EE535: OPTICS & PHOTONICS

LECT. 03 2025 01 21
LECTURE OUTLINE

Topics
• Some math:
• A glimpse of calculus of variation
• Space metric
• Ch. 1: Ray Optics
• 1.2:Simple Optical Components
• 1.3:Graded-Index Optics
• 1.4: Transfer matrices for ray optics
Notes
• Class time and days:
• 14:30 – 15:45
• Some of your colleagues want to change the days from UT to TR due to some teaching
assignments.
1.2: SIMPLE OPTICAL COMPONENTS

 Let’s remind ourselves of the physical condition to apply ray optics; what is it?
 The wavelength is very small compared to the structure dimension; i.e. 𝜆 → 0.
 In this case, the light can be considered geometrically as a ray.
 What are the needed variables to describe rays?
 Direction and starting point

 Today, we will consider the following optical components and apply ray optics treatments to them,
assuming that the above condition is satisfied:
 Mirrors
 Planar boundaries (reflection and refraction)
 Spherical boundaries and lenses
 Light guides (next class)
1.2: SIMPLE OPTICAL COMPONENTS

 Last class, we discussed multiple simple optical systems:

 Mirrors
 Planar boundaries
 Prisms
 Beamsplitters
 Beam directors
 Spherical boundaries and lenses

 In all analyses, we assumed paraxial approximation.

1.2: SIMPLE OPTICAL COMPONENTS
SPHERICAL BOUNDARIES AND LENSES

 Again, the previous results assume paraxial approximation.

Things are different in real life.
 Obviously, different curvatures and shapes can be used.
Also, the curves can be convex or concave.
1.2: SIMPLE OPTICAL COMPONENTS
LIGHT GUIDES

 Light may be guided from one location to another by using of a set of optical components.
 Based on what we learned last time, can you suggest some ways to guide light?
 By lenses
 By mirrors
 Utilizing total internal reflection

 Guiding light by the total internal reflection:

 An ideal mechanism needs two media.
 Rays are reflected repeatedly without
undergoing refraction.
 Which refractive index should be larger and why?
 The inner to force an internal refraction.
1.2: SIMPLE OPTICAL COMPONENTS
LIGHT GUIDES

 An optical fiber is a light conduit made of two concentric glass cylinders.

 The inner is called the core with a refractive index 𝑛1 .
 The outer is called the cladding with a refractive index 𝑛2 .

 What are the conditions to achieve TIR?

 𝑛1 > 𝑛2
 The incident angle at the interface between
core and cladding should be greater than 𝜃𝑐 .
 But we have another interface between the core and air, it
should ensure having 𝜃𝑐 at the other interface.
 This condition is applied to the angle of acceptance 𝜃𝑎 .

 After some math: sin 𝜃𝑎 = 𝑛12 − 𝑛22 = NA (numerical aperture)

SOME MATHEMATICS
CALCULUS OF VARIATION: THE CONCEPT

 Let’s consider this, if we have a functional 𝐽: 𝑉 → ℝ

𝐵 𝑥𝐵
𝑑𝑦 𝑑𝑦
𝐽 = න 𝑓 𝑥, 𝑦, 𝑑𝑥 ≡ න 𝑓 𝑥, 𝑦, 𝑑𝑥
𝐴 𝑑𝑥 𝑥𝐴 𝑑𝑥
 But, in nature, we usually deal with having minimum functional:
 Shortest optical path
 Lowest state energy
 Cost function in machine learning
 Expectation values in probabilistic treatments
𝐵 𝑑𝑦
 So, we need 𝛿𝐽 = 𝛿 ‫𝑥 𝑓 𝐴׬‬, 𝑦, 𝑑𝑥 = 0
𝑑𝑥
 The question becomes then what is 𝑦(𝑥) that extremize 𝐽.
SOME MATHEMATICS
CALCULUS OF VARIATION: THE CONCEPT

 For that, we assume that there is some 𝑦 𝑥 that extermize 𝐽.

