2 Chapter 1. Preface

The most famous lens with a gradient refractive index is the Spherical Luneb-urg 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 withspherical geometry with normalized unitary radius and stigmatic property thatfocuses a sphere into a sphere [7]. As a particular example, Luneburg solved theproblem for incident rays at the anterior surface coming from infinity (infinitesphere) focused at the opposite side surface of the spherical lens (sphere withradius r = 1).

The importance of this lens in visual optics is due that four decades ago itwas proposed that the human lens can be studied as a Luneburg lens [8]. It isvery important to say that the Maxwell’s Fisheye Lens also has been proposedfor represent the GRIN of the human lens [9]. However, the Maxwell’s Fisheyehas the drawback that only points within or on the surface of the lens are sharplyimaged while that the lens proposed by Luneburg images sharply every parallelbundle of rays incident on the outer surface of the lens [10].

In this work we propose a more realistic theoretical schematic eye using theidea that the human lens can be represented as a composite asymmetric Luneburglens calculating a continuos GRIN function from the anterior to the posteriorconicoid faces. Our model takes into account the obliquity of the rays comingfrom the cornea illuminated by a source at infinity. We have used the biometricparameters provided by A. M. Rosen, et al. [11] but our method allows obtainingcustom made GRIN distributions given the corresponding biometric parameters.

This work is divided in six chapters: The Chapter 1 is this introduction. In theChapter 2 we make a analysis for the gradient mediums with planar and cylindricalgeometry. Right after, the spherical gradient index medium is explained and westudy fully the classical Luneburg lens and the Elliptical Luneburg Lens in theChapter 3. The Chapter 4 explains the human eye as an optical system, wherethe optical components are described. Also, in this Chapter the most importantsschematic human eye models are presented and its properties and its drawbacksare studied. In the Chapter 5 we introduce a new model of the human lensand 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 ourinvestigations.

Chapter 2

Gradient Index Media

Gradient index media have many applications in telecommunications to modelslim antennas and in Visual Optics to model the human eye lens. Because ofthis, it is necessary to start by defining: what is a gradient index medium?. Thegradient index medium is an inhomogeneous medium in which the refractive indexvaries from point to point [12].

The gradient index type depends on the geometry of the medium is the gradientindex type. The three most important types of gradient index are: The lineargradient, 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 GradientRefractive index, the rays are represented as curves. The shape of these curvesare given by the solution of the Ray Equation.

For this reason, we will begin this section by describing Fermat’s Principle inorder to define the ray equation so that we will be able to analyze the differenttypes GRIN types.

2.1 The Ray Equation

If we take any two points P1 and P2, these points can be connected with aninfinite number of curves. If each curve travels a distance ds with a velocityv = c/n (x, y, z), where n (x, y, z) is the space dependent refractive index in agiven point and c represents the speed of light in free space. The time taken totransverse the geometrical path ds in a medium of refractive index n (x, y, z) is

6 Chapter 2. Gradient Index Media

given by

dt =n (x, y, z) ds

c(2.1)

where the distance ds is a small part of a curve between P1 and P2, as show inthe Fig. 2.1.

Now, if t represents the total time taken by the ray to transverse the pathfrom P1 to P2 along the curve, we have

t =1

c

∑i

ni (x, y, z) dsi (2.2)

dsi represents the ith ds along of a curve, to each dsi corresponds only oneni(x, y, z). We can observe that this equation represents the time for a discreterefractive 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 chang-ing by each point, the equation can be rewritten as

t =1

c

∫ P2

P1

n (x, y, z) ds (2.3)

2.1. The Ray Equation 7

from this equation is very easy see that the path length of ray is

L =

∫ P2

P1

n (x, y, z) ds. (2.4)

Once obtained this result we can define Fermat’s Principle: ”The actual raypath between two points is the one for which the optical path length is stationarywith respect to variations of the path". This represents that

δ

∫ P2

P1

n (x, y, z) ds = 0. (2.5)

where δ variation of the integral means that it is a variation of the path of theintegral such that the endpoints P1 and P2 are fixed.

The equation (2.5) represents that the ray path is an extremum and it may bemaxima, minimum or stationary. For example, if n (x, y, z) is a constant at eachpoint, i e, a homogenous medium, the rays are straight lines that correspond to aminimum value of the optical path connecting two points in the medium.

The trajectory of rays can be compared with the trajectory of the particles inclassical mechanics. The Hamilton’s principle in classical mechanics establishesthat the trajectory of a particle between times t1 and t2 is such that

δ

∫ t2

t1

Ldt = 0. (2.6)

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 isover the space. This difference can be eliminated if the infinitesimal arclenght dsis rewritten as

ds =

√(dx)2 + (dy)2 + (dz)2 = dz

√(dx

dz

)2

+

(dy

dz

)2

+ 1 (2.7)


if x = dx/dz and y = dy/dz, then

ds = dz√x2 + y2 + 1 (2.8)

where dz is equivalent to dt in the Eq. (2.6) and in this case, Eq. (2.5)

δ

∫ P2

P1

n (x, y, z)√

1 + x2 + y2dz = 0. (2.9)

From this equation we can define an optical Lagrangian as

L (x, y, x, y, z) = n (x, y, z)√

1 + x2 + y2. (2.10)

As in classical mechanics, the solution of this problem must satisfy the Euler-Lagrange equations, in this case these equations are given by

d

dz

(∂L∂x

)=∂L∂x

(2.11)

andd

dz

(∂L∂y

)=∂L∂y. (2.12)

If we substitute Eq. (2.10) into Eq (2.11), we obtain

d

dz

[n (x, y, z) x√1 + x2 + y2

]=√

1 + x2 + y2∂n (x, y, z)

∂x(2.13)

from Eq. (2.7) is easy to see that Eq. (2.13) can be reduced to

d

ds

[n (x, y, z)

dx

ds

]=∂n (x, y, z)

∂x(2.14)

becaused

ds=

1√1 + x2 + y2

d

dz. (2.15)

2.1. The Ray Equation 9

Similarly for y and z components, we have

d

ds

[n (x, y, z)

dy

ds

]=∂n (x, y, z)

∂y(2.16)

andd

ds

[n (x, y, z)

dz

ds

]=∂n (x, y, z)

∂z(2.17)

If we define ~r as a position vector, we can obtain a vector equation which containsequations (2.14), (2.16) and (2.17), i e,

d

ds

[n (x, y, z)

d~r

ds

]= ∇n (x, y, z) (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 describesthe phasefront of a wave. In order to obtain the Ray Equation from the Eikonalequation, 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 functiondescribing the phasefront of the wave. Now, if we defined a unit vector normal tothe phase fronts and tangent to the light ray s, i.e., along the ray, then

∇S (x, y, z) = n (x, y, z) s (2.20)

where s = d~r/ds, hence

d

ds[∇S (x, y, z)] =

d

ds

[n (x, y, z)

d~r

ds

](2.21)


since as d/ds = s · ∇, this equation can be represented as

d

ds

[n (x, y, z)

d~r

ds

]= (s · ∇)∇S (2.22)

We want to know the direction of the rays, hence

∇[[∇S (x, y, z)]2

]= ∇n2 (x, y, z) (2.23)

using∇[[∇S (x, y, z)]2

]= 2n (x, y, z) (s · ∇)∇S (2.24)

and∇n2 (x, y, z) = 2n (x, y, z)∇n (x, y, z) (2.25)

we obtain(s · ∇)∇S = ∇n (x, y, z) (2.26)

from the Eq. (2.22)d

ds

[n (x, y, z)

d~r

ds

]= ∇n (x, y, z) (2.27)

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 indexmedium 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 continuousway 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 isshow 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

n2 (x) =

n2

0 − αx x > 0

n20 x < 0

(2.28)

and in the y-axis by

n2 (y) =

n2

0 − 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 tosolve the ray equation, which is

d

ds

[n (x, y, z)

d~r

ds

]= ∇n (x, y, z) (2.30)

this equation can be rewritten as a set of three equations, shown in equation


(2.31).dds

[n (x, y, z) dx

ds

]= ∂n(x,y,z)

∂xdds

[n (x, y, z) dy

ds

]= ∂n(x,y,z)

∂ydds

[n (x, y, z) dz

ds

]= ∂n(x,y,z)

∂z.

(2.31)

If the refractive index does not depend on z, we have

d

ds

[n (x, y)

dz

ds

]=∂n (x, y)

∂z= 0. (2.32)

implying that

n (x, y)dz

ds= β (2.33)

where β is an invariant of the ray path. From Fig. 2.3 we can observe that

Figure 2.3: An arc lenght along the ray path.

dz

ds= cos θ (x, y) (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 tothe 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 13

asd2x

dz2=

1

2β2 (x, y)

∂n2 (x, y)

∂x(2.36)

d2y

dz2=

1

2β2 (x, y)

∂n2 (x, y)

∂y(2.37)

dβ (x, y)

dz=

1

2β (x, y)

∂n2 (x, y)

∂z= 0. (2.38)

If the linear refractive index is given by Eq. (2.28), then Eq. (2.36), for x > 0,becomes

d2x

dz2= − α

2β2(2.39)

and the general solution is

x (z) = − α

4β2z2 + C (2.40)

where C = Constant, β = n0 cos θ0 and θ0 = θ (0, 0). In this case, θ0 is theincident 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 constantC, takes the value of

C = z tan θ0 (2.41)

thereforex (z) = − α

4β2z2 + z tan θ0. (2.42)

We can observe in the Fig. 2.4 that if x < 0 the trajectory of the ray is astraight 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 thex-axis, for this reason, it is not necessary solve it.


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 maximumat the cylinder axis and decreases continuously from the axis to the periphery alongthe transverse direction in such a way that the isoindicial surfaces are concentriccylinders 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 15

n2 (r) =

n21

[1− 2∆

r2

a2

]= n2

1

[1− 2∆

x2 + y2

a2

]0 < r < a

n22 = n1 [1− 2∆] r > a

(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 equationis 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

d2xdz2

+ Γ2x = 0d2ydz2

+ Γ2y = 0(2.44)

where Γ = n1

√2∆

aβand 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 isnecessary to impose initial launching conditions [13], like these

x (z = 0) = 0

y (z = 0) = 0dxdz|z=0 = 0

dydz|z=0 = 0.

(2.46)

These launching conditions make that the meridional rays are confined in thesagittal plane; we may mention that the meridional rays are defined such that theyare confined in a plane and intersect the sagittal plane. Using these conditions


the solution is of the form

x = aβ

n1

√2∆

tan θ0 sin[ √

2∆a cos θ0

z]

= a sin θ0√2∆

sin[ √

2∆a cos θ0

z]

y = 0(2.47)

where θ0 is the incident angle of the ray at the point x = 0 and y = 0. Thesolution 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 cylin-drical gradient index and the linear gradient in the same plane is that in the firstthe gradient is given in an interval of [−a, a] while in the second one the gradientis in an interval of [0, a], where a is any number, for this reason the solutions aredifferent.

