# Mathematical Methods of Optical Coherence Tomography

Reference work entry

## Abstract

In this chapter a general mathematical model of Optical Coherence Tomography (OCT) is presented on the basis of the electromagnetic theory. OCT produces high-resolution images of the inner structure of biological tissues. Images are obtained by measuring the time delay and the intensity of the backscattered light from the sample considering also the coherence properties of light. The scattering problem is considered for a weakly scattering medium located far enough from the detector. The inverse problem is to reconstruct the susceptibility of the medium given the measurements for different positions of the mirror. Different approaches are addressed depending on the different assumptions made about the optical properties of the sample. This procedure is applied to a full field OCT system and an extension to standard (time and frequency domain) OCT is briefly presented.

## 1 Introduction

Optical Coherence Tomography (OCT) is a noninvasive imaging technique producing high-resolution images of biological tissues. OCT is based on Low (time) Coherence Interferometry and takes into account the coherence properties of light to image microstructures with resolution in the range of few micrometers. Standard OCT operates using broadband and continuous wave light in the visible and near-infrared spectrum. OCT images are obtained by measuring the time delay and the intensity of backscattered or back-reflected light from the sample under investigation. Since it was first established in 1991 by Huang et al. [24], the clinical applications of OCT have been greatly improved. Ophthalmology remains the dominant one, initially applied in 1993 [17, 41]. The main reason is that OCT has limited penetration depth in biological tissues, but high resolution. The theory of OCT has been analyzed in details in review papers [14, 16, 32, 36, 44] in book chapters [15, 19, 42] and in books [4, 5, 10].

To derive a mathematical model for the OCT system, the scattering properties of the sample need to be described. There exist several different approaches to model the propagation of light within the sample: the radiative transfer equation with scattering and absorption coefficients [9, 38, 45], Lambert–Beer’s law with the attenuation coefficient [39, 46], the equations of geometric optics with the refractive index [7], and Maxwell’s equations with the susceptibility (or the refractive index) as optical parameters of the medium [6, 12, 27, 37, 43]. Also statistical approaches using Monte Carlo simulations are used [2, 11, 26, 31, 40].

This chapter describes the propagation of the electromagnetic wave through the sample using Maxwell’s equations and adopts the analysis based on the theory of electromagnetic fields scattered by inhomogeneous media [8, 20]. The sample is hereby considered as a linear dielectric medium (potentially inhomogeneous and anisotropic). Moreover, the medium is considered weakly scattering so that the first-order Born approximation can be used and, as it is usually assumed in OCT, the backscattered light is detected far enough from the sample so that the far field approximation is valid. Starting from this model, different reconstruction formulas for special cases regarding the inner structure of the sample are presented.

This chapter is organized as follows. In Sect. 2, the principles of OCT and different variants of OCT systems are presented. Section 3 describes the solution of Maxwell’s equations and an appropriate formula for the measurements of OCT is derived. Given the initial field and the optical properties (the susceptibility) of the sample, the solution of the direct problem is obtained in Sect. 4. An iterative scheme is derived in the last section for the reconstruction of the unknown susceptibility, which is the inverse problem of OCT.

## 2 Basic Principles of OCT

OCT is used to gain information about the light scattering properties of an object by illuminating it with some short laser pulse and measuring the backscattered light.

The name “Optical Coherence Tomography” is motivated by the way the scattering data are measured: To get more precise measurements, the backscattered light is not directly detected, but first superimposed with the original laser pulse and then the intensity of this interference pattern is measured (this means that one measures the “coherence” of these two light beams).

Experimentally, this is done by separating the incoming light at a beam splitter into two identical beams which travel two different paths. One beam is simply reflected by a mirror and sent back to the beam splitter, while the other beam is directed to the sample. At the beam splitter, the beam reflected by the mirror and the backscattered light from the sample are recombined and sent to the detector [16, 25, 44]. See Fig. 1 for an illustration of this procedure.
There exist different variants of the OCT regarding the way the measurements are done:
Time and frequencydomain OCT:

In time domain OCT, the position of the mirror is varied and for each position one measurement is performed. On the other hand, in frequency domain OCT, the reference mirror is fixed and the detector is replaced by a spectrometer. Both methods provide equivalent measurements which are connected by a Fourier transform.

Standard and fullfield OCT:

In standard OCT, the incoming light is focused through objective lenses to one spot in a certain depth in the sample and the backscattered light is measured in a point detector. This means that to obtain information of the whole sample, a transversal-lateral scan has to be performed (by moving the light beam over the frontal surface of the sample). In full field OCT, the entire frontal surface of the sample is illuminated at once and the single point detector is replaced by a two-dimensional detector array, for instance by a charge-coupled device (CCD) camera.

Polarization-sensitiveOCT:

In classical OCT setups, the electromagnetic wave is simply treated as a scalar quantity. In polarization-sensitive OCT, however, the illuminating light beams are polarized and the detectors measure the intensity of the two polarization components of the interfered light.

There are also further modifications such as Doppler OCT and quantum OCT, which are not addressed here. In this chapter, the focus is mainly on time domain full field OCT, but also the others are discussed.