 But, since we don’t know 𝑦(𝑥), we can introduce a deviation of it.
 𝜂(𝑥) is the difference function:
 𝜂 𝑥𝐴 = 𝜂 𝑥𝐵 = 0, why?
 No deviation at the end points

 So, any path can be describe by: 𝑦 𝑥, 𝛼 = 𝑦 𝑥 + 𝛼 𝜂(𝑥)

 Clearly, 𝑦 𝑥 = 𝑦(𝑥, 0)
𝜕𝐽
 The problem becomes then trackable. The condition for extreme value: ቚ = 0.
𝜕𝛼 𝛼=0
𝑥𝐵 𝑥𝐵
𝜕𝐽 𝜕 𝑑𝑦 𝜕𝑓 𝜕𝑦 𝜕𝑓 𝜕𝑦𝑥
= න 𝑓 𝑥, 𝑦, 𝑑𝑥 = න + 𝑑𝑥
𝜕𝛼 𝜕𝛼 𝑥𝐴 𝑑𝑥 𝑥𝐴 𝜕𝑦 𝜕𝛼 𝜕𝑦𝑥 𝜕𝛼
SOME MATHEMATICS
CALCULUS OF VARIATION: THE CONCEPT
𝑑𝑦 𝑥,𝛼 𝑑𝑦 𝑥 𝑑𝜂 𝑥
 But, 𝑦 𝑥, 𝛼 = 𝑦 𝑥 + 𝛼 𝜂 𝑥 ⟹ = +𝛼
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝜕𝑦 𝜕𝑦𝑥 𝜕 𝑑𝑦 𝑑𝜂 𝑥

𝜕𝛼
= 𝜂(𝑥) and 𝜕𝛼
= 𝜕𝛼 𝑑𝑥
= 𝜂𝑥 𝑥 = 𝑑𝑥

𝜕𝐽 𝑥 𝜕𝑓 𝜕𝑦 𝜕𝑓 𝜕𝑦𝑥 𝑥 𝜕𝑓 𝜕𝑓 𝑑𝜂 𝑥
 So, = ‫𝐵 𝑥׬‬ + 𝜕𝑦 𝑑𝑥 = ‫𝐵 𝑥׬‬ 𝜂(𝑥) + 𝜕𝑦 𝑑𝑥
𝜕𝛼 𝐴 𝜕𝑦 𝜕𝛼 𝑥 𝜕𝛼 𝐴 𝜕𝑦 𝑥 𝑑𝑥

 The 2nd term can be integrated by parts:

𝑥
𝑥𝐵 𝜕𝑓 𝑑𝜂 𝑥 𝜕𝑓 𝐵 𝑥 𝑑 𝜕𝑓 𝑥 𝑑 𝜕𝑓
‫𝑥𝑑 𝑦𝜕 𝑥׬‬ 𝑑𝑥 = 𝜂 𝑥 𝜕𝑦𝑥
ฬ − ‫𝑥 𝜂 𝐵 𝑥׬‬ 𝑑𝑥 𝜕𝑦𝑥
𝑑𝑥 = − ‫𝑥 𝜂 𝐵 𝑥׬‬ 𝑑𝑥 𝜕𝑦𝑥
𝑑𝑥
𝐴 𝑥 𝐴 𝐴
𝑥𝐴

𝜕𝐽 𝑥 𝜕𝑓 𝑑 𝜕𝑓 𝑥 𝜕𝑓 𝑑 𝜕𝑓
 So, = ‫𝐵 𝑥׬‬ 𝜂 𝑥 −𝜂 𝑥 𝑑𝑥 = ‫𝑥 𝜂 𝐵 𝑥׬‬ − 𝑑𝑥 𝑑𝑥 = 0
𝜕𝛼 𝐴 𝜕𝑦 𝑑𝑥 𝜕𝑦𝑥 𝐴 𝜕𝑦 𝜕𝑦𝑥

𝜕𝑓 𝑑 𝜕𝑓
 The only way for this to be generally true is that: − 𝑑𝑥 =0
𝜕𝑦 𝜕𝑦𝑥
𝑘
σ∞ 𝑘 𝑑 𝜕𝑓
 If higher derivatives appear in the functional: 𝑘=0 −1 =0
𝑑𝑥 𝑘 𝜕 𝑑 𝑘 𝑦
𝑥
SOME MATHEMATICS

 Mathematically, the trajectory of a ray is described parametrically

by 𝑥 𝜉 , 𝑦 𝜉 , & 𝑧 𝜉 where 𝜉 is an independent variable of choice.
 Common choices of 𝜉, time 𝑡, displacement 𝑠, and a coordinate.
 What is an infinitesimal displacement 𝑑𝑠 of an arc length?
2 2 2
2 2 2 2 𝑑𝑥 𝑑𝑦 𝑑𝑧
 𝑑𝑠 = 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧 = + + 𝑑𝜉 2
𝑑𝜉 𝑑𝜉 𝑑𝜉

2 2 2
𝑑𝑠 𝑑𝑥 𝑑𝑦 𝑑𝑧
 = + +
𝑑𝜉 𝑑𝜉 𝑑𝜉 𝑑𝜉

 Let’s see the implications of different independent variables:

𝑑𝑠
 If 𝜉 = 𝑡: = 𝑣𝑥2 + 𝑣𝑦2 + 𝑣𝑦2
𝑑𝑡

𝑑𝑠 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
 If 𝜉 = 𝑠: =1= + +
1.3: GRADED-INDEX OPTICS

 A graded-index (GRIN) material has a refractive index that varies with position in
accordance with a continuous function 𝑛(𝒓).
 Should the optical light follow straight paths in GRIN?
 No, in general.
 We need to have the trajectory of light rays.