The most complete solution is when we impose the initial launching conditionsas

x (z = 0) = a′

y (z = 0) = 0dxdz|z=0 = 0

dydz|z=0 = tan θ′

(2.48)

where a′ is the distance on the x axis, which was launched the ray in the y−z planeand θ′ is the angle that the ray makes with the z axis. With these conditions, oneobtains

x = a′ cos Γz

y = a′ sin Γz(2.49)

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 remainconfined to a plane and represent the solution for the radial cylindrical gradientindex medium, as show in Fig. 2.7.

2.4 Conclusions

The importance of this Chapter is that we have reviewed the theory of propagationof the rays in gradient media and with this we have presented the ray equationwhich 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 ainhomogenous lens with a gradient index that can be represented as a sphericalgradient index.

Chapter 3

The Spherical Luneburg Lens

This chapter provides a description of light propagation through the sphericalgradient medium. We will analyse the spherical Luneburg lens and Maxwell’sFish Eye who are the most famous examples of lenses with spherical gradientindex. The difference between the spherical Luneburg lens and the Maxwell’sFisheye is the representation of the mathematical approach of the refractive indexand that the Maxwell fish eye lens has a drawback; only points within or on thesurface of the lens are sharply imaged. The lens proposed by Luneburg imagessharply 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 toknow 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 beenproposed to represent the refractive gradient index of the crystalline. It is veryimportant to mention that it has been proposed, but for the case of the Luneburglens does not exist a complete analysis.

For this reason, in this chapter we will focus our attention to study the spher-ical 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 indexis constant. The smaller sphere has the highest refractive index and it decreasesfrom the inner sphere to a sphere of radius a, then we can say that we have aspherical 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 mathematicalexpressions. An interesting case is when we define the spherical gradient index as

n2 (r) =

C +1

r0 < r < a

n22 r > a

(3.1)

where r =√x2 + y2 + z2, C is a constant, n2 = C − 1

a, 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 representedin the Fig. 3.2 for different values of C. Notice that the spherical gradient indexhas a singularity at r = 0, for this reason, the radius is r > 0 in Eq. (3.1). Thesingularity at r = 0 is not a problem for the ray tracing in this medium, becausewe will find a ray integral equation that depend on a new function ρ (r). Thisfunction 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 sectionwe will answer this question and later we will solve the refractive gradient index

3.1. The Spherical Gradient Index Medium 21

Figure 3.2: The Index Variation.


given by the Eq. (3.1).

3.2 The Ray Integral Equation

We know from the literature about media with radial symmetry that the raysare confined to a single plane [13]. In this work we are interested in knowing thesolutions the a x− y plane, for this reason it is possible to reduce this problem ofthree variables into a two variables problem.

Now, if we consider a continuous medium with a refractive index that hasradial symmetry n (r), as shown in the Fig. 3.3, we can easily know the rayintegral equation.

Figure 3.3: Medium with radial symmetry

The ray integral equation is obtained from the Eikonal equation, then we canrewrite 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 formx = r cos θ

y = r sin θ(3.3)

andr =

√x2 + y2 (3.4)

we obtained (∂ψ

∂r

)2

+1

r2

(∂ψ

∂θ

)2

= n2(r) (3.5)

This equation can be solved with the method of Separation of Variables, then theEq. (3.5) can be rewritten as

∂ψ

∂θ= r

√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∫dψ(θ) =

∫Kdθ∫

dψ(θ) = K∫dθ

ψ(θ) = Kθ

(3.7)


and the right of the equation does not depend on θ, we have

K = r

√n2(r)−

(dψ(r)dr

)2

K2 = r2

[n2(r)−

(dψ(r)dr

)2]

K2

r2− n2(r) = −

(dψ(r)dr

)2

dψ(r)dr

=√n2(r)− K2

r2

dψ(r) =√n2(r)− K2

r2dr∫

dψ(r) =∫ rr0

√n2(r)− K2

r2dr

ψ(r) =∫ rr0

√n2(r)− K2

r2dr

(3.8)

if the solution is given byψ = ψ(θ)± ψ(r) (3.9)

therefore

ψ = Kθ ±∫ r

r0

√n2(r)− K2

r2dr (3.10)

where K is for the moment an arbitrary constant.

We know that the Eikonal equation describes the phase front of a wave, andthe rays are lines that are normal to these wavefronts, showing the direction ofenergy flow at one particular point, for this reason it is necessary to apply theJacobi Theorem to obtain the integral ray equation. This theorem is representedby

∂ψ

∂K= α (3.11)


applying the Jacobi theorem, we have

∂ψ∂K

= ∂∂K

[Kθ ±

∫ rr0

√n2(r)− K2

r2dr

]∂ψ∂K

= ∂∂K

[Kθ]± ∂∂K

∫ rr0

√n2(r)− K2

r2dr

∂ψ∂K

= θ ∂K∂K±∫ rr0

∂∂K

√n2(r)− K2

r2dr

∂ψ∂K

= θ ±∫ rr0

∂∂K

[(n2(r)− K2

r2

)1/2]

∂∂K

[n2(r)− K2

r2

]dr

∂ψ∂K

= θ ±∫ rr0

(n2(r)− K2

r2

)−1/2 [2Kr2

]dr

∂ψ∂K

= θ ±∫ rr0

2Kr2√

n2(r)−K2

r2

dr

∂ψ∂K

= θ ±∫ rr0

2K

r2√n2(r)−K2

r2

dr

∂ψ∂K

= θ ± 2K∫ rr0

drr2

r

√n2(r)r2−K2

∂ψ∂K

= θ ± 2K∫ rr0

dr

r√n2(r)r2−K2

(3.12)

α = θ ± 2K∫ rr0

dr

r√n2(r)r2−K2

α− θ = ±2K∫ rr0

dr

r√n2(r)r2−K2

α2− θ

2= ±K

∫ rr0

dr

r√n2(r)r2−K2

(3.13)

we definedα2

= θθ2

= θ0

(3.14)

the rays of light in the x− y plane are obtained by the following equation

θ − θ0 = ±K∫ r

r0

dr

r√n2(r)r2 −K2

(3.15)

The latter equation is known as the integral ray equation. The integration con-stants θ0 and r0 are the initial coordinates of the ray on the spherical refractivegradient index.

In this analysis, the constant K is very important because it represents theray direction at the point (r0, θ0), then we need to know the value of K. In the


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 withgradient index n = n (r) are planar curves in a plane containing the origin. Wecan consider a position vector −→r along the path of the curve that represents a lightray inside the spherical refractive gradient index, and a unit vector −→s tangent tothe 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 wederive with respect to s, we obtain

d

ds[−→r × [n (r)−→s ]] =

d−→rds× [n (r)−→s ] +−→r × d

ds[n (r)−→s ] (3.16)

butd−→rds

= −→s (3.17)

and from Eq. (2.27)d

ds

(nd−→rds

)=−→5−→n (3.18)


therefored

ds[−→r × [n (r)−→s ]] = −→s × [n (r)−→s ] +−→r ×

−→5−−→n (r) (3.19)

it is very easy to see that the first term is 0 and the equation reduces to

d

ds[−→r × [n (r)−→s ]] = −→r ×

−→5−−→n (r) (3.20)

as n (r) is a function that only depend on r

−→5−−→n (r) =

dn (r)

dr

−→rr

(3.21)

from this equation it is clear to see

d

ds[−→r × [n (r)−→s ]] = 0 (3.22)

this leads us to−→r × [n (r)−→s ] = cte (3.23)

a vector product that can be expressed as

−→r × [n (r)−→s ] =‖ −→r ‖‖ n (r)−→s ‖ sinϕ (3.24)

where ϕ is the angle between the vectors −→r and −→s , and as the vector −→s is a unitvector, we have

‖ −→r ‖= r

‖ n (r)−→s ‖= n (r)(3.25)

if we substitute these equations into Eq. (3.25)

−→r × [n (r)−→s ] = rn (r) sinϕ (3.26)

hencern (r) sinϕ = cte (3.27)


and this constant is K, i.e.,K = rn (r) sinϕ (3.28)

The Eq. (3.28) is known as the generalized Snell law for inhomogeneous mediawhit spherical symmetry. This demonstrates that the path of rays in a mediumwith gradient index n = n (r) are planar curves because K is always a constantwith 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 maximunpoint of incidence can be obtained when r = 1, n = 1 and ϕi = π

2. From the

generalized Snell’s law, we have that K = 1 and the minimum point is when weare on the axis of propagation, which leads us to have ϕi = 0 and for consequenceK = 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 ison the negative side on the same axis. We observe that the incident angle relativeto the normal is given by θ − π, i.e., θ0 = π, ϕ = α0, and K = n0r0 sinα0, by

3.3. Rays in a medium with n (r) =√C + 1

r29

substituting these values in Eq. (3.15), we obtain

θ − π = ±n0r0 sinα0

∫ r

r0

dr

r√n2(r)r2 − n2

0r20 sin2 α0

(3.30)

for α0 <π2, we have

θ − π = +n0r0 sinα0

∫ r

r0

dr

r√n2(r)r2 − n2

0r20 sin2 α0

(3.31)

and for α0 >π2, we have

θ − π = −n0r0 sinα0

∫ r

r0

dr

r√n2(r)r2 − n2

0r20 sin2 α0

(3.32)

If we considered that α0 is given from 0 to π2, we obtain that

0 ≤ K ≤ n0r0 (3.33)

3.3 Rays in a medium with n (r) =√C + 1

r

We have that the integral ray equation is

θ − θ0 = ±K∫ r

r0

dr

r√n2(r)r2 −K2

(3.34)

we can define a function ρ (r), that has the form

ρ(r) = n(r)r. (3.35)

ifn2(r) = C +

1

r(3.36)

thenρ2(r) = Cr2 + r (3.37)


by substituting Eq. (3.37) into Eq. (3.34) and we obtain

θ − π = −K∫ r

r0

dr

r√Cr2 + r −K2

(3.38)

in this case θ0 = π, because we are considering that origin of ray is on the hor-izontal axis. We can find the solution of the integral when we rewrite the lowerterm as

√Cr2 + r −K2 = r

√C +

1

4K2

√√√√[1−(Kr− 1

2K

)2

C + 14K2

](3.39)

and by making a change of variable

z =Kr− 1

2K√C + 1

4K2

(3.40)

dr = −r2√C + 1

4K2

Kdz (3.41)

we observe that

θ − π = −∫ z

z0

Kr2√C+ 1

4K2

Kdz

r2

√C + 1

4K2

√1− z2

(3.42)

thereforeθ − π = −

∫ z

z0

dz√1− z2

(3.43)

the solution is very easy if we assume

z = sinu (3.44)

dz = cosudu (3.45)

then ∫dz√

1− z2=

∫cosudu√1− sin2 u

=

∫du = u = arcsin z (3.46)


r31

henceθ − π = −

∫ z

z0

dz√1− z2

= arcsin z0 − arcsin z (3.47)

Now, if arcsin z0 = β − π2is 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 =2K2

1 +√

4CK2 + 1 cos (θ − β)(3.49)

the latter equation determines the type of curves and these curves can be easilyseen if us assume that

β = 0

x = r cos (θ)

y = r sin (θ)

r =√x2 + y2

(3.50)

with equations (3.50) we find from the Eq. (3.49)

4C2

(x− 1

2C

√1 + 4CK2

)2

− C

K2y2 = 1 (3.51)

ory2 = K2

[4Cx2 − 4x

√1 + 4K2C + 4K2

]. (3.52)

Equation 3.51 tell us that the curves are conic sections and that a given C allthese conics have the same principal axes A = 1/2C. From this equation we canobserve that the eccentricity is [6]

e =1

2C

√1 + 4K2C =

√1

4C2+K2

C. (3.53)

The type of these conics is determined by the value of C, i.e.:


If C > 0: Equation 3.51 represents hyperbolas with the same principal axisA = 1

2C, 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 focalpoint, as shown in Fig. 3.7.