## 3 The Direct Scattering Problem

To derive a mathematical model for an OCT system, one has to describe on one hand the propagation and the scattering of the laser beam in the presence of the sample and on the other hand the way how this scattered wave is measured at the detectors. For the first part, the interaction of the incoming light with the sample can be modeled with Maxwell’s macroscopic equations.

### Maxwell’s Equations

Maxwell’s equations in matter consist of the partial differential equations
$$\displaystyle\begin{array}{rcl} {\mathrm{div}}_{x}D(t,x) = 4\pi \rho (t,x),\qquad \qquad \qquad \qquad t \in \mathbb{R},\;x \in \mathbb{R}^{3},& &{}\end{array}$$
(1a)
$$\displaystyle\begin{array}{rcl} {\mathrm{div}}_{x}B(t,x) = 0,\quad \qquad \qquad \qquad \qquad t \in \mathbb{R},\;x \in \mathbb{R}^{3},& &{}\end{array}$$
(1b)
$$\displaystyle\begin{array}{rcl} {\mathrm{curl}}_{x}E(t,x) = -\frac{1} {c} \frac{\partial B} {\partial t} (t,x),\quad \qquad \qquad \qquad \qquad t \in \mathbb{R},\;x \in \mathbb{R}^{3},& &{}\end{array}$$
(1c)
$$\displaystyle\begin{array}{rcl} {\mathrm{curl}}_{x}H(t,x) = \frac{4\pi } {c}J(t,x) + \frac{1} {c} \frac{\partial D} {\partial t} (t,x),\qquad \qquad t \in \mathbb{R},\;x \in \mathbb{R}^{3},& &{}\end{array}$$
(1d)
relating the following physical quantities (at some time $$t \in \mathbb{R}$$ and some location $$x \in \mathbb{R}^{3}$$):
 Speed of light c $$\mathbb{R}$$ External charge density ρ(t, x) $$\mathbb{R}$$ External electric current density J(t, x) $$\mathbb{R}^{3}$$ Electric field E(t, x) $$\mathbb{R}^{3}$$ Electric displacement D(t, x) $$\mathbb{R}^{3}$$ Magnetic induction B(t, x) $$\mathbb{R}^{3}$$ Magnetic field H(t, x) $$\mathbb{R}^{3}$$

Maxwell’s equations do not yet completely describe the propagation of the light (even assuming that the charge density ρ and the current density J are known, there are only 8 equations for the 12 unknowns E, D, B, and H).

Additionally to Maxwell’s equations, it is therefore necessary to specify the relations between the fields D and E as well as between B and H.

Let $$\Omega \subset \mathbb{R}^{3}$$ denote the domain where the sample is located. It is considered as a nonmagnetic, dielectric medium without external charges or currents, this means that for all $$t \in \mathbb{R}$$ and all $$x \in \Omega$$ the electric and magnetic fields fulfil the relations
$$\displaystyle\begin{array}{rcl} D(t,x)& =& E(t,x) +\int _{ 0}^{\infty }\chi (\tau,x)E(t-\tau,x){\mathrm{d}}\tau,{}\end{array}$$
(2a)
$$\displaystyle\begin{array}{rcl} B(t,x)& =& H(t,x),{}\end{array}$$
(2b)
$$\displaystyle\begin{array}{rcl} \rho (t,x)& =& 0,{}\end{array}$$
(2c)
$$\displaystyle\begin{array}{rcl} J(t,x)& =& 0,{}\end{array}$$
(2d)
where the function $$\chi: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3\times 3}$$ (for convenience, χ is also defined for negative times by χ(t, x) = 0 for t < 0, $$x \in \mathbb{R}^{3}$$) is called the (electric) susceptibility and is the quantity to be recovered. The time dependence of χ hereby describes the fact that a change in the electric field E cannot immediately cause a change in the electric displacement D. Since this delay is quite small, it is sometimes ignored and χ(t, x) is then replaced by δ(t)χ(x). Moreover, the medium is often considered to be isotropic, which means that χ is a multiple of the identity matrix.
The sample is situated in vacuum and the assumptions (2) are modified by setting for all $$t \in \mathbb{R}$$ and all $$x \in \mathbb{R}^{3}\setminus \Omega$$
$$\displaystyle\begin{array}{rcl} D(t,x) = E(t,x),& &{}\end{array}$$
(3a)
$$\displaystyle\begin{array}{rcl} B(t,x) = H(t,x),& &{}\end{array}$$
(3b)
$$\displaystyle\begin{array}{rcl} \rho (t,x) = 0,& &{}\end{array}$$
(3c)
$$\displaystyle\begin{array}{rcl} J(t,x) = 0.& &{}\end{array}$$
(3d)
This simply corresponds to extend the Eq. (2) to $$\mathbb{R} \times \mathbb{R}^{3}$$ and to assume χ(t, x) = 0 for all $$t \in \mathbb{R}$$, $$x \in \mathbb{R}^{3}\setminus \Omega$$.
In this case of a nonmagnetic medium, Maxwell’s equations result into one equation for the electric field E. To get rid of the convolution in (2a), it is practical to consider the Fourier transform with respect to time. In the following, the convention
$$\displaystyle{\hat{f}(\omega,x) =\int _{ -\infty }^{\infty }f(t,x){\mathrm{e}}^{{\mathrm{i}}\omega t}{\mathrm{d}}t,}$$
for the Fourier transform of a function f with respect to t is used.