 This is very handy and it allows using GRIN as a conventional optical component, such as
a prism or lens.
 But as a ray, the light should satisfy Fermat’s principle; the light takes the shortest optical
path between two points.
1.3: GRADED-INDEX OPTICS
THE RAY EQUATION

 For this, we need the calculus of variation:

𝐵
 The optical pathlength: 𝐿Opt = ‫𝑠𝑑 𝒓 𝑛 𝐴׬‬
 This is a functional.

 For the space metric, let’s use the arc-length 𝑠 as the independent variable:

𝑑𝑠 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
 =1= + + ⟹ 𝑑𝑠 → + + 𝑑𝑠
𝑑𝑠 𝑑𝑠 𝑑𝑠 𝑑𝑠 𝑑𝑠 𝑑𝑠 𝑑𝑠

𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
 For collective estimation, + + is set to 1.
𝑑𝑠 𝑑𝑠 𝑑𝑠

 When considering local variation of 𝑑𝑠 w.r.t. a function depending on coordinates, we keep the full form.

𝐵 𝐵 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
 So, the optical pathlength: 𝐿Opt = ‫𝑛 𝐴׬‬ 𝒓 1 𝑑𝑠 = ‫𝑛 𝐴׬‬ 𝒓 + + 𝑑𝑠
𝑑𝑠 𝑑𝑠 𝑑𝑠
1.3: GRADED-INDEX OPTICS
THE RAY EQUATION

 By Fermat’s principle, the light takes the shortest optical path between two points.

𝐵 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
 The shortest path is an extremum of 𝐿Opt . So, 𝛿 ‫𝑛 𝐴׬‬ 𝒓 + + 𝑑𝑠 = 0
𝑑𝑠 𝑑𝑠 𝑑𝑠

𝜕(𝑛 ⋯ ) 𝑑 𝜕 𝑛 ⋯
 The highest derivative order is 1. So, − =0
𝜕𝑞 𝑑𝑠 𝜕𝑞𝑠

𝑑𝑥
𝜕𝑛 𝑑 2 𝑑𝑠 𝑑 𝑑𝑥 𝜕𝑛
 For 𝑞 = 𝑥: ⋯ − 𝑛 =0⟹ 𝑛 =
𝜕𝑥 𝑑𝑠 2 ⋯ 𝑑𝑠 𝑑𝑠 𝜕𝑥

𝑑 𝑑𝑦 𝜕𝑛 𝑑 𝑑𝑧 𝜕𝑛
 Similarly, for 𝑞 = 𝑦 and 𝑞 = 𝑧, one obtains 𝑛 = and 𝑛 =
𝑑𝑠 𝑑𝑠 𝜕𝑦 𝑑𝑠 𝑑𝑠 𝜕𝑧

𝑑 𝑑𝒓
 In a full vectoral form, the above 3 equations are combined in: 𝑛 = ∇𝑛(𝒓)
𝑑𝑠 𝑑𝑠
1.3: GRADED-INDEX OPTICS
THE RAY EQUATION: THE PARAXIAL RAY EQUATION

 In paraxial approximation, the rays are almost parallel to the optical axis.
 Let’s use 𝑧-axis as the optical axis.

𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2 𝑑𝑥 2 𝑑𝑦 2
 So, we need to set 𝜉 = 𝑧, 𝑑𝑠 = + + , 𝑑𝑧 = + + 1 𝑑𝑧
𝑑𝑧 𝑑𝑧 𝑑𝑧 𝑑𝑧 𝑑𝑧

𝑑 𝑑𝑥 𝜕𝑛 𝑑 𝑑𝑦 𝜕𝑛
 By applying the calculus of variation, we obtain 𝑛 = & 𝑛 =
𝑑𝑧 𝑑𝑧 𝜕𝑥 𝑑𝑧 𝑑𝑧 𝜕𝑦
1.3: GRADED-INDEX OPTICS
GRADED-INDEX OPTICAL COMPONENTS: GRADED-INDEX SLAB