If C < 0: Equation 3.51 represents ellipses with the same principal axis A =1

2|C| and x = y = 0 as common focal point, as shown in Fig. 3.8. In this case, thelight rays cannot penetrate into the region r > − 1

2C.

In this section our aim was to find the shape of the rays, but it is important tosee that the refractive index given by Eq. (3.36) can be transformed into anothervital refractive index. By recalling the Eikonal equation

ψ2x + ψ2

y + ψ2z = n2(r) (3.54)

and substituting the Eq. (3.36) in cartesian coordinates into the Eq. (3.54), weobtain

ψ2x + ψ2

y + ψ2z = C +

1√x2 + y2 + z2

(3.55)


r33

Figure 3.7: The rays are parabolas when C = 0.

Figure 3.8: The rays are ellipses when C < 0.


using Legendre transformations

ψ = ψ (x, y, z)

ω = ω (ξ, η, ς)

ψ + ω = xξ + yη + zς

(3.56)

whereξ = ψx

η = ψy

ς = ψz

x = ωξ

y = ωη

z = ως

(3.57)

it is very easy to see that the Eq. (3.55) is transformed into

ξ2 + η2 + ς2 = C +1√

ω2ξ + ω2

η + ω2ς

(3.58)

and this equation can be rewritten as

ω2ξ + ω2

η + ω2ς =

(1

−C + ξ2 + η2 + ς2

)2

(3.59)

if C = −1, so this equation is reduced to the Eikonal equation with a refractiveindex given by

n (ξ, η, ς) =1

1 + ξ2 + η2 + ς2(3.60)

orn (r) =

1

1 + r2. (3.61)

This optical medium is know as Maxwell’s Fisheye. A medium with thisrefractive index will be studied in next section.

3.4. Maxwell’s Fisheye 35

3.4 Maxwell’s Fisheye

The Maxwell’s Fisheye has been proposed to represent the refractive gradientindex of the crystalline [9], for this reason is important to study it. The gradientrefractive index of Maxwell’s fisheye is generally represented as

n =a

(b2 + 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 integralray equation

θ − π = K

∫dr

r√n2r2 −K2

(3.64)

substituting the gradient refractive index into Eq. (3.64) and by doing somealgebra, we obtain

θ − π =

∫K (1 + r2)

r√r2 −K2 (1 + r2)2

dr (3.65)

this integral can be solve by using

f(r) = arcsin

(r2 − 1

r

K√1− 4K2

)(3.66)

wheredf(r) =

K (1 + r2)

r√r2 −K2 (r2 + 1)2

dr (3.67)

we can observe that

θ − π =

∫K (1 + r2)

r√r2 −K2 (1 + r2)2

dr =

∫df(r) (3.68)


and the solution is

θ − π = arcsin

(r2 − 1

r

K√1− 4K2

)+ C (3.69)

where C is the integration constant. Equation (3.69) is rewritten as

r2 − r√

1− 4K2

Ksin (θ − π − C)− 1 = 0 (3.70)

C can be defined asC = β − 3π

2(3.71)

substituting the value of C and using the trigonometric identities sin(π2− θ)

=

cos θ and, cos θ = cos (−θ) into Eq. (3.70), we have

r2 − r√

1− 4K2

Kcos (θ − β)− 1 = 0 (3.72)

if β = 0, then

r2 − r√

1− 4K2

Kcos θ − 1 = 0 (3.73)

if we introduce cartesian coordinates into this equation, we have(x−√

1− 4K2

2K

)2

+ y2 =1

4K2(3.74)

and if R = 12K

, then (x−√R2 − 1

)2

+ y2 = R2. (3.75)

The form of the rays is given by Eq. (3.75). We can observe that this equationrepresents 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, asshown in the Fig 3.9.

So far, we have not said where the rays were originated, we only know that therays have a circular path. Now, we will consider that all rays are originated in a


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 cartesiancoordinates, we have(

x−√R2 − 1 cos β

)2

+(y −√R2 − 1 sin β

)2

= R2 (3.76)

where R = 12K

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, andthis leads us to the equation

x2 − 2x√R2 − 1cosβ − 1 = 0. (3.77)

It can be solved if we use

x0,1 =b±√b2 − 4ac

2a(3.78)

if a = 1, b = −2√R2 − 1 cos β and c = −1, then

x0,1 =√R2 − 1 cos β ±

√(R2 − 1) cos2 +1 (3.79)


Figure 3.10: Maxwell Solution

and the solutions are

x0 =√R2 − 1 cos β +

√(R2 − 1) cos2 +1

x1 =√R2 − 1 cos β −

√(R2 − 1) cos2 +1

(3.80)

the relationship between x0 and x1 is given by

x0x1 = −1 (3.81)

we can observe that this relationship is independent of the parameters R and βand this tells us that all rays that originate at x0 are intersected at a point x1,where

x1 = − 1

x0

. (3.82)

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


Figure 3.11: Maxwell Solution Off-Axis

P1 = (x1, y1), wherex1 = −x0

r20

y1 = −y0r20

(3.83)

the solution is sketched in Fig. 3.11.

From the latter solution we can observe that all the rays are focusing in oneonly 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 = − r1

r0.

These solutions are interesting, because the curves that represent the rays areinside the medium, i.e., the Maxwell’s Fisheye is not a lens, for this reason, itcannot be proposed to represent the refractive gradient index of the human lens.

The well known perfect optical Instrument is the Luneburg lens which is ourmain interest in this work and it will study in the next section.


3.5 The Luneburg Lens

The Luneburg lens is the answer to the problem posed by R. K. Luneburg in1944, this problem is stated as follows: If we have a refractive index with radialsymmetry, where the maximum radius is 1 and it is submerged in a refractiveindex n = 1 (air), What should be the refractive index of the medium where allincident rays focus at a single point?.

In order to solve this problem, we begin by considering a sphere with maximumradius 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 areconsidering some boundary conditions, i.e., the maximum radius (rm) is 1 and alsowe have n (rm) = 1. The first boundary condition (rm = 1) does not generate anyloss of generality of this problem solution, because r is converted into a normalizedvariable. The second boundary condition (n (rm) = 1) does not mean that thismedium is air or vacuum, this means that the refractive index of the elementconsidered is normalized with respect to the refractive index where the sphere issubmerged [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 pointr′0 and the point r′1 is the point where all rays are focused after passing throughthe sphere. The point P0 indicates the point where the incident ray enters at thesphere and the point P1 is the point at which the ray exits the same sphere. Theradius 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 r′0

to P0 and the ray P1 to r′1, because they are in a medium with constant refractiveindex.

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 angleθs. In the point P0 we have that

π = α0 + Ψi + π − θiθi = α0 + Ψi + π − π

θi = α0 + Ψi

(3.85)

and K in this point isK = 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 = arcsinK(3.87)

we observe thatΨi + Ψ = π

Ψi + arcsinK = π

Ψi = π − arcsinK

(3.88)


substituting the value of Ψi into Eq. (3.85)

Ψi + α0 + π − θi = π

Ψi + α0 − θi = 0

π − arcsinK + α0 − θi = 0

(3.89)

thereforeθi = π − arcsinK + α0. (3.90)

In the outter point P1, we have

θs + α1 + π −Ψs = π

θs = Ψs − α1 + π − πθs = Ψs − α1

(3.91)

but from the Generalized Snell’s Law in this point

Ψs = arcsinK (3.92)

and substituting the value of Ψs into Eq. (3.91)

θs = arcsinK − α1. (3.93)

The angular variation inside the sphere is given by

θs − θi = arcsinK − α1 − [π − arcsinK + α0]

θs − θi = arcsinK − α1 − π + arcsinK − α0

θs − θi = 2 arcsinK − π − [α0 + α1]

(3.94)

using the trigonometric identity

arcsin k + arccos k =π

2(3.95)


we haveθs − θi = − [2 arccosK − (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 themedium. With this equation, we can rewrite Eq. (3.96) as

θs − θi = − [2 arccosK + αT (K)] . (3.98)

From Eq. (3.33) is easy to find the K value range, because we know the value ofn0 and r0, then

0 ≤ K ≤ 1. (3.99)

It is possible to obtain from the ray integral equation, the variation from θi toθs, i.e.,

θs − θi = K

∫ r1

r0

dr

r√ρ2 −K2

(3.100)

to evaluate the integral is necessary to divide the variation of the angle into twovariations. The first variation is from r0 to r∗ and the second is r∗ to r1, this leadsto

θs − θi = K

∫ r∗

r0

dr

r√ρ2 −K2

−K∫ r1

r∗

dr

r√ρ2 −K2

(3.101)

but r0 and r1 are equal to 1, then

θs − θi = −2K

∫ 1

r∗

dr

r√ρ2 −K2

. (3.102)

Since Eq. (3.96) is equal to Eq. (3.102) we have

K

∫ 1

r∗

dr

r√ρ2 −K2

= arccosK +1

2αT (K) (3.103)


where 0 ≤ K ≤ 1.