### Proposition 1.

Let E, D, B, and H fulfil Maxwell’s equations (1) . Moreover, let assumptions (2) and (3) be satisfied. Then the Fourier transform$$\hat{E}$$of E fulfils the vector Helmholtz equation
$$\displaystyle{ {\mathrm{curl}}_{x}{\mathrm{curl}}_{x}\hat{E}(\omega,x) - \frac{\omega ^{2}} {c^{2}}(\mathbb{1} +\hat{\chi } (\omega,x))\hat{E}(\omega,x) = 0,\quad \omega \in \mathbb{R},\;x \in \mathbb{R}^{3}. }$$
(4)

### Proof.

Applying the curl to (1c) and using (1d) with the assumptions B = H and J = 0, yields
$$\displaystyle{ {\mathrm{curl}}_{x}{\mathrm{curl}}_{x}E(t,x) = -\frac{1} {c} \frac{\partial {\mathrm{curl}}_{x}B} {\partial t} (t,x) = -\frac{1} {c^{2}} \frac{\partial ^{2}D} {\partial t^{2}} (t,x). }$$
(5)
The Fourier transform of (2a) and (3a) and the Fourier convolution theorem (recall that χ is set to zero outside $$\Omega$$) imply that
$$\displaystyle{\hat{D}(\omega,x) = (\mathbb{1} +\hat{\chi } (\omega,x))\hat{E}(\omega,x),\quad \text{for all}\quad \omega \in \mathbb{R},\;x \in \mathbb{R}^{3}.}$$
Therefore, the Eq. (4) follows by taking the Fourier transform of (5). □

### Initial Conditions

The sample is illuminated with a laser beam described initially (before it interacts with the sample) by the electric field $$E^{(0)}: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$$ which is (together with some magnetic field) a solution of Maxwell’s equations (1) with the assumptions (3) for all $$x \in \mathbb{R}^{3}$$. Then, it follows from the proof of the 1, for χ = 0, that
$$\displaystyle{ {\mathrm{curl}}_{x}{\mathrm{curl}}_{x}\hat{E}^{(0)}(\omega,x) - \frac{\omega ^{2}} {c^{2}}\hat{E}^{(0)}(\omega,x) = 0,\quad \omega \in \mathbb{R},\;x \in \mathbb{R}^{3}. }$$
(6)

Moreover, it is assumed that E(0) does not interact with the sample until the time t = 0, which means that supp $$E^{(0)}(t,\cdot ) \cap \Omega =\emptyset$$ for all t ≤ 0.

The electric field $$E: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$$ generated by this incoming light beam in the presence of the sample is then a solution of Maxwell’s equations (1) with the assumptions (2) and the initial condition
$$\displaystyle{ E(t,x) = E^{(0)}(t,x)\quad \text{for all}\quad t \leq 0,\;x \in \mathbb{R}^{3}. }$$
(7)

Since Maxwell’s equations for E in 1 are reformulated as an equation for the Fourier transform $$\hat{E}$$, it is helpful to rewrite the initial condition in terms of $$\hat{E}$$.

### Proposition 2.

Let E (together with some magnetic field H) fulfil Maxwell’s equations (1) with the assumptions (2) and (3) and with the initial condition (7) .

Then the Fourier transform of E − E(0)fulfils that the function$$\omega \mapsto \hat{E}(\omega,x) -\hat{ E}^{(0)}(\omega,x)$$, defined on$$\mathbb{R}$$, can be extended to a square integrable, holomorphic function on the upper half plane$$\mathbb{H} =\{\omega \in \mathbb{C}\mid \mathfrak{I}{\mathrm{m}}(\omega ) > 0\}$$for every$$x \in \mathbb{R}^{3}$$.

### Proof.

From the initial condition (7) it follows that E(t, x) − E(0)(t, x) = 0 for all t ≤ 0. Thus, the result is a direct consequence form the Paley–Wiener theorem, which is based on the fact that in this case
$$\displaystyle{\hat{E}(\omega,x) -\hat{ E}^{(0)}(\omega,x) =\int _{ 0}^{\infty }(E - E^{(0)})(t,x){\mathrm{e}}^{{\mathrm{i}}\omega t}{\mathrm{d}}t}$$
is well defined for all $$\omega \in \mathbb{H}$$ and complex differentiable with respect to $$\omega \in \mathbb{H}$$. □

Remark that the electric field E is uniquely defined by (4) and 2.

### The Measurements

The measurements are obtained by the combination of the backscattered field from the sample and the back-reflected field from the mirror. In practice, see Fig. 1, the sample and the mirror are in different positions. However, without loss of generality, a placement of them around the origin is assumed in the proposed formulation, in order to avoid rotating the coordinate system. To do so, the simultaneously illumination of the sample and the mirror is suppressed and two different illumination schemes are considered. The gain is to keep the same coordinate system but the reader should not be confused with illumination at different times.