 Let’s consider a slab:

 We need to choose the geometrical description in a way
simplifying the math.
 Let 𝑦-axis be the thin direction
 𝑥- and 𝑧- directions will be symmetrically equivalent.
 Then, let the plane containing the ray trajectory be 𝑦𝑧 plane as shown in the figure.
𝜕 𝑑𝑥
 So, nothing is change in the 𝑥-direction, = 0 & 𝑑𝑧 = 0
𝜕𝑥

 Structurally, the refractive index changes only in

the thin direction, i.e 𝑛(𝒓) ≡ 𝑛(𝑦).
 The paraxial ray equation is reduced to:
𝑑 𝑑𝑦 𝑑𝑛 𝑑2 𝑦 1 𝑑𝑛
𝑛 = ⟹ 2=
𝑑𝑧 𝑑𝑧 𝑑𝑦 𝑑𝑧 𝑛 𝑦 𝑑𝑦
1.3: GRADED-INDEX OPTICS
GRADED-INDEX OPTICAL COMPONENTS: GRADED-INDEX SLAB

 Example 1.3-1: Slab with Parabolic Index Profile, 𝑛2 𝑦 = 𝑛02 1 − 𝛼 2 𝑦 2

 This profile is known as self focus index (SelFoc).
 𝛼 is very small to ensure that 𝛼 2 𝑦 2 ≪ 1 for all 𝑦 within the slab.
𝑑2𝑦 1 𝑑𝑛 1 𝑛0 𝛼 2 𝑦 𝛼2𝑦
 The ray equation: = =− = − ≈ −𝛼 2 𝑦
𝑑𝑧 2 𝑛 𝑦 𝑑𝑦 𝑛0 1−𝛼 2 𝑦 2 1−𝛼 2 𝑦 2 1−𝛼 2 𝑦 2

 What is the solution of this differential equation?

 Two sinusoidals: 𝑦 𝑧 = 𝑎 cos 𝛼𝑧 + 𝑏 sin 𝛼𝑧
𝜃0
 By applying the boundary conditions: 𝑎 = 𝑦0 & 𝑏 =
𝛼
 The travelling angle is calculated by:
𝑑𝑦
𝜃 𝑧 ≈ tan 𝜃 𝑧 = = −𝑦0 𝛼 sin 𝛼𝑧 + 𝜃0 cos 𝛼𝑧
𝑑𝑧
 Clearly, there is some 𝑦max (SelFoc)
1.3: GRADED-INDEX OPTICS
GRADED-INDEX OPTICAL COMPONENTS: GRADED-LNDEX FIBERS

 What if the structure is cylindrical and the refractive index decreases with the radius:
𝑛2 (𝒓) ≡ 𝑛2 (𝑥, 𝑦) = 𝑛02 1 − 𝛼 2 𝑥 2 + 𝑦 2 .
 The 𝑧-axis is assumed to be the direction of propagation.
 Again, 𝛼 2 𝑥 2 + 𝑦 2 ≪ 1

 Following the same math and approximations, we

reach to the following decoupled equations:
𝑑2 𝑥
 ≈ −𝛼 2 𝑥: 𝑥 𝑧 ≈ 𝑎𝑥 cos 𝛼𝑧 + 𝑏𝑥 sin 𝛼𝑧
𝑑𝑧 2
𝑑2 𝑦
 ≈ −𝛼 2 𝑦: 𝑦 𝑧 ≈ 𝑎𝑦 cos 𝛼𝑧 + 𝑏𝑦 sin 𝛼𝑧
𝑑𝑧 2
1.3: GRADED-INDEX OPTICS
THE EIKONAL EQUATION

 The ray trajectories are often characterized by the surfaces to which they are normal.
 Let 𝑆(𝒓) be a scalar function such that its equilevel surfaces.
 Clearly, if we know 𝑆 𝒓 , one can determine the ray trajectories. How?
 The rays at any point will be in the direction of ∇𝑆 at that point.
 The full derivation is based on Hamiltonian mehcnaics.
 Eikonal equation:
2 2 2
2
𝜕𝑆 𝜕𝑆 𝜕𝑆
∇𝑆 = + + = 𝑝𝑥2 + 𝑝𝑦2 + 𝑝𝑧2 = 𝑛2
𝜕𝑥 𝜕𝑦 𝜕𝑧
 The optical pathlength can then be obtained directly for 𝑆(𝒓):
𝐵 𝐵 𝐵
𝑆 𝒓𝐵 − 𝑆 𝒓𝐴 = න 𝑑𝑆 = න ∇𝑆 𝑑𝑠 = න 𝑛 𝒓 𝑑𝑠 = 𝐿Opt
𝐴 𝐴 𝐴
1.4: TRANSFER MATRICES FOR RAY OPTICS

 So far, we‘ve been dealing with ray optics.