Since the Luneburg lens is a stigmatic system is possible to find the deflexionfunction [7]. From the triangle given by the points r′0, P0, and the origin, we canobtain the value α0

1sinα0

=r′0

sin Ψi

sinα0 = sin Ψi

r′0

α0 = arcsin[

sin Ψi

r′0

]α0 = arcsin

[Kr′0

] (3.104)

and from the triangle given by the points r′1, P1, and the origin; we have that theα1 value is

1

sinα1

=r′1

sin (π −Ψs)(3.105)

butsin (π ± x) = ∓ sinx (3.106)

thensin (π −Ψs) = sin Ψs (3.107)

we obtain1

sinα1=

r′1sin Ψs

sinα1 = sin Ψs

r′1

α1 = arcsin[

sin Ψs

r′1

]α1 = arcsin

[Kr′1

] (3.108)

α0 and α1 only depend on K, then

α0 (K) = arcsin[Kr′0

]; α1 (K) = arcsin

[Kr′1

](3.109)

therefore, the deflexion function is given by

αT (K) = arcsin

[K

r′0

]+ arcsin

[K

r′1

]. (3.110)


Substituting this equation into Eq. (3.103),

K

∫ 1

r∗

dr

r√ρ2 −K2

= arccosK +1

2

[arcsin

(K

r′0

)+ arcsin

(K

r′1

)](3.111)

iff (K) = arccosK +

1

2

[arcsin

(K

r′0

)+ arcsin

(K

r′1

)](3.112)

we have

K

∫ 1

r∗

dr

r√ρ2 −K2

= f (K) . (3.113)

Since we want to know the refractive index of the medium, we will consider achange of variable of the form

τ = log r (3.114)

wheredτ =

dr

r(3.115)

we know that ρ is a function of r, but with the change of variable, ρ becomes afunction of τ , i.e., ρ = ρ (τ). With this change of variable, we have

K

∫ 0

−∞

dτ√ρ2 (τ)−K2

= f (K) . (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 withthe 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)

this leads us toτ (ρ) = log r (ρ) (3.118)


ifΩ (ρ) = τ (ρ) = log r (ρ) (3.119)

anddΩ (ρ)

dρ=dτ (ρ)

dρ=

1

r (ρ)

dr (ρ)

dρ(3.120)

we can rewrite this derivative as

Ω′ (ρ) dρ =dr (ρ)

r (ρ)(3.121)

whereΩ′ (ρ) =

dΩ (ρ)

dρ(3.122)

Equation (3.116) is transformed with these equations in

K

∫ 1

K

Ω′ (ρ) dρ√ρ2 −K2

= f (K) (3.123)

from the denominator we can see that the highest value of ρ is 1 and the smallestvalue is K, because 0 ≤ K ≤ 1. This is an of Abel’s type integral where thesolution is known [16], but we will find its solution using the Luneburg theoremwhich tells us [15]:

If the function f (K) is defined by the integral

K

∫ λ

K

Ω′ (ρ) dρ√ρ2 −K2

= f (K) (3.124)

in the interval 0 ≤ K ≤ λ, then Ω (ρ) is determined by the integral

Ω (λ)− Ω (ρ) =2

π

∫ 1

ρ

f (K)√K2 − ρ2

dK (3.125)

in the interval 0 ≤ ρ ≤ λ.


Using the Luneburg theorem we can solve the integral of Eq. (3.123), then

log (1)−log (r) =2

π

∫ 1

ρ

arccosK√K2 − ρ2

dK+1

π

∫ 1

ρ

[arcsin K

r0+ arcsin K

r1

]√K2 − ρ2

dK (3.126)

but as log (1) = 0

− log (r) ==2

π

∫ 1

ρ


dK +1

π

∫ 1

ρ

[arcsin K

r0+ arcsin K

r1

]√K2 − ρ2

dK. (3.127)

The first integral is easy to solve when we define

f (K) = arccosK (3.128)

where

f (K) = K

∫ 1

K

dρ

ρ√K2 − ρ2

= K

∫ 1

K

ddρ

log ρdρ√K2 − ρ2

= K

∫ 1

K

d log ρ√K2 − ρ2

(3.129)

and applying the Luneburg theorem, we have

− log ρ =2

π

∫ 1

ρ


(3.130)

therefore

− log (r) = − log ρ+1

π

∫ 1

ρ

[arcsin K

r′0+ arcsin K

r′1

]√K2 − ρ2

dK. (3.131)

We introduce a new function given by

ω (ρ, a) =1

π

∫ 1

ρ

arcsin ta√

t2 − ρ2dt (3.132)


we obtainlog

ρ

r= ω (ρ, r′0) + ω (ρ, r′1) (3.133)

but knowing that ρ = nr, then we will have that

n = e[ω(ρ,r′0)+ω(ρ,r′1)] (3.134)

andr = ρe−[ω(ρ,r′0)+ω(ρ,r′1)] (3.135)

Finally, if the solution for the ω (ρ, a) function is known, it is possible to solvethe system of equations given by Eqs. (3.134) and (3.135), and we will be able toknow the refractive index n (r).

For example, if we have a system where r′0 =∞ and r′1 = 1, the solutions forthe ω (ρ, a) function when a = r′0 and a = r′1 are

ω (ρ,∞) =1

π

∫ 1

ρ

arcsin t∞√

t2 − ρ2dt = 0 (3.136)

and

ω (ρ, 1) =1

π

∫ 1

ρ

arcsin t√t2 − ρ2

dt =1

2log[1 +

√1− ρ2

]. (3.137)

Substituting these expressions into the Eqs. (3.134) and (3.135), we have that

n =√

1 +√

1− ρ2

r = ρn

(3.138)

If we solve this system of equations, we obtain the refractive index n (r) given by

n (r) =√

2− r2 (3.139)

where r =√x2 + y2.

A sphere with this refractive index makes that all rays from infinity focusedon the point r′1 = 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 TheLuneburg Lens.

Figure 3.13: The Luneburg Lens

The Luneburg lens can be generalized to focus the rays outside or inside thelens because Luneburg’s original lens is a limiting case in which r′0 is at infinityand r′1 is on the surface of the lens. We are interested in the first case because inthe eye all rays focus on the retina rather than inside the crystalline. Recallingthat 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 "GeneralSolutions of the Luneburg Lens Problem", where he demonstrates the possibilityof 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-


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 amore complete and explicit analysis of the generalized Luneburg’s problem [17].

The initial conditions to solve this problem are: a refractive index n (r) mustexist in the interval 0 ≤ r ≤ 1, and the external foci are given by r′0 > 1 andr′1 > 1.

To find the refractive index n (r) we start from Eq. (3.113),

K

∫ 1

r∗

dr

r√ρ2 −K2

= f (K) . (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 r′0 will be totally reflected. Then,the full aperture of the lens will not be used [10]. To avoid this problem, let usintroduce a new refractive index in the interval a ≤ r ≤ 1, which is given by

n (r) =P (r)

r≥ 1

r(3.141)

if P (r) is known and a ≤ 1, the solution of the Eq. (3.140) is

K

∫ a

r∗

dr

r√ρ2 −K2

= f (K)− F (K) (3.142)

3.6. The Generalized Luneburg Lens 51

where

F (K) = K

∫ 1

a

dr

r√P 2 (r)−K2

(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 theΩ (ρ) function (Eq. (3.119)) to solve Eq. (3.142), i.e., the integral reduces to

K

∫ 1

K

Ω′ (ρ) dρ√ρ2 −K2

= f (K)− F (K) (3.144)

using the Luneburg theorem, the solution of this integral is

− loga

r=

2

π

∫ 1

ρ

f (K)− F (K)√K2 − ρ2

(3.145)

it can rewrite as

− loga

r=

2

π

∫ 1

ρ

f (K)√K2 − ρ2

− 2

π

∫ 1

ρ

F (K)√K2 − ρ2

(3.146)

from Eqs. (3.130) and (3.132) is very easy to see that

2

π

∫ 1

ρ

f (K)√K2 − ρ2

= ω (ρ, r′0) + ω (ρ, r′1)− log ρ (3.147)

and we define a new Υ (ρ) function given by

Υ (ρ) =2

π

∫ 1

ρ

F (K)√K2 − 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

n = e[ω(ρ,r′0)+ω(ρ,r′1)−Υ(ρ)] (3.149)


andr =

ρ

n(3.150)

with these two equations we can find a refractive index for any value of r′0 and r′1.For example, if a = 1 is very easy to see that Υ (ρ) = 0 and leading to Luneburgsolution, i.e., when r′0 =∞ and r′1 = 1.

Morgan established a condition for the existence of a solution, that is, in orderfor a solution to exist for any value of r′0 and r′1 it is necessary that the followingcondition is satisfied

arcsin

(1

r′0

)+ arcsin

(1

r′1

)≥ 2

∫ 1

a

dr

r√P 2 (r)− 1

. (3.151)

Nowadays, Luneburg lens and the generalized Luneburg lens have great in-terest in various areas. For example; in telecommunications, the Luneburg lensis studied because it eliminates the aberrations caused by the antennas and thescanners [19]. Also, this lens provides excellent steering capabilities for incidentwide angles [20, 21].

The Luneburg lens have a spherical geometry, this geometry can be changedto a ellipsoidal geometry when we make a linear transformation. The new lensis named as the elliptical Luneburg lens and it has the same property of theLuneburg 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 ellipticalgeometry will be made.

3.7 The Elliptical Luneburg Lens

We can define the Luneburg lens as a marvellous optical lens because it is aaberrations free lens. In telecommunications is extremely difficult to be appliedin any practical antenna system due to its large spherical shape.

Currently, it has been proposed a transformation that reduces the profile ofthe 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 appli-cation [21].

The Luneburg lens and The ellipsoidal Luneburg lens have been designed ex-perimentally using Polymeric nanolayered. The first lens is presented as a devel-oping application of the nanolayered polymer technology and the second lens isused to model a human crystalline lens using the anterior and posterior shape ofthe crystalline for a age=5 years. The importance of this work is that the anteriorand posterior GRIN lenses were assembled into a bio-inspired GRIN human eyelens through which a clear imaging was possible [22, 23].

Due of these experimental advances is necessary to analyze the EllipticalLuneburg lens.

