## Geometrical Optics

While Maxwell’s equations can solve light propagation in a rigorous way, the exact solutions can be found in fairly limited cases, and most practical examples require approximations.

Based on the specific method of approximation, optics has been broadly divided into two categories, namely:

i. Geometrical Optics (ray optics), treated in the first half of the class

- Emphasizes on finding the light path

e Especially useful for studying the optical behavior of the system which has length scale much larger than the wavelength of light, such as: - designing optical instruments,
- tracing the path of propagation in inhomogeneous media.

ii. Wave Optics (physical optics)- - Emphasizes on analyzing interference and diffraction
- Gives more accurate determination of light distributions, because both the amplitude and phase of the light are considered.

A. Geometrical Light Rays Geometrical optics is an intuitive and efficient approximation:

Radios and mobile phones make use of same Maxwell Equations to transfer information as carried by light waves, but our perception is quite different. Why?

## Matrix Methods in Paraxial Optics

**Optical Invariant**

-What happens to an arbitrary “axial” ray that originates from the axial intercept of the object, after passing through a series of lenses?

If we make use of the relationship between launching angle and the imaging conditions, we have:

$$

\begin{array}{c}

\theta_{\text {in }}=\frac{x_{\text {in }}}{s_{o}} \text { and } \theta_{\text {out }}=-\frac{x_{\text {in }}}{s_{i}} \

\frac{\theta_{\text {in }}}{\theta_{\text {out }}}=-\frac{s_{i}}{s_{o}}=\frac{h_{i}}{h_{o}}

\end{array}

$$

Rearranging, we obtain:

$$

\theta_{\text {in }} h_{o}=\theta_{\text {out }} h_{i}

$$

We see that the product of the image height and the angle with respect to the axis (the components of the ray vector!) remains a constant. Indeed a more general result, $n h_{o} \sin \theta_{\text {in }}=n^{\prime} h_{i} \sin \theta_{\text {out }}$ is a constant (often referred as a Lagrange invariant in different textbooks) across any surface of the imaging system. The invariant may be used to deduce other quantities of the optical system, without the necessity of certain intermediate ray-tracing calculations. You may regard it as a precursor to wave optics: the angles are approximately proportional to lateral momentum of light, and the image height is equivalent to separation of two geometric points. For two points that are separated far apart, there is a limiting angle to transmit their information across the imaging system.

**A few limiting cases:**

a) Parallel beams from the left: $s_{i 2}$ is the back-focal length (BFL)

$$

\frac{1}{\mathrm{BFL}}=\left(\frac{1}{f_{1}}+\frac{1}{f_{2}}\right)-\frac{d}{\left(d-f_{1}\right) f_{1}}

$$

b) collimated beams to the right: $s_{o 1}$ is the front-focal length (FFL)

$$

\frac{1}{\mathrm{FFL}}=\left(\frac{1}{f_{1}}+\frac{1}{f_{2}}\right)-\frac{d}{\left(d-f_{2}\right) f_{2}}

$$

The composite lens does not have the same apparent focusing length in front and back end!

c) $d=f_{1}+f_{2}$ : parallel beams illuminating the composite lens will remain parallel at the exit; the system is often called afocal. This is in fact the principle used in most telescopes, as the object is located at infinity and the function of the instrument is to send the image to the eye with a large angle of view. On the other hand, a point source located at the left focus of the first lens is imaged at the right focus of the second lens (the two are called conjugate points). This is often used as a condenser for illumination.

## EM Wave Equations

**From time domain to frequency domain**

Continuous wave laser light field understudy are often mono-chromatic. These problems are mapped in the Maxwell equations by expanding complex time signals to a series of time harmonic components (often referred to as “single” wavelength light):

e.g. $\vec{E}(\vec{r}, t)=\int_{-\infty}^{\infty} \overrightarrow{\boldsymbol{E}}(\vec{r}, \omega) \exp (-i \omega t) d \omega$

Advantage:

$\frac{\partial}{\partial t} \vec{E}(\vec{r}, t)=\int_{-\infty}^{\infty} \overrightarrow{\boldsymbol{E}}(\vec{r}, \omega) \frac{\partial}{\partial t} \exp (-i \omega t) d \omega=\int_{-\infty}^{\infty}[-\boldsymbol{i} \omega \overrightarrow{\boldsymbol{E}}(\overrightarrow{\boldsymbol{r}}, \boldsymbol{\omega})] \exp (-i \omega t) d \omega$ (6)

So, we can replace all time derivatives $\frac{\partial}{\partial t}$ by $-i \omega$ in frequency domain:

(Faraday’s Law:)

$$

\vec{\nabla} \times \vec{E}=+i \omega \vec{B}

$$

(Ampere-Maxwell’s Law:)

$$

\vec{\nabla} \times \vec{H}=\vec{J}-i \omega \vec{D}

$$

The other two equations remain unaltered.