 Optical rays travel in 2D manifolds. In paraxial approximation, where rays travel around the
optical axis, the 2D manifolds are planes.
 In this section, we will present a handy matrix transfer method to trace rays.
 For that, what are the needed
quantities to describe rays
mathematically in a 𝑦𝑧 plane?
 Position: 𝑦(𝑧)
𝑑𝑦
 Angle: 𝜃 𝑧 ≈ tan 𝜃 𝑧 =
𝑑𝑧
1.4: TRANSFER MATRICES FOR RAY OPTICS
THE RAY-TRANSFER MATRIX

 So, to trace a ray as it travels around the optical axis (𝑧-axis),

we need to check how 𝑦(𝑧) and 𝜃(𝑧) change.
 In principle, we should be able to relate 𝑦 𝑧1 , 𝜃 𝑧1
to 𝑦 𝑧2 , 𝜃 𝑧2 by considering the optical medium
between 𝑧1 & 𝑧2 .
 𝑦 𝑧2 = 𝑓 𝑦 𝑧1 , 𝜃 𝑧1

 𝜃 𝑧2 = 𝑔 𝑦 𝑧1 , 𝜃 𝑧1

 If the medium is linear, the relations are reduced to:

𝑦 𝑧2 = 𝐴 𝑦 𝑧1 , +𝐵 𝜃 𝑧1 𝑦 𝐴 𝐵 𝑦
ቋ⟹ 𝑧2 = 𝑧 ⟹ 𝒘 𝑧2 = 𝐌 𝒘 𝑧1
𝜃 𝑧2 = 𝐶 𝑦 𝑧1 , +𝐷 𝜃 𝑧1 𝜃 𝐶 𝐷 𝜃 1
1.4: TRANSFER MATRICES FOR RAY OPTICS
MATRICES OF SIMPLE OPTICAL COMPONENTS

1. Propagation over a distance 𝑑:

 No change in the angle: 𝜃2 = 𝜃1
 What about 𝑦?
 Clearly, 𝑦2 = 𝑦1 + Δ𝑦 = 𝑦1 + 𝑑 tan 𝜃1 ≈ 𝑦1 + 𝑑 𝜃1
1 𝑑
 So, 𝐌 =
0 1
2. Refraction at a Planar Boundary:
 Clearly, 𝑦2 = 𝑦1 .
𝑛1
 By Snell’s law: 𝑛2 sin 𝜃2 = 𝑛1 sin 𝜃1 → 𝜃2 ≈ 𝜃
𝑛2 1

1 0
 So, 𝐌 = 0 𝑛1
𝑛2
1.4: TRANSFER MATRICES FOR RAY OPTICS
MATRICES OF SIMPLE OPTICAL COMPONENTS

3. Refraction at a Spherical Boundary:

 Let’s start with the easy part. Clearly, 𝑦2 = 𝑦1 .
 Previously, we considered spherical boundaries and found that:
𝑛1 𝑛1 𝑛1 𝑛2 − 𝑛1 𝑦
𝜃2 ≈ 𝜃1 + − 1 𝜃0 ≈ 𝜃1 −
𝑛2 𝑛2 𝑛2 𝑛2 𝑅

1 0
 So, 𝐌 = − (𝑛2−𝑛1) 𝑛1
𝑛2 𝑅 𝑛2

 What if the boundary becomes flat?

 𝑅→∞

1 0 1 0
 (𝑛2 −𝑛1 ) 𝑛1 → 0 𝑛1
− 𝑛2
𝑛2 𝑅 𝑛2
1.4: TRANSFER MATRICES FOR RAY OPTICS
MATRICES OF SIMPLE OPTICAL COMPONENTS

4. Transmission Through a Thin Lens:

 Again, the easy part; 𝑦2 = 𝑦1 .
𝑦
 As learned last Tuesday: 𝜃2 = 𝜃1 −
𝑓

1 0
 So, 𝐌 = − 1 1
𝑓

5. Reflection from a Planar Mirror:

 What is changing?
 The optical axis got reflected.
 In this case, 𝑦 and 𝜃 remained unchanged.

1 0
 So , 𝐌 =
0 1