3.7.1 Linear Transformation of The Spherical Luneburg

Lens

We will demonstrate that from a linear transformation of a circle we can generatean 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 propertyof

CtC = x2c + y2

c = 1 (3.152)

and also has a imaging vector given by

C ′ =

[xc′

yc′

]

whereC ′ = AC (3.153)


thenC = A−1C ′

Ct = (A−1C ′)t (3.154)

from Eq. (3.152)CtC = (A−1C ′)

t(A−1C ′)

(A−1C ′)t(A−1C ′) = 1

(3.155)

if

A =

[a b

c d

]

At =

[a c

b d

]

A−1 =1

detA

[d −b−c a

]=

[d′ −b′

−c′ a′

](A−1

)t=

[d′ −c′

−b′ a′

]now, we have

(A−1C ′

)=

[d′ −c′

−b′ a′

][xc′

yc′

]=

[d′xc′ − c′yc′−b′xc′ + a′yc′

](A−1C ′

)t=[d′xc′ − c′yc′ −b′xc′ + a′yc′

](A−1C ′

)t (A−1C ′

)=[d′xc′ − c′yc′ −b′xc′ + a′yc′

] [ d′xc′ − c′yc′−b′xc′ + a′yc′

](A−1C ′

)t (A−1C ′

)=[

(d′xc′ − c′yc′)2 + (−b′xc′ + a′yc′)2]

(d′xc′ − c′yc′)2+ (−b′xc′ + a′yc′)

2= 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 =

[α 0

0 1

]

then

(A)−1 =1

α

[1 0

0 α

]=

[1α

0

0 1

]is clear to see that if a′ = 1, b′ = 0, c′ = 0 y d′ = 1

α, then Eq. (3.156) becomes

(1

αxc′

)2

+ y2c′ = 1 (3.157)

and if

A =

[1 0

0 α

]Equation (3.156) becomes

x2c′ +

(1

αyc′

)2

= 1. (3.158)

The parameter α is very important because it represents a compression or aexpansion 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 lensand 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) =√

2− r2 (3.159)

where r =√x2 + y2 is the the radius on one point inside the lens and 0 ≤ r ≤ 1.


Now, if we defined a new radius given by

re =

√(1

αx

)2

+ y2 (3.160)

and we substitute it into Eq. (3.159), we have

n (re) =

√√√√2−

[(1

αx

)2

+ y2

]. (3.161)

This equation represents the refractive index of the elliptical Luneburg lens. Ifwe consider that the α parameter is α < 1, we have an elliptical refractive indexwith 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 lineartransformation. Now, we need to know how the rays propagate inside this lens.

3.8. Ray Tracing 57

3.8 Ray Tracing

If we know the refractive index n (r), it is possible to obtain the rays inside ofthis refractive index from Eq. (3.15). But in this section, we will use the theoryof J. A. Grzesik published on the paper entitled "Focusing properties of a three-parameter class of oblate, Luneburg-like inhomogeneous lenses", because it is alot easier to understand and also because this theory can apply to both sphericaland elliptical Luneburg Lenses [24].

The refractive index of spherical or elliptical Luneburg lenses can be repre-sented as

n (x, y) =√nc − (nc − ns) (µ2x2 + y2) (3.162)

where√nc and

√ns is the central and surface refractive index of the lens, respec-

tively, and µ is a compression factor along x-axis. If µ−1 < 1 we obtain an oblateellipsoidal 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) generatesthe same refractive index that in Eq. (3.161) if α = 1

µ. 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 theray equation given by Eq. (2.14) in order to see how the rays propagate in the


medium. Equation (2.14) with some minor manipulation, adopts the form

d

dx

[n2 (x, y)

1 + y2

]=∂n2 (x, y, z)

∂x(3.163)

free from any square root entanglement. To solve this equation, we will consideronly two parameters α0 and P0. The parameter α0 is the angle that the ray makeswith respect to the direction of compression and P0 = (x0, y0) is the point wherethe 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

(y0Λ−1

2

)± µ−1

[sin−1

(xΛ−1

1

)− sin−1

(x0Λ−1

1

)]](3.164)

whereΛ2

1 =ns cos2 (α0) + µ2 (nc − ns)x2

0

µ2 (nc − ns)(3.165)

and

Λ22 =

ns sin2 (α0) + (nc − ns) y20

(nc − ns). (3.166)

The ± sign in Eq. (3.164) provides the possibility to have an ascendent or adescent ray, as dictated by the sign of injection angle α0.

The ray tracing with this solution is very easy to obtain. For example, if wehave nc = 2, ns = 1, µ = 1 and the rays come from infinity, i.e. α0 = 0, the Λ1

and Λ2 parameters areΛ1 =

√1 + x2

0 (3.167)

andΛ2 = y0. (3.168)

With these parameters and if x goes from x0 to 1 for each ray, we obtain the raytracing 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 caseµ = 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

Λ1 =

√1

6+ x2

0 (3.169)

andΛ2 = y0. (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


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 pa-rameters (nc, ns and µ) and we must choose very carefully the values of theseparameters so that the rays are focused.

For example, if nc = 2, ns = 1, µ = 2 and α0 = 0, we obtain a lens which onlyfocuses 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 gradientindex medium; its refractive index and how the rays propagate inside that medium.We have given several examples using the ray equation, the ray integral equationand the Grzesik’s theory to have a better understanding of the ray propagationin 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 fundamentto build our proposed crystalline. But, before we move on, we need to know theanatomy of the eye and the geometrical parameters (measurements) in order touse them in our model.

Chapter 4

The Human Eye As An OpticalSystem

The human visual system is composed, in one hand, by the optics of the eye, andin the other one, by a signal processing system wired to the brain. In this work weare interested in the optics of the eye, for this reason, the signal processing systemwill not be treated here. However, there is a considerable number of informationabout this area in the literature that can be consulted [25, 26, 27].

The eye is an optical system with an extraordinary complexity. The descriptionor modeling of the eye is done by organizing average measurements and propertiesinto simplified models called schematic eyes [25, 27]. These models are useful tosystematically study the properties and performance of individual components ofthe eye.

The complexity of the eye is due to the fact that its refractive surfaces arenot strictly spherical and its lens has a gradient refractive index. Also, the eye iscomplex 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 schematicmodels do not include the lens or the gradient refractive index of it, and it ischanged by a constant refractive index.

In this chapter, the components of the human eye are studied and also wedescribe 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 ofthe eye is divided into three compartments [28, 29]:

• The anterior chamber, between the cornea and the iris, which contains theaqueous fluid.

• The posterior chamber, between the iris, the ciliary body and the lens, whichcontains the aqueous fluid.

• The vitreous chamber, between the lens and the retina, which contains atransparent colourless and gelatinous mass called the vitreous humour orvitreous 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 opticalsystem, 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 withmore detailed in an anatomical description sense in many specialized books in thisarea [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 thelens. These elements are very different in their geometrical shape and in their re-fractive index, but in order to provide a good quality retinal image, these elementsmust be transparent and have appropriate curvatures and refractive indices [28].

Given the importance of these two elements in the human eye, the cornea andthe lens are described in this section.

4.1.1.1 Cornea

The cornea is the curved transparent front surface of the eye. It has approximatelyspherical shape with a radius of curvature of about 8 mm.

The anterior surface of the cornea is protected by the sclera which is theoutermost structure of the human eye. It is a dense, white, opaque, fibrous tissuethat is mainly protective in function and is approximately spherical with a radiusof curvature of about 12 mm. The centres of curvature of the sclera and corneaare separated by about 5 mm [28, 29].

The posterior surface of the cornea is in contact with the aqueous humour thatis a colorless liquid and is composed by a 98% of water. The aqueous humour isa liquid with constant refractive index. The value of its refractive index is welldefined 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 visionsystem.

The refractive index of the cornea is usually taken as 1.376 and this value isconsidered constant over the entire cornea. However, the anterior corneal surface


is not a smooth optical surface due to its cellular structure, although an opticallysmooth surface is provided by the very thin tear film which covers it. While thishas an index less than 1.376, for most optical calculations this tear film can beregarded as a very thin optical element consisting of two concentric surfaces ofalmost equal radii of curvature and therefore has negligible power [1].

In some works, the refractive index of the cornea is considered a gradientrefractive index [31, 32], because the cornea is composed by many layers, as shownin 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 thecornea is usually taken as 1.376.

Since the cornea is the first and most accessible optical element in the humaneye and by its important optical power makes it the object of several investigationsand treatments [33]. And it is probably also the most measured.


4.1.1.2 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 irisand the aqueous humour. The posterior surface is in contac with the vitreoushumour that is a clear gel that occupies the posterior segment of the eye and itsrefractive index can be considered equal to aqueous humour, 1.336.

The geometric shape of the lens and its refractive index are very difficult tomeasure in vivo, because the lens is a dynamic lens responsible for adjusting thefocusing distance. This changed of focusing distances is called accommodation.Also, as the crystalline lens ages, its geometric shape changes, and it becomes lessflexible, and consequently a person’s ability to accommodate is lost slowly withtime [25]. It is very important to say that also its refractive index changes withage.

The lens is supported by tiny muscular fibers, called ciliary muscles that pullthe lens. During accommodation, when the eye needs to change focus from distantto 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 surfacemoves slightly forward. These changes result in an increase in the equivalentpower 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 invitro, exist [11]. These show that the lens is not axially symmetric, as can be seenin Fig. 4.3, and the lens due to its biological nature and protein distributions isan 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 isrepresented by a gradient refractive index where the iso-indicial surfaces are ellip-soidal curves.

Throught the years, the gradient index of the lens has been represented by


Figure 4.3: The geometric shape of the lens. za and zp are differents.

two mathematical equations, one equation for anterior lens and one equationfor the posterior lens [3, 4]. These models present drawbacks because they arediscontinuous at the equator. These discontinuities can be classified into two maingroups:

• In the first group the gradient index variation in the equator is discontin-uaous between the anterior lens and posterior lens [3].

• In the second group the discontinuities are between the derivatives of firstand second order with respect to the iso-indical surface geometry [34, 4].

These problems are generated by the accommodation of the lens, because itreshapes is required to change its internal gradient index distribution imposing aconstraint to stablish a definitive expression for modeling its gradient refractiveindex [35]. In other words, there cannot exists an unique mathematical expressionto simulate a living biological crystalline.

The paper of A. M. Rosen, et al. entitled "In vitro dimensions and curvatures


of human lenses" provides excellent experimental data for the geometric shapeand refractive index of the lens [11]. In this paper, the refractive index rangesfrom nc = 1.4181 ± 0.075 in the core to ns = 1.3709 ± 0.0039 at the boundarysurface of the lens. And also, it provides the equations for the parameters of thelens 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 at0.70. Measurements were made on 37 human lenses ranging in age from 20 to 99years.

Numerous studies and measurements of the defects of the cornea and crys-talline have given the basis to the hypothesis that the crystalline compensates thedefects of the cornea by minimizing or balancing them. A discussion on this canbe found in a review by Artal [33].

These and other important characteristics of the lens will be studied with moredetail 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 asthe pupil. The pupil size is determined by two antagonistic muscles, which areunder autonomic (reflex) control [28]:

• The sphincter pupillae, which is a smooth muscle forming a ring around thepupillary margin of the iris. When it contracts, the pupil constricts. It isinnervated 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-epithelialcells that extend radially from the sphincter into the ciliary body. It dilates


the pupil and is innervated by sympathetic nerve fibres, which synapse inthe superior cervical ganglion and enter the eye by way of the short andlong ciliary.

The diameter of the pupil may vary from about 2-3 mm at high illuminationto about 8 mm in darkness.