Thus, the electric field E, which is obtained by illuminating the sample with the initial field E(0) (that is E solves (4) with the initial condition (7)), is combined with Er which is the electric field obtained by replacing the sample by a mirror and illuminating with the same initial field E(0).

The mirror is placed orthogonal to the unit vector e3 = (0, 0, 1) through the point re3. As in (7), it is assumed that supp E(0)(t, ⋅ ) does not interact with the mirror for t < 0, so that
$$\displaystyle{ E_{r}(t,x) = E^{(0)}(t,x)\quad \text{for all}\quad t < 0,\;x \in \mathbb{R}^{3}. }$$
(8)
Then the resulting electric field $$E_{r}: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$$ is given as the solution of the same equations as E (Maxwell’s equations (1) together with the assumptions (2) and initial condition (8)) with the susceptibility χ replaced by the susceptibility χr of the mirror at position r. One sort of (ideal) mirror can be described via the susceptibility χr(t, x) = 0 for x3 > r and $$\chi _{r}(t,x) = C\delta (t)\mathbb{1}$$ for x3 ≤ r with an (infinitely) large constant C > 0.
The intensity Ir of each component of the superposition of the electric fields E and Er averaged over all time is measured at some detector points. The detectors are positioned at all points on the plane
$$\displaystyle{\mathcal{D} =\{ x \in \mathbb{R}^{3}\mid x_{ 3} = d\}}$$
parallel to the mirror at a distance d > 0 from the origin. The mirror and the sample are both located in the lower half plane of the detector surface with some minimal distance to $$\mathcal{D}$$. Moreover, the highest possible position R ∈ (δ, d − 2δ) of the mirror shall be by some distance δ > 0 closer to the detector than the sample, this means (see Fig. 2)
$$\displaystyle{ \sup _{x\in \Omega }x_{3} < R -\delta \quad \text{and}\quad r \in (-\infty,R). }$$
(9)
To simplify the argument, let us additionally assume that the incoming electric field E(0) does not influence the detector after the time t = 0, meaning that
$$\displaystyle{ E^{(0)}(t,x) = 0\quad \text{for all}\quad t \geq 0,\;x \in \mathcal{D}. }$$
(10)
At the detector array, the data are obtained by measuring
$$\displaystyle{ I_{r,j}(x) =\int _{ 0}^{\infty }\vert E_{ j}(t,x) + E_{r,j}(t,x)\vert ^{2}{\mathrm{d}}t,\quad x \in \mathcal{D},\;j \in \{ 1,2,3\}. }$$
(11)
In standard OCT, the polarization is usually ignored. In this case, only the total intensity $$I_{r} =\sum _{ j=1}^{3}I_{r,j}$$ needs to be measured, see Sect. 5 for the reconstruction formulas in the isotropic case.
In this measurement setup, it is easy to acquire besides the intensity Ir also the intensity of the two waves E and Er separately by blocking one of the two waves E and Er at a time. Practically, it is sometimes not even necessary to measure them since the intensity of the reflected laser beam Er can be explicitly calculated from the knowledge of the initial beam E(0), and the intensity of E is usually negligible compared with the intensity Ir (because of the assumption (10), the field E contains only backscattered light at the detector after the measurement starts). Therefore, one can consider instead of Ir the function
$$\displaystyle{ M_{r,j}(x) = \frac{1} {2}\left (I_{r,j} -\int _{0}^{\infty }\vert E_{ j}(t,x)\vert ^{2}{\mathrm{d}}t -\int _{ 0}^{\infty }\vert E_{ r,j}(t,x)\vert ^{2}{\mathrm{d}}t\right ) }$$
(12)
for r ∈ (−, R), j ∈ { 1, 2, 3}, and $$x \in \mathcal{D}$$ as the measurement data.

### Proposition 3.

Let the initial conditions (7) and (8) and the additional assumption (10) be satisfied. Then, for all$$x \in \mathcal{D}$$, r ∈ (−∞,R), and j ∈{ 1,2,3} the measurements Mr, defined by (12), fulfil
$$\displaystyle\begin{array}{rcl} M_{r,j}(x)& =& \int _{-\infty }^{\infty }(E_{ j} - E_{j}^{(0)})(t,x)(E_{ r,j} - E_{j}^{(0)})(t,x){\mathrm{d}}t{}\end{array}$$
(13a)
$$\displaystyle\begin{array}{rcl} =\int _{ -\infty }^{\infty }(\hat{E}_{ j} -\hat{ E}_{j}^{(0)})(\omega,x)\overline{(\hat{E}_{ r,j} -\hat{ E}_{j}^{(0)})(\omega,x)}{\mathrm{d}}\omega,& &{}\end{array}$$
(13b)