There are total of 12 unknowns $(\boldsymbol{E}, \boldsymbol{H}, \boldsymbol{D}, \boldsymbol{B})$ but so far we only obtained 8 equations from the Maxwell equations ( 2 vector form $x 3+2$ scalar forms ) so more information needed to understand the complete wave behavior!

Generally we may start to construct the response of a material by applying a excitation field $\mathrm{E}$ or $\mathrm{H}$ in vacuum. Therefore it is more typical to consider the $\mathrm{E}, \mathrm{H}$ field as input and D, B fields as output. In common optical materials, we may enjoy the following simplification of local (i.e. independent of neighbors) and linear relationship:

$$

\begin{array}{r}

\overrightarrow{\boldsymbol{D}}(\omega)=\varepsilon(\omega) \varepsilon_{0} \overrightarrow{\boldsymbol{E}}(\omega) \

\overrightarrow{\boldsymbol{B}}(\omega)=\mu(\omega) \mu_{0} \overrightarrow{\boldsymbol{H}}(\omega)

\end{array}

$$

The so called (electric) permittivity $\varepsilon(\omega)$ and (magnetic) permeability $\mu(\omega)$ are unitless parameters that depend on the frequency of the input field. In the case of anisotropic medium, both $\varepsilon(\omega)$ and $\mu(\omega)$ become $3 \times 3$ dimension tensor.

Now we have 6 more equations from material response, we can include them together with Maxwell equations to obtain a complete solution of optical fields with proper boundary condition.

## Eikonal Equations, Gradient Index Lenses, Hamiltonian Optics

How can we obtain Geometric optics picture such as ray tracing from wave equations? Now let’s go back to real space and frequency domain (in an isotropic medium but with spatially varying permittivity $\varepsilon(x, z)$, for example).

$$

\frac{\partial^{2}}{\partial x^{2}} E_{x}+\frac{\partial^{2}}{\partial z^{2}} E_{x}+\varepsilon(x, z)\left(\frac{\omega^{2}}{c_{0}^{2}}\right) E_{x}=0

$$

To see the connection to geometric optics, we decompose the field $\mathbf{E}(\mathbf{r}, \omega)$ into two forms: $\mathrm{a}$ fast oscillating component $\exp \left(\mathrm{ik}*{0} \Phi\right)$, $\mathrm{k}*{0}=\omega / c_{0}$ and a slowly varying envelope $\mathbf{E}*{0}(\mathbf{r})$ as Furthermore, if the envelope of field varies slowly with wavelength (e.g. $\frac{1}{k} \frac{\partial}{\partial x} E*{0} \ll 1, \frac{1}{k} \frac{\partial}{\partial z} E_{0} \ll 1$ )

then we can convert wave equations to the wellknown Eikonal equation:

$\left(\frac{\partial \Phi}{\partial x}\right)^{2}+\left(\frac{\partial \Phi}{\partial z}\right)^{2}=\varepsilon(x, z)=n^{2}(x, z)$

**Path of Light in an Inhomogeneous Medium**

The best known example of this kind is probably the Mirage effect in desert or near a seashore, and we heard of the explanation such as the refractive index increases with density (and hence decreases with temperature at a given altitude). With the picture in mind, now can we predict more accurately the ray path and image forming processes?

Image of Mirage effect removed due to copyright restrictions.

Starting from the Eikonal equation and we assume $n^{2}(x, z)$ is only a function of $x$, then we find:

$$

\left(\frac{\partial \Phi}{\partial x}\right)^{2}+\left(\frac{\partial \Phi}{\partial z}\right)^{2}=n^{2}(x)

$$

Since there is the index in independent of $\mathrm{z}$, we may assume the slope of phase change in $z$ direction is linear:

$$

\left(\frac{\partial \Phi}{\partial z}\right)=C(\text { const })

$$

This allows us to find

$$

\frac{\partial \Phi}{\partial x}=\sqrt{n^{2}(x)-C^{2}}

$$

From Fermat’s principle, we can visualize that direction of rays follow the gradient of phase front:

z-direction:

$$

\begin{array}{r}

n \frac{d \vec{r}}{d l}=\nabla \Phi \

n(x) \frac{d z}{d t}=C

\end{array}

$$

$\mathrm{x}$ -direction:

$n(x) \frac{d x}{d t}=\sqrt{n^{2}(x)-C^{2}}$

Therefore, the light path $(x, z)$ is determined by:

$$

\frac{d z}{d x}=\frac{c}{\sqrt{n^{2}(x)-C^{2}}}

$$

Hence

$$

z-z_{0}=\int_{x_{0}}^{x} \frac{C}{\sqrt{n^{2}(x)-C^{2}}} d x

$$

Without loss of generality, we may assume a quadratic index profile along the $\mathrm{x}$ direction, such as found in gradient index optical fibers or rods:

$$

\begin{array}{c}

n^{2}(x)=n_{0}^{2}\left(1-\alpha x^{2}\right) \

z-z_{0}=\int_{x_{0}}^{x} \frac{C}{\sqrt{n_{0}^{2}\left(1-\alpha x^{2}\right)-C^{2}}} d x

\end{array}

$$

To find the integral explicitly we may take the following transformation of the variable $\mathrm{x}$ :

$$

x=\sqrt{\frac{n_{0}^{2}-C^{2}}{n_{0}^{2} \alpha}} \sin \theta

$$

Therefore,

$$

\begin{array}{c}

z-z_{0}=\int_{\theta_{0}}^{\theta} \frac{C}{n_{0} \sqrt{\alpha}} d \theta \

z=z_{0}+\frac{C}{n_{0} \sqrt{\alpha}}\left(\theta-\theta_{0}\right)

\end{array}

$$

## Fraunhofer Diffraction

The field is affected by 3 contributions:

a) The illumination source $E(x, y)$

b) The transmission function through an aperture $\mathrm{t}(\mathrm{x}, \mathrm{y})$

c) (Dipole or Huygens source) radiation at location $(\mathrm{x}, \mathrm{y})$, propagating to screen $\left(x^{\prime}, y^{\prime}\right): \quad h\left(x^{\prime}-x, y^{\prime}-y, z\right)$

Note: The step c) is known as Huygens principle: every point along a wave-front emits a spherical wave that interferes with all others. Several scientists, including Kirchhoff, and Bethe-Boukamp (1946) attempted to quantify this idea based on Maxwell Equations, but the strength and orientation of the source in metallic holes at optical wavelength is now a hot topic under debate, since Ebbesen’s experiments in 1998 .

Here we take the simplest case of sphere waves:

$$

h\left(x^{\prime}-x, y^{\prime}-y, z\right)=\frac{\exp (i k r)}{r}

$$

where

$$

r=\sqrt{\left(x^{\prime}-x\right)^{2}+\left(y^{\prime}-y\right)^{2}+z^{2}}

$$

The resulting field is then a convolution of all the three factors:

$$

E\left(x^{\prime}, y^{\prime}\right)=\iint h\left(x^{\prime}-x, y^{\prime}-y, z\right) t(x, y) E(x, y) d x d y

$$

o Fraunhoffer diffraction: Far field $\left(z>>x^{\prime}, y^{\prime}, x, y\right)$

$$

\begin{array}{c}

r=\sqrt{\left(x^{\prime}-x\right)^{2}+\left(y^{\prime}-y\right)^{2}+z^{2}} \

r \approx z\left(1+\frac{\left(x^{\prime}-x\right)^{2}+\left(y^{\prime}-y\right)^{2}}{2 z^{2}}\right) \

\exp (i k r) \approx \exp \left(i k z+i k \frac{\left(x^{\prime}-x\right)^{2}+\left(y^{\prime}-y\right)^{2}}{2 z}\right) \

\approx \exp (i k z) \exp \left(-i k \frac{x x^{\prime}+y y^{\prime}}{z}\right) \exp \left(i k \frac{x^{\prime 2}+y^{\prime 2}+x^{2}+y^{2}}{2 z}\right)

\end{array}

$$

(up to here, it is Fresnel condition for diffraction)

Now if we further assume the quadratic terms are negligible:

(Fraunhofer condition, difficult to achieve!)

$$

\frac{k\left(x^{\prime 2}+y^{\prime 2}\right)}{2 z} \ll 1, \frac{k\left(x^{2}+y^{2}\right)}{2 z} \ll 1

$$

Then we have a set of plane waves (rays) launched at $\mathrm{x}, \mathrm{y}$ :

$$

\begin{array}{r}

\exp \left(-i k \frac{x x^{\prime}+y y^{\prime}}{z}\right) \approx \exp \left(-i k\left(\theta_{x^{\prime}} x+\theta_{y^{\prime}} y\right)\right) \

\theta_{x^{\prime}} \approx \frac{x^{\prime}}{z^{\prime}}, \theta_{y^{\prime}} \approx \frac{y^{\prime}}{z}, \text { or } k_{x} \approx k \frac{x^{\prime}}{z^{\prime}} k_{y} \approx k \frac{y_{\prime}}{z}

\end{array}

$$

What we