If we considered identical illumination conditions is possible to observe thatthe diameter is different for each person. This is due to pupil size decreases withincreasing 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 4.8mm, while at 45 will be of 4.0 mm, and 3.4 mm at 80 years. For an eye in totaldarkness the most common diameters are 7.6 mm at 10 years, 6.2 mm at 45, and5.2 mm at 80 years [29].

4.1.3 Retina

The retina is the ocular surface where images are projected and is the light-sensitive tissue of the eye. It is composed of a number of cellular and pigmentedlayers, and a nerve fibre layer, i.e., the retina consists of ten layers, as shown inFig. 4.4.

The study by direct observation of this membrane is known as Retinoscopy.This procedure has undergone significant evolution since the adaptive optics sys-tems were first implemented [36].

The retina is especially interesting since it is directly connected to the centralnervous system and some experts even consider it to be part of the brain sinceneural cells can be found in it [37]. Here is where specialized cells that can convertlight into nervous impulses can be found. There are two main types of cornealcells which are responsible for light conversion. The cones are the ones responsiblefor color discrimination, they come in three types, each one capable of detecting aspecific frequency band associated to color blue, green, and red, respectively. Therods are responsible for night vision, since they are very sensitive to light. Thishigh sensitivity is somehow related to the wiring between rods which combines


Figure 4.4: The layers at the back of the human eye [28].


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 millionsof rods and as much as a million nervous fibers. The fovea is the region of theretina where a higher cone density is found.

4.2 Schematics Eye

During the last decades different schematic eye models have been proposed. Theparaxial schematic models are the most relevant [28, 1]. However, more compli-cated schematic eye models containing a Gradient Refractive Index lens have beenpublished in recent years [3, 38].

The paraxial schematic eye models have a problem; they use a homogeneousindex lens and a limited numbers of surfaces. In the Emsley reduced model, theeye is represented only by a single refractive surface; being the anterior surface ofthe 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 vitreousrefractive index were modified up to 1.416 when it is well known that the realvalues are 1.336 [28].

The Le Grand full theoretical model is represented by four surfaces; two forthe 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 acladding, the core has a high refractive index limited by the interior surfaces andthe cladding has lower refractive index limited by the exterior surfaces. The lowerrefractive index is surrounding the core, i. e., Gullstrand represented the lens asa lens inside other lens, but it is very important to say that it does not representa GRIN lens, because both lenses, independently, have a homogeneous refractiveindex.

In 1974, R. G. Zainullin and et al. in the paper entitled "The crystalline lensas a Luneburg lens" proposed to use the Luneburg lens to represent the crystalline


lens of the eye of vertebrates [8]. We say that they proposed, because they do notmake a schematic eye in that paper.

Using the same idea of R. G. Zainullin and et al., Campbell and Hughespublished a paper entitled "An analytic, gradient index schematic lens and eye forthe rat which predicts aberrations for finite pupils", where the lens is representedas a Luneburg lens [2]. But, there is a problem in this paper, it can not be usedto represent a schematic human eye since the anterior and posterior faces of thelens are considered to be symmetric, as shown in Fig. 4.8.

Figure 4.8: The schematic eye is analytically derived from the refractive indexprofile of the crystalline lens and anatomical measurements of a rat eye. [2].

Also, the Maxwell fish eye has been considered to build a schematic humaneye, however, its design is a spherically symmetric lens, and the index of refractionvaries symmetrically about a point [9]. For this reason, the schematic eye havethe 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 ageneralized Maxwell’s Fisheye, and the Maxwell’s Fisheye is not a lens (see Section3.4), 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 andits 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 theidea 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 havepresented the optical characteristics of each element of the human eye.

Also, we have studied the different schematic eyes, in which we have observedtheir problems to represent a realistic schematic eye model.

With the characteristics of the human eye given in this chapter, we may be ableto build a schematic human eye using a new lens that we will call the compositemodified Luneburg lens, which will be discussed in the next chapter. This lensis based on the elliptical Luneburg lens and the new model can be modified for


different eye conditions, as we will show in Chapter 5.

Chapter 5

Schematic Eye with CompositeLuneburg Crystalline

The original Luneburg lens is a sphere with gradient index that has the propertyof being spherical aberrations free [6, 17]. For every pair of object and imageconjugate points the internal refractive index distribution is different presentinga similitude with the human lens [8].

Now, it has been demonstrated that it is possible to perform a geometrictransformation in such a way that the original spherical Luneburg lens can betransformed into an oblate spheroidal shape maintaining its aberration free prop-erty [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 analternative model will be proposed having an asymmetric bi-spheroidal shape butassuming that the gradient refractive index can be represented with a given ana-lytical expression [41, 42]. As discussed above, the crystalline shape and refractiveindex are being modified continuously as the vision point changes, so a realisticmodel of the crystalline must consider the dynamics of both of them.

In this chapter we propose a dynamic model of the crystalline constructed withtwo separate spheroidal hemispheres with variable curvatures for the anterior andposterior sections and considering a varying Gradient Refractive Index (GRIN).Its imaging properties are investigated based on the Luneburg lens theory modifiedaccordingly to the proposed composite lens. We impose the condition in our modelthat the isoindical lines be continuous as well as their first order derivatives at theequatorial plane. This condition creates a perfect match of the gradient refractiveindex 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 schematiceye using the composite modified Luneburg lens. This schematic eye is based inbiometric parameters reported in Ref. [11].

5.1 Composite Modified Luneburg Lens

As mentioned above, the problem of the spherical Luneburg lens after a simpletransformation can be reformulated into that of a spheroid maintaining its aberra-tion free property [17, 21]. The index in the spherical Luneburg lens is describedby a function n (r) where r =

√x2 + y2 + z2 and x, y and z are the Cartesian

coordinates. We set ρ =√x2 + y2 and by symmetry around the z-axis we will

work only with y. As r defines a spherical shape, the transformation to obtainthe desired spheroidal shape can be either

(1

sy′)2

+ z′2 = r2 (5.1)

or

y′2 +

(1

sz′)2

= r2. (5.2)

In these equations s is a constant parameter. By properly choosing the value ofs we can compress or expand the sphere along the y or z axis as needed. The modelof crystalline will be approximated by two spheroidal hemispheres [21, 24, 41, 42].

We now proceed to construct the model by imposing the condition that thegradient indices and the axial derivatives of the anterior and posterior hemispheresmatch at every point on the equatorial plane. For this purpose the refractiveindices, na (r) for the anterior and np (r) for the posterior hemisphere, must be ofthe form

5.1. Composite Modified Luneburg Lens 77

n(y′, z′) =

na(y

′, z′) =

√n2c −

[(y′

R

)2+(

1Rsa

z′)2]

z′ ≤ 0

np(y′, z′) =

√n2c −

[(y′

R

)2+(

1Rsp

z′)2]

z′ ≥ 0

(5.3)

where nc is the refractive index at the center of the lens, R is the radius of the lensmeasured on the equatorial plane. The scaling factor is s(a,p) = z(a,p)/R

√n2c − n2

s

where za is the anterior vertex and zp the posterior vertex. For na (r) the variablez′ is negative and for np (r) the variable z′ is positive. Figure 5.1 displays theprojection on a meridional plane of the constructed crystalline lens with the bi-elliptical iso-indical lines showing continuity at the equatorial plane.

Figure 5.1: Geometry of the bi-spherical model showing the continuity of theisoindical lines at the equator plane.


For the numerical example under study we used as a reference the biometricparameters 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 theequations given by Eq. (4.1).

For a lens aged 35 years the corresponding parameters are

R = 4.4005mmza = 1.8215mmzp = 2.5890mm.

(5.4)

The curvatures of the anterior and posterior hemispheres are such that the ratiobetween 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 proceduresimilar to that described in Ref. [24]. this is, to modified it accordingly to fit thenew geometry and the imposed continuity conditions. We assumed that the lensis embedded in a medium with refractive index equals to 1.336 and by keepingconstant the surface refractive index ns.

The first case that will be studied simulates a relaxed crystalline assuming anobject at infinity so that the rays impigne parallel to its anterior surface. Thechosen geometrical parameters and refractive indices produce practically a per-fect focus at the image plane placed at a distance of 63.05mm from the posteriorsurface of the lens as shown in Fig. 5.2 a). For the next two cases we modifythe vertex ratio za/zp and calculate the refractive index that is necessary to forma point image at the same plane of a point object placed at 250 mm from theanterior surface. In Fig. 5.2 b) we see that for a reduced ratio the CompositeModified Luneburg (CML) lens is thinner resulting in a GRIN distribution suchthat the central refractive index has been increased. The opposite case of mak-ing thicker the CML lens with a larger ratio za/zp = 0.7185 results in a GRINredistribution with a reduced refractive index at its center, as is shown in Fig 5.2c). 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-


Figure 5.2: Ray tracing through the proposed CML lens embedded in a mediumwith refractive index of 1.336. a) Rays incident from an infinite distance. Raysincidents from a finite distance of 250 mm with b) za/zp = 0.6872 and c) za/zp =0.7185.


tribution change. Although in the present model we have used perfect geometricsurfaces, implementing minor modifications to our model, by using real biometricdata 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 changethe ratio za/zp.

Figure 5.3: The GRIN distribution and the GRIN profiles for each lens withdifferent ratio za/zp.

We have said the central refractive index increases when za/zp decreases andthe central refractive index decreases when za/zp decreases, this is much easier tosee 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 dueto the GRIN is almost imperceptible appearing as if the refractive index washomogeneous. However, from Fig. 5.5 we observe that the rays are planar curves


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 changeon 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 lowerrefractive 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 keepingthe 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 indexequals to 1.336.

In these computational simulations, the only parameter of the lens that can bemodified is the central refractive index nc, because we decided that the parametersns and za/zp to be constants. A parameter that also can be modified, but thisis not a parameter of the lens, is the plane of a point object, these planes wereplaced 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-


Figure 5.5: Trajectory of the inner rays for each CML lens with different ratioza/zp.

tically a perfect focus at the image plane placed at a distance of 63.05 mm fromthe posterior surface of the lens.

From Fig. 5.6, we can observe that the central refractive index decreases whenwe placed the point object plane near the anterior surface of the lens. This meansthat the refractive power of the lens is lower when the objects are closer to thelens.

These results give us a better idea of the changes in the refractive index of thelens, because we are not altering the geometry of the lens and due to this fact, wedo not make assumptions of changes in the ratio za/zp for different object pointdistances. The change of the refractive index is much easier to see when all GRINprofiles 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, itis very difficult to measure this change in vivo, because the lens is dynamic and


Figure 5.6: The GRIN distribution and the GRIN profiles for each lens with thesame ratio za/zp and different central refractive index.


Figure 5.7: The GRIN profiles for each lens with the same ratio za/zp and differentcentral refractive index.

there are no measurements of this change.