### Proof.

Expanding the function Ir, j, given by (11), gives
$$\displaystyle{I_{r,j}(x) =\int _{ 0}^{\infty }\big(\vert E_{ j}(t,x)\vert ^{2} + \vert E_{ r,j}(t,x)\vert ^{2} + 2E_{ j}(t,x)E_{r,j}(t,x)\big){\mathrm{d}}t.}$$
Thus, by the Definition (12) of Mr, it follows that
$$\displaystyle{M_{r,j}(x) =\int _{ 0}^{\infty }E_{ j}(t,x)E_{r,j}(t,x){\mathrm{d}}t,}$$
which, using the assumption (10), can be rewritten in the form
$$\displaystyle{M_{r,j}(x) =\int _{ 0}^{\infty }(E_{ j} - E_{j}^{(0)})(t,x)(E_{ r,j} - E_{j}^{(0)})(t,x){\mathrm{d}}t.}$$
Then, since E and Er coincide with E(0) for t < 0, see (7) and (8), the integration can be extended to all times. This proves the formula (13a) for Mr. The second formula follows from Plancherel’s theorem. □

## 4 Solution of the Direct Problem

In this section the solution of the direct problem, to determine the measurements Mr, defined by (13a), from the susceptibility χ, is derived using Born and far field approximation for the electric field.

### Proposition 4.

Let E be a solution of the Eqs. (4) and (7) . Then, the Fourier transform$$\hat{E}$$solves the Lippmann–Schwinger integral equation
$$\displaystyle{ \hat{E}(\omega,x) =\hat{ E}^{(0)}(\omega,x) + \left ( \frac{\omega ^{2}} {c^{2}}\mathbb{1} + {\mathrm{grad}}_{x}{\mathrm{div}}_{x}\right )\int _{\mathbb{R}^{3}}G(\omega,x - y)\hat{\chi }(\omega,y)\hat{E}(\omega,y){\mathrm{d}}y, }$$
(14)
where G is the fundamental solution of the Helmholtz equation given by
$$\displaystyle{G(\omega,x) = \frac{{\mathrm{e}}^{{\mathrm{i}} \frac{\omega } { c}\vert x\vert }} {4\pi \vert x\vert },\quad x\neq 0,\,\omega \in \mathbb{R}.}$$

### Proof.

Equation (4) can be rewritten in the form
$$\displaystyle{{\mathrm{curl}}_{x}{\mathrm{curl}}_{x}\hat{E}(\omega,x) - \frac{\omega ^{2}} {c^{2}}\hat{E}(\omega,x) =\phi (\omega,x)}$$
with the inhomogeneity
$$\displaystyle{ \phi (\omega,x) = \frac{\omega ^{2}} {c^{2}}\hat{\chi }(\omega,x)\hat{E}(\omega,x). }$$
(15)
Using that $$\hat{E}^{(0)}$$ solves (6), the difference $$\hat{E} -\hat{ E}^{(0)}$$ satisfies the inhomogeneous vector Helmholtz equation
$$\displaystyle{ {\mathrm{curl}}_{x}{\mathrm{curl}}_{x}(\hat{E} -\hat{ E}^{(0)})(\omega,x) - \frac{\omega ^{2}} {c^{2}}(\hat{E} -\hat{ E}^{(0)})(\omega,x) =\phi (\omega,x). }$$
(16)
The divergence of this equation, using that $${\mathrm{div}}_{x}{\mathrm{curl}}_{x}(\hat{E} -\hat{ E}^{(0)}) = 0,$$ implies
$$\displaystyle{ {\mathrm{div}}_{x}(\hat{E} -\hat{ E}^{(0)})(\omega,x) = -\frac{c^{2}} {\omega ^{2}} {\mathrm{div}}_{x}\phi (\omega,x). }$$
(17)
Applying the vector identity
$$\displaystyle{{\mathrm{curl}}_{x}{\mathrm{curl}}_{x}(\hat{E} -\hat{ E}^{(0)}) = {\mathrm{grad}}_{ x}{\mathrm{div}}_{x}(\hat{E} -\hat{ E}^{(0)}) - \Delta _{ x}(\hat{E} -\hat{ E}^{(0)})}$$
in (16) and using (17) yields
$$\displaystyle{\Delta _{x}(\hat{E} -\hat{ E}^{(0)})(\omega,x) + \frac{\omega ^{2}} {c^{2}}(\hat{E} -\hat{ E}^{(0)})(\omega,x) = -\frac{c^{2}} {\omega ^{2}} {\mathrm{grad}}_{x}{\mathrm{div}}_{x}\phi (\omega,x) -\phi (\omega,x).}$$
This is a Helmholtz equation for $$\hat{E} -\hat{ E}^{(0)}$$ and the general solution which is (with respect to ω) holomorphic in the upper half plane (equivalent to (7) by 2) is given by, see [8]
$$\displaystyle\begin{array}{rcl} (\hat{E} -\hat{ E}^{(0)})(\omega,x)& =& -\frac{c^{2}} {\omega ^{2}} \int _{\mathbb{R}^{3}}G(\omega,x - y)\left ( \frac{\omega ^{2}} {c^{2}}\mathbb{1} + {\mathrm{grad}}_{y}{\mathrm{div}}_{y}\right )\phi (\omega,y){\mathrm{d}}y {}\\ & =& -\frac{c^{2}} {\omega ^{2}} \left ( \frac{\omega ^{2}} {c^{2}}\mathbb{1} + {\mathrm{grad}}_{x}{\mathrm{div}}_{x}\right )\int _{\mathbb{R}^{3}}G(\omega,x - y)\phi (\omega,y){\mathrm{d}}y. {}\\ \end{array}$$
For the last equality, integration by parts and $${\mathrm{grad}}_{x}G(\omega,x - y) = -{\mathrm{grad}}_{y}G(\omega,x - y)$$ were used. The Lippmann–Schwinger equation (14) follows from the last expression inserting the expression (15) for ϕ.  □