The propagation of the rays are shown in Fig. 5.8, in which can be observedthat the rays are focused on an image plane placed at 63.05 mm from the posteriorsurface of the lens. The incident rays have been cut, because the distances of wherethe 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 indexappear to be equal to each other, however, the rays and the gradient refractiveindex 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 whenwe have a lower gradient refractive index, but this is not true, because we mustremember that we are changing the incidence point on the lens. From Fig. 3.16is easy to see that the angle α0 is larger when the point of incidence approachesto the lens. This implies that the constant K given by generalized Snell law forinhomogeneous media with spherical symmetry is changing when the angle α0 ischanging, 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,


Figure 5.8: Ray tracing through the proposed CML lens embedded in a mediumwith a refractive index of 1.336 and a constant ratio za/zp = 0.7036. Rays inci-dents from a finite distance of a) Infinity, b) 2000 mm, c)1000 mm, d) 500 mm,and e)250 mm.


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 dependon age. Then, it is very interesting to make an analysis of second case when wehave a human lens of a different age.

For example, the Fig. 5.10 represents the GRIN profiles of a human lens thathas 20 years old for different object distances. The different with a human lens of35 years old is the increment of 0.0002 in the central refractive index and a imagedistance of 62.35 mm. The increment in the central refractive index is due to thelens 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: Itis always possible to find a distance where practically all the rays are focused, butthis distance depend on the geometrical parameters and the refractive index ofthe lens. In the case where the lens has 20 years old, it is impossible to find arefractive 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 thicknessis constant at 0.70 for all age". It is very interesting because we have the same


Figure 5.10: The GRIN profiles for each lens with the same ratio za/zp anddifferent central refractive index. The human lens is 20 years old.


ratio for all age and different Za and Zp, for this reason, we have different imagedistances 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 dis-tances.

We can observe the Fig. 5.2 and the question arises: why in the Fig. 5.2 waspossible 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. Observethat the ratio Za/Zp is different to 0.70 as we said in previous paragraphs. Tofind 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 fixedthe geometrical parameters or when we fixed the image distance and the centralrefractive index.

With the carried out analysis of the CML lens is possible to construct a newSchematic Model of the Human Eye using the CML lens as a human lens. Thismodel is studied in next section.

5.2 Schematic Luneburg Eye

In the previous section, we constructed a CML lens where its geometrical parame-ters and refractive indices were chosen to produce practically a perfect focus. It iswoeth to note that, the human lens does not have those central refractive indicesused in Section 5.1.

In this section, we are interested in understanding the behavior of the lightinside the complete eye, but first we want to know about the behavior of the lightwhen 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 parametersand refractive indices of the eye, we can generate the ray tracing along of this newCML 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 inthe eye.

We can see in Fig. 5.11 that rays inside the lens are planar curves and alsowe can see that the lens produce a negative spherical aberration when rays areimpigne 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 perfectlyspherical, because the cornea has a positive spherical aberration as it is shownin Fig. 5.12. The positive aberration from the cornea can be suppressed bythe negative aberration from the crystalline [43] when we choose an appropriaterefractive index.

For example, if we have nc = 1.41262, ns = 1.3709, za/zp = 0.7036, andrays that impigne from infinity distance, then the aberration from the cornea issuppressed by the lens aberration as is shown in Fig. 5.13. In this figure, we cansee that the rays focused on the retina.

With this model, by knowing the curvature it is possible to know the refractiveindex or if knowing the refractive index it is possible to know the curvatures ofthe lens. Also, we can change the distance from which the rays are coming fromand we can construct a new lens that can suppress the cornea positive spherical


Figure 5.12: Positive spherical aberration due to the cornea. Rays incident frominfinity distance, the anterior and posterior surfaces of the cornea are immersedin a refractive index of 1 and 1.336, respectively.

Figure 5.13: Schematic eye model with CML lens.


aberration. The parameters of the human lens for different distances are given inTable 5.1 and the propagations are presented in Fig. 5.14.

InfinityRatio za/zp nc ns

0.7036 1.41262 1.3709250 mm

Ratio za/zp nc ns0.7185 1.41477 1.37090.7323 1.41249 1.37090.7450 1.41088 1.37090.7568 1.40897 1.37090.7677 1.40803 1.3709

500 mmRatio za/zp nc ns

0.7185 1.41255 1.37090.7323 1.41092 1.37090.7450 1.40939 1.37090.7568 1.40801 1.37090.7677 1.40669 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 surfacefixed, then the ratio between vertex, za/zp , is change, and a suitable refractiveindex 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 doesaffect 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 comingfrom a distance of 250 mm, we can observe that the refractive index is bigger thanwhen the ratio is of 0.7036 and rays come from infinity. This means that for closerdistances, 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


Figure 5.14: Examples of Schematic model eyes with CML lens: a) Incident rayscoming from a 250 mm distance and b) Incident rays coming from a 500 mmdistance. Both Schematic eyes have a relation of 0.7450.


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 areoff-axis, the rays reach the retina with a coma aberration. This aberration islarger when the rays are farther from the axis as shown in Fig. 5.15.

This schematic model can be modified for different object distances, differentrefractive indices and different curvatures of the crystalline; this is to make it asversatile as possible. Also, it is possible to change the corneal topography to get amore realistic scenario, since every single eye is different. The lens accommodationis obtained by reshaping the lens, resulting in the necessity of a modification inits gradient refractive index.

As can be seen from the simulations, the rays inside the crystalline are planarcurves.

The analysis of the off-axis case shows that when the rays enter with a certainangle with respect to the optical- or z-axis into the eye, a coma aberration ispresented. As this angle increase the resultant coma aberration increase.


5.3 Conclusions

In conclusion, a schematic model eye based on a composite modified Luneburglens 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 newschematic human eye using a composite modified Luneburg lens.

This lens has been proposed as a bi-spheroidal model of the human lens basedon the gradient index Luneburg lens. Using biometrical data reported in theliterature our model accurately predicts the imaging properties of a human lens,where we have observed that the composite modified Luneburg lens has the samephysiological properties of the human lens. It was also obtained that is necessarya modification in its gradient refractive index when simulating accommodation byreshaping the lens it is necessary a modification in its gradient refractive indeximplying that it is not possible to establish a definitive mathematical expressionfor the gradient refractive index of a biological crystalline.

Also, with this lens model it was possible to create a schematic GRIN eyethat emulates the human lens behavior. With this model, we did an analysis thathad never been done, i.e., this model allows the evaluation of different imagingsituations; i.e., the effects of the accommodation in the human lens due to thechanged in distance from which rays impigne.

The analysis predicts that the accommodation of the human lens is accompa-nied by a change in the gradient refractive index of this lens, which depends onthe ratio za/zp.

This analysis was completed when an off-axis analysis was made. This demon-strated that the rays off-axis reach the retina with a coma aberration, due tothe inclination of the rays. In the specialized literature, the coma aberrationis obtained when the lens has a inclination on the optical axis, however, in ourschematic 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 ofthe human lens; also, our model can be modified for different topographies of thecornea, i.e., it can be modified fully to emulate any experimental data.

The work presented in this thesis has opened various research topics in theoptical area. For example, in Visual Optics , it will be of interest to study the raypropagation in a three-dimensional elliptical Luneburg lens, the lens is shown inFig. 6.1.

Figure 6.1: Three-dimensional elliptical Luneburg lens.

The importance of considering the study of the elliptical Luneburg lens isbecause 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, inwhich 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 ellipticalLuneburg lens can be used to replace conventional antenna systems.

In the fiber integrated optics, it is possible to produce a loss-free waveguideusing 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 doan analysis using the off-axis rays to obtain a loss-free waveguide that does notdepend on the angle of incidence of the rays. Note that this idea generates acritical angle of π.

In general, we have introduced a model of the human lens built as a compositeLuneburg lens whose gradient index function and its derivatives are continuous atthe equatorial plane. This allowes the construction of a more realistic schematicmodel of the eye. We demonstrated its imaging capabilities of the whole schematiceye for sources at infinity. In our model we have used conicoid surfaces but it canbe generalized to any surface with the condition of having continuos derivativesat the equator plane. Our method allows us to obtain custom made GRIN distri-butions once the biometric parameters are given.

List of Figures

2.1 Fermat’s Principle. . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 The linear gradient index medium. . . . . . . . . . . . . . . . . . 112.3 An arc lenght along the ray path. . . . . . . . . . . . . . . . . . . 122.4 Solution for the linear gradient index medium. . . . . . . . . . . . 142.5 The radial gradient index medium. . . . . . . . . . . . . . . . . . 142.6 Solution for the Sagittal plane. . . . . . . . . . . . . . . . . . . . . 162.7 Solution for the radial gradient index medium. . . . . . . . . . . . 17

3.1 The spherical gradient. . . . . . . . . . . . . . . . . . . . . . . . . 203.2 The Index Variation. . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Medium with radial symmetry . . . . . . . . . . . . . . . . . . . . 223.4 The path of rays in a medium with gradient index . . . . . . . . . 263.5 Rays on Horizontal axis . . . . . . . . . . . . . . . . . . . . . . . 283.6 The rays are hyperbolas when C > 0. . . . . . . . . . . . . . . . . 323.7 The rays are parabolas when C = 0. . . . . . . . . . . . . . . . . . 333.8 The rays are ellipses when C < 0. . . . . . . . . . . . . . . . . . . 333.9 Maxwell Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.10 Maxwell Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.11 Maxwell Solution Off-Axis . . . . . . . . . . . . . . . . . . . . . . 393.12 Luneburg Sphere Parameters . . . . . . . . . . . . . . . . . . . . . 413.13 The Luneburg Lens . . . . . . . . . . . . . . . . . . . . . . . . . . 493.14 The Generalized Luneburg Lens . . . . . . . . . . . . . . . . . . . 503.15 The Elliptical Luneburg Lens . . . . . . . . . . . . . . . . . . . . 563.16 Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.17 Ray tracing of The Luneburg Lens . . . . . . . . . . . . . . . . . 593.18 Ray tracing of The Elliptical Luneburg Lens . . . . . . . . . . . . 593.19 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 indexprofile of the crystalline lens and anatomical measurements of a rateye. [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 theisoindical lines at the equator plane. . . . . . . . . . . . . . . . . . 77

5.2 Ray tracing through the proposed CML lens embedded in a mediumwith refractive index of 1.336. a) Rays incident from an infinitedistance. Rays incidents from a finite distance of 250 mm with b)za/zp = 0.6872 and c) za/zp = 0.7185. . . . . . . . . . . . . . . . . 79

5.3 The GRIN distribution and the GRIN profiles for each lens withdifferent 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 ratioza/zp. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

5.6 The GRIN distribution and the GRIN profiles for each lens withthe same ratio za/zp and different central refractive index. . . . . 83

5.7 The GRIN profiles for each lens with the same ratio za/zp anddifferent central refractive index. . . . . . . . . . . . . . . . . . . . 84

5.8 Ray tracing through the proposed CML lens embedded in a mediumwith 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

List of Figures 101

5.9 Trajectory of the inner rays for each CMLL with constant ratioza/zp = 0.7036. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

5.10 The GRIN profiles for each lens with the same ratio za/zp anddifferent central refractive index. The human lens is 20 years old. 87

5.11 Negative spherical aberration from the human lens. . . . . . . . . 895.12 Positive spherical aberration due to the cornea. Rays incident from

infinity distance, the anterior and posterior surfaces of the corneaare immersed in a refractive index of 1 and 1.336, respectively. . . 90

5.13 Schematic eye model with CML lens. . . . . . . . . . . . . . . . . 905.14 Examples of Schematic model eyes with CML lens: a) Incident rays

coming from a 250 mm distance and b) Incident rays coming froma 500 mm distance. Both Schematic eyes have a relation of 0.7450. 92

5.15 Schematic eye when rays are off-axis. . . . . . . . . . . . . . . . . 93

6.1 Three-dimensional elliptical Luneburg lens. . . . . . . . . . . . . . 966.2 Three-dimensional human lens. . . . . . . . . . . . . . . . . . . . 97

Bibliography

[1] David A. Atchison and George Smith. The Eye and Visual Optical Instru-ments. Cambridge University Press, USA, 1997. (Cited on pages 1, 64and 70.)

[2] M.C.W. Campbell and A. Hughes. An analytic, gradient index schematiclens and eye for the rat which predicts aberrations for finite pupils. VisionRes., 21(7):1129–1148, 1981. (Cited on pages 1, 72 and 100.)

[3] Alexander V. Goncharov and Chris Dainty. Wide-field schematic eye modelswith gradient-index lens. J. Opt. Soc. Am. A, 24(8):2157–2174, 2007. (Citedon pages 1, 66 and 70.)

[4] Fernando Palos Rafael Navarro and Luis González. Adaptive model of thegradient index of the human lens. i. formulation and model of aging ex vivolenses. J. Opt. Soc. Am. A, 24(8):2175–2185, 2007. (Cited on pages 1 and 66.)

[5] R. Meder C. E. Jones, D. A. Atchison and J. M. Pope. Refractive indexdistribution and optical properties of the isolated human lens measured usingmagnetic resonance imaging (mri). Vision Res., 45:2352–2366, 2005. (Citedon page 1.)

[6] R. K. Luneburg. Mathematical Theory of Optics. University of CaliforniaPress, Los Angeles California, 1964. (Cited on pages 2, 31 and 75.)

[7] Rudolf Kingslake and R. Barry Johnson. Lens Design Fundamentals. Aca-demic Press (SPIE Press), USA, 2010. (Cited on pages 2 and 44.)

[8] A. B. Kravtsov R. G. Zainullin and E. P. Shaitor. The crystalline lens as aluneburg lens. Biofizica, 19(5):913–915, 1974. (Cited on pages 2, 72 and 75.)

[9] Hao Lv Yun Wu, Aimei Liu and et al. Finite schematic eye model withmaxwell fish-eye spherical lens. In Symposium on Photonics and Optoelec-tronic (SOPO) 2010, pages 1–4, 2010. (Cited on pages 2, 35, 72, 73 and 100.)

104 Bibliography

[10] Erich W. Marchand. Gradient Index Optics. Academic Press, New York,1978. (Cited on pages 2, 19, 49 and 50.)

[11] V. Fernandez A. M. Rosen, D. B. Denham and et al. In vitro dimensions andcurvatures of human lenses. Vision Res., 46(6):1002–1009, 2006. (Cited onpages 2, 65, 67, 76 and 78.)

[12] María Victoria Pérez Carlos Gómez-Reino and Carmen Bao. Gradient-IndexOptics Fundamentals and Applications. Springer, New York, 2002. (Cited onpages 5 and 14.)

[13] Ajoy K. Ghatak Vasudevan Lakshminarayanan and K. Thyagarajan. La-grangian Optics. Kluwer Academid Publishers, Massachusetts USA, 2002.(Cited on pages 15 and 22.)

[14] Max Born and Emil Wolf. Principles of Optics: Electromagnetic Theory ofPropagation, Interference and Diffraction of Light. Cambridge UniversityPress, UK, 7th edition, 1978. (Cited on pages 9 and 26.)

[15] J. R. Flores. Estudio de Elementos Ópticos de Gradiente de Índice de SimetríaEsférica. PhD thesis, Universidade de Santiago, España, 1992. (Cited onpages 40 and 46.)

[16] A. I. Kiselev M. L. Krasnov and G. I. Makarenko. A book of problems inordinary differential equations. MIR, URSS, 1981. (Cited on page 46.)

[17] Samuel P. Morgan. General solution of the luneburg lens problem. J. Appl.Phys., 29(9):1358–1368, 1958. (Cited on pages 49, 50, 75 and 76.)

[18] G. Toraldo di Francia. Il problema matematico del sistema ottico concentricostigmatico. Ann. Mat. Pura Appl., 44:35–44, 1957. (Cited on page 50.)

[19] Ning Wang Yoke Leng Loo, Yarong Yang and et al. Broadband microwaveluneburg lens made of gradient index metamaterials. J. Opt. Soc. Am. A,29(4):426–430, 2012. (Cited on page 52.)

Bibliography 105

[20] Nathan Kundtz and David R. Smith. Extreme-angle broadband metamateriallens. Nature Materials, 9:129–132, 2010. (Cited on page 52.)

[21] Angela Demetriadou and Yang Hao. Slim luneburg lens for antenna appli-cations. Opt. Express, 19(21):19925–19934, 2011. (Cited on pages 52, 53, 75and 76.)

[22] Matthew Mackey Shanzuo Ji, Kezhen Yin and et al. Polymeric nanolay-ered gradient refractive index lenses: technology review and introduction ofspherical gradient refractive index ball lenses. Opt. Eng., 52(11):112105–1–112105–13, 2013. (Cited on page 53.)

[23] Richard S. Lepkowicz Shanzuo Ji, Michael Ponting and et al. A bio-inspiredpolymeric gradient refractive index (grin) human eye lens. Opt. Express,20(24):26746–26754, 2012. (Cited on page 53.)

[24] J. A. Grzesik. Focusing properties of a three-parameter class of oblate,luneburg-like inhomogeneous lenses. J. of Electromagn. Waves and Appl.,19(8):1005–1019, 2005. (Cited on pages 57, 75, 76 and 78.)

[25] Michael P. Keating. Geometric, Physical and Visual Optics. Butterworth-Heinemann, Boston, 2ed edition, 2002. (Cited on pages 61 and 65.)

[26] E Dalimier and Science Faculty. Adaptive Optics Correction of Ocular Higher-Order Aberrations and the Effects on Functional Vision. PhD thesis, Univer-sidade de Santiago, España, 2007. (Cited on page 61.)

[27] Steven H. Schwartz. Geometrical and Visual Optics, a clinical approach.McGraw-Hill, 2002. (Cited on page 61.)

[28] David A. Atchison and George Smith. Optics of the Human Eye.Butterworth-Heinemann, Oxford, 2000. (Cited on pages 62, 63, 64, 65, 67,69, 70 and 100.)

106 Bibliography

[29] María Cinta Puell Marín. Óptica Fisiológica: El sistema óptico del ojo y lavisión binocular. E-Prints Complutense (Universidad Complutense Madrid),Madrid, 2006. (Cited on pages 62, 63, 65 and 68.)

[30] J. A. Alvarado M. J. Hogan and J. E. Weddell. Histology of the Human Eye.Saunders and Co., 1971. (Cited on page 63.)

[31] Sergio Barbero. Refractive power of a multilayer rotationally symmetricmodel of the human cornea and tear film. J. Opt. Soc. Am. A, 23(7):1578–1585, 2006. (Cited on page 64.)

[32] A. Díaz del Rio M. T. Flores-Arias and et al. Description of gradient-indexhuman eye by a first-order optical system. J. Opt. A: Pure Appl. Opt., 11(12),2009. (Cited on page 64.)

[33] Pablo Artal and Juan Tabernero. The eye’s aplanatic answer. Nature Pho-tonics, 2:586–589, 2008. (Cited on pages 64, 67 and 70.)

[34] Enrique Gambra Alberto de Castro, Sergio Ortiz and et al. Three-dimensionalreconstruction of the crystalline lens gradient index distribution from octimaging. Opt. Express, 18(21):21905–21917, 2010. (Cited on page 66.)

[35] Barbara K. Pierscionek and Justyn W. Regini. The gradient index lens ofthe eye: An opto-biological synchrony. Progress in Retinal and Eye Research,31(4):332–349, 2012. (Cited on pages 66 and 78.)

[36] David R. Williams Junzhong Liang and Donald T. Miller. Supernormal visionand high-resolution retinal imaging through adaptive optics. J. Opt. Soc. Am.A, 14(11):2884–2892, 1997. (Cited on page 68.)

[37] G. J. Augustine D. Purves and D. Fitzpatrick (Editors). Neurociencia. Edi-torial Médica Panamericana, Madrid, 3rd edition, 2008. (Cited on pages 68and 70.)

Bibliography 107

[38] Barbara Pierscionek Mehdi Bahrami, Masato Hoshino and et al. Opticalproperties of the lens: An explanation for the zones of discontinuity. Experi-mental Eye Research, 124:93–99, 2014. (Cited on pages 70 and 80.)

[39] H. H. Emsley. Visual Optics. Butterworth-Heinemann, Oxford, 1952. (Citedon page 70.)

[40] J. P. C. Southall. Helmholtz?s Treatise on Physiological Optics. OpticalSociety of America, New. York, 1924. (Cited on page 71.)

[41] B. K. Pierscionek G. Smith and D. A. Atchison. The optical modelling of thehuman lens. Ophthal. Physiol. Opt., 11(4):359–369, 1991. (Cited on pages 75and 76.)

[42] C. Bao M. A. Rama, M. V. Pérez and et al. Gradient-index crystalline lensmodel: A new method for determining the paraxial properties by the axialand field rays. Optics Communications, 249(4):595–609, 2005. (Cited onpages 75 and 76.)

[43] Antonio Benito Pablo Artal and Juan Tabernero. The human eye is anexample of robust optical design. Journal of Vision, 6(1):1–7, 2006. (Citedon page 89.)

[44] G. P. Tsironis M. M. Mattheakis and V. I. Kovanis. Luneburg lens waveguidenetworks. J. Opt., 14(11):1–8, 2012. (Cited on page 96.)

complete gradient refractive index lens schematic human ... · complete gradient refractive index...

Documents