This integral equation uniquely defines the electric field E. The reader is referred to [1, 8] for the isotropic case and to [33] for an anisotropic medium.

### Born and Far Field Approximation

To solve the Lippmann–Schwinger equation (14), the medium is assumed to be weakly scattering, which means that $$\hat{\chi }$$ is sufficiently small (implying that the difference $$E -\hat{ E}^{(0)}$$ becomes small compared to E(0)) so that the Born approximationE(1), defined by
$$\displaystyle\begin{array}{rcl} \hat{E}^{(1)}(\omega,x)& =& \hat{E}^{(0)}(\omega,x) + \left ( \frac{\omega ^{2}} {c^{2}}\mathbb{1} + {\mathrm{grad}}_{x}{\mathrm{div}}_{x}\right ) \\ & & \int _{\mathbb{R}^{3}}G(\omega,x - y)\hat{\chi }(\omega,y)\hat{E}^{(0)}(\omega,y){\mathrm{d}}y, {}\end{array}$$
(18)
is considered a good approximation for the electric field E, see [3]. To describe multiple scattering events, one considers higher order Born approximations. For different linearization techniques, the reader is referred to [1, 23]. Moreover, since the detector in OCT is typically quite far away from the sample, one can simplify the expression (18) for the electric field at the detector array by replacing it with its asymptotic behavior for | x | → , that is replace the formula for E(1) by its far field approximation (the far field approximation could also be applied to the solution E of the Lippmann–Schwinger equation (14)).

### Proposition 5.

Consider, for a given function$$\phi: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$$with compact support and some parameter$$a \in \mathbb{R},$$the function
$$\displaystyle{g: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3},\quad g(x) =\int _{ \mathbb{R}^{3}} \frac{{\mathrm{e}}^{{\mathrm{i}}a\vert x-y\vert }} {\vert x - y\vert }\phi (y){\mathrm{d}}y.}$$
Then, it follows, asymptotically for ρ →∞ and uniformly in$$\vartheta \in S^{2},$$that
$$\displaystyle{ (a^{2} + {\mathrm{grad}}_{ x}{\mathrm{div}}_{x})g(\rho \vartheta ) \simeq -\frac{a^{2}{\mathrm{e}}^{{\mathrm{i}}a\rho }} {\rho } \int _{\mathbb{R}^{3}}\vartheta \times (\vartheta \times \phi (y)){\mathrm{e}}^{-{\mathrm{i}}a\left <\vartheta,y\right >}{\mathrm{d}}y }$$
(19)

### Proof.

Consider the function
$$\displaystyle{\Gamma: \mathbb{R}^{3} \rightarrow \mathbb{C},\quad \Gamma (x) = \frac{{\mathrm{e}}^{{\mathrm{i}}a\vert x\vert }} {\vert x\vert }.}$$
Then
$$\displaystyle\begin{array}{rcl} \frac{\partial ^{2}\Gamma } {\partial x_{j}\partial x_{k}}(x)& =& \frac{\partial } {\partial x_{j}}\left [\left ( \frac{{\mathrm{i}}a} {\vert x\vert ^{2}} - \frac{1} {\vert x\vert ^{3}}\right )x_{k}{\mathrm{e}}^{{\mathrm{i}}a\vert x\vert }\right ] {}\\ & =& \left [\left ( \frac{{\mathrm{i}}a} {\vert x\vert ^{2}} - \frac{1} {\vert x\vert ^{3}}\right )\delta _{jk} + \left ( \frac{{\mathrm{i}}a} {\vert x\vert ^{2}} - \frac{1} {\vert x\vert ^{3}}\right )\frac{{\mathrm{i}}ax_{j}x_{k}} {\vert x\vert } \right. {}\\ & & +\left.\left (-2 \frac{{\mathrm{i}}a} {\vert x\vert ^{3}} + 3 \frac{1} {\vert x\vert ^{4}}\right )\frac{x_{j}x_{k}} {\vert x\vert } \right ]{\mathrm{e}}^{{\mathrm{i}}a\vert x\vert }. {}\\ \end{array}$$
Therefore, writing x in spherical coordinates: $$x =\rho \vartheta$$ with ρ > 0, $$\vartheta \in S^{2}$$, for ρ →  uniformly in $$\vartheta,$$ it can be seen that
$$\displaystyle{ \frac{\partial ^{2}\Gamma } {\partial x_{j}\partial x_{k}}(\rho \vartheta ) = -\frac{a^{2}{\mathrm{e}}^{{\mathrm{i}}a\rho }} {\rho } \vartheta _{j}\vartheta _{k} + \mathcal{O}\left (\frac{1} {\rho ^{2}} \right ),}$$
The approximation (locally uniformly in $$y \in \mathbb{R}^{3})$$
$$\displaystyle{\vert \rho \vartheta - y\vert =\rho \sqrt{\vert \vartheta \vert ^{2 } - \frac{2} {\rho } \left < \vartheta,y\right > + \frac{1} {\rho ^{2}} \vert y\vert ^{2}} =\rho -\left < \vartheta,y\right > + \mathcal{O}\left (\frac{1} {\rho } \right ),}$$
implies that (again uniformly in $$\vartheta \in S^{2}$$)
$$\displaystyle{ \frac{\partial ^{2}\Gamma } {\partial x_{j}\partial x_{k}}(\rho \vartheta -y) = -\frac{a^{2}{\mathrm{e}}^{{\mathrm{i}}a(\rho -\left <\vartheta,y\right >)}} {\rho } \vartheta _{j}\vartheta _{k} + \mathcal{O}\left (\frac{1} {\rho } \right ).}$$
Now, considering the compact support of ϕ and using that $$x \in \mathbb{R}^{3}\setminus {\mathrm{supp}}\phi$$
$$\displaystyle\begin{array}{rcl} ({\mathrm{grad}}_{x}{\mathrm{div}}_{x}g)_{j}(x)& =& \sum _{k=1}^{3} \frac{\partial } {\partial x_{j}}\int _{\mathbb{R}^{3}} \frac{\partial \Gamma } {\partial x_{k}}(x - y)\phi _{k}(y){\mathrm{d}}y {}\\ & =& \int _{\mathbb{R}^{3}}\sum _{k=1}^{3} \frac{\partial ^{2}\Gamma } {\partial x_{j}\partial x_{k}}(x - y)\phi _{k}(y){\mathrm{d}}y. {}\\ \end{array}$$
Asymptotically for | x | →  (again using the compact support of ϕ) one obtains
$$\displaystyle{a^{2}g_{ j}(\rho \vartheta ) + ({\mathrm{grad}}_{x}{\mathrm{div}}_{x}g)_{j}(\rho \vartheta ) \simeq a^{2}\int _{ \mathbb{R}^{3}}\sum _{k=1}^{3}\frac{{\mathrm{e}}^{{\mathrm{i}}a(\rho -\left <\vartheta,y\right >)}} {\rho } \left (\delta _{jk} -\vartheta _{j}\vartheta _{k}\right )\phi _{k}(y){\mathrm{d}}y.}$$
The approximation (19) follows from the vector identity $$\vartheta \times (\vartheta \times \phi ) = \left < \vartheta,\phi \right >\vartheta -\vert \vartheta \vert ^{2}\phi$$ and $$\vert \vartheta \vert = 1.$$ □
The application of both the far field and the Born approximation, this means 5 for the expression (18) of E(1), that is setting a = ωc and $$\phi = \tfrac{1} {4\pi }\hat{\chi }\hat{E}^{(0)}$$ in 5, imply the asymptotic behavior
$$\displaystyle{ \hat{E}^{(1)}(\omega,\rho \vartheta ) \simeq \hat{ E}^{(0)}(\omega,\rho \vartheta ) -\frac{\omega ^{2}{\mathrm{e}}^{{\mathrm{i}} \frac{\omega } {c}\rho }} {4\pi \rho c^{2}} \int _{\mathbb{R}^{3}}\vartheta \times \big (\vartheta \times (\hat{\chi }(\omega,y)\hat{E}^{(0)}(\omega,y))\big){\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}\left <\vartheta,y\right >}{\mathrm{d}}y. }$$
(20)

### The Forward Operator

To obtain a forward model for the measurements described in Sect. 3, the (approximative) formula (20) is considered as a model for the solution of the scattering problem. To make this formula concrete, one has to plug in a function E(0) describing the initial illumination (recall that E(0) has to solve (6)).

The specific illumination is a laser pulse propagating in the direction − e3, orthogonal to the detector surface $$\mathcal{D} =\{ x \in \mathbb{R}^{3}\mid x_{3} = d\}$$, this means
$$\displaystyle{ E^{(0)}(t,x) = f(t + \tfrac{x_{3}} {c} )p, }$$
(21)
which solves Maxwell’s equations (1) with the assumptions (3) for some fixed vector $$p \in \mathbb{R}^{3},$$ with $$p_{3} = \left < p,e_{3}\right > = 0,$$ describing the polarization of the initial laser beam.

### Proposition 6.

The function E(0), defined by (21) with$$\left < p,e_{3}\right > = 0$$, solves together with the magnetic field H(0), defined by
$$\displaystyle{H^{(0)}(t,x) = f(t + \tfrac{x_{3}} {c} )p \times e_{3},}$$
Maxwell’s equations (1) in the vacuum, that is with the additional assumptions (3) .

### Proof.

The four equations of (1) can be directly verified:
$$\displaystyle\begin{array}{rcl} {\mathrm{div}}_{x}E^{(0)}(t,x)& =& \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )\left < e_{3},p\right > = 0, {}\\ {\mathrm{div}}_{x}H^{(0)}(t,x)& =& \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )\left < e_{3},p \times e_{3}\right > = 0, {}\\ {\mathrm{curl}}_{x}E^{(0)}(t,x)& =& \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )e_{3} \times p = -\frac{1} {c} \frac{\partial H^{(0)}} {\partial t} (t,x), {}\\ {\mathrm{curl}}_{x}H^{(0)}(t,x)& =& \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )e_{3} \times (p \times e_{3}) = \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )p = \frac{1} {c} \frac{\partial E^{(0)}} {\partial t} (t,x). {}\\ \end{array}$$
□
To guarantee that the initial field E(0) (and also the magnetic field H(0)) does not interact with the sample or the mirror for t ≤ 0 and neither contributes to the measurement at the detectors for t ≥ 0 as required by (8) and (10) the vertical distribution $$f: \mathbb{R} \rightarrow \mathbb{R}$$ should satisfy (see Fig. 2)
$$\displaystyle{ {\mathrm{supp}}f \subset (\tfrac{R} {c}, \tfrac{d} {c}). }$$
(22)
In the case of an illumination E(0) of the form (21), the electric field Er produced by an ideal mirror at the position r is given by
$$\displaystyle{ E_{r}(t,x) = \left \{\begin{array}{@{}l@{\quad }l@{}} \big(f(t + \frac{x_{3}} {c} ) - f(t + \frac{x_{3}} {c} + 2\,\frac{r-x_{3}} {c} )\big)p\quad &\text{if}\;x_{3} > r,\\ 0 \quad &\text{if} \;x_{ 3} \leq r. \end{array} \right. }$$
(23)
This just corresponds to the superposition of the initial wave with the (orthogonally) reflected wave, which travels additionally the distance $$2\tfrac{x_{3}-r} {c}$$. The change in polarization of the reflected wave (from p to − p) comes from the fact that the tangential components of the electric field have to be continuous across the border of the mirror.

The following proposition gives the form of the measurements Mr, described in Sect. 3, on the detector surface $$\mathcal{D}$$ for the specific illumination (21).

### Proposition 7.

Let E(0)be an initial illumination of the form (21) satisfying (22) . Then, the equations for the measurements Mrfrom3are given by
$$\displaystyle\begin{array}{rcl} M_{r,j}(x)& =& -p_{j}\int _{-\infty }^{\infty }(E_{ j} - E_{j}^{(0)})(t,x)f(t + \tfrac{2r-x_{3}} {c} ){\mathrm{d}}t,{}\end{array}$$
(24a)
$$\displaystyle\begin{array}{rcl} & =& -\frac{p_{j}} {2\pi } \int _{-\infty }^{\infty }(\hat{E}_{ j} -\hat{ E}_{j}^{(0)})(\omega,x)\hat{f}(-\omega ){\mathrm{e}}^{{\mathrm{i}} \frac{\omega } {c} (2r-x_{3})}{\mathrm{d}}\omega {}\end{array}$$
(24b)
for all j ∈{ 1,2,3}, r ∈ (−∞,R), and$$x \in \mathcal{D}$$.

### Proof.

Since the electric field Er reflected on a mirror at vertical position r ∈ (−, R) is according to (23) given by
$$\displaystyle{E_{r}(t,x) =\big (f(t + \tfrac{x_{3}} {c} ) - f(t + \tfrac{2r-x_{3}} {c} )\big)p\quad \text{for all}\quad t \in \mathbb{R},\;x \in \mathcal{D},}$$
the measurement functions Mr (defined by (12) and computed with (13a)) are simplified, for the particular initial illumination E(0) of the form (21), to (24a) for $$x \in \mathcal{D}.$$
Since formula (24a) is just a convolution, the electric field EE(0) can be rewritten, in terms of its Fourier transform, in the form
$$\displaystyle{M_{r,j}(x) = -\frac{p_{j}} {2\pi } \int _{-\infty }^{\infty }\int _{ -\infty }^{\infty }(\hat{E}_{ j} -\hat{ E}_{j}^{(0)})(\omega,x){\mathrm{e}}^{-{\mathrm{i}}\omega t}f(t + \tfrac{2r-x_{3}} {c} ){\mathrm{d}}\omega {\mathrm{d}}t.}$$
Interchanging the order of integration and applying the Fourier transform $$\hat{f}$$ of f, it follows Eq. (24b). □
In the limiting case of a delta impulse as initial wave, that is for f(ξ) = δ(ξξ0) with some constant $$\xi _{0} \in (\frac{R} {c}, \frac{d} {c})$$ satisfying (22), the measurements provide directly the electric field. Indeed, it can be seen from (24a) that
$$\displaystyle{M_{r,j}(x) = -p_{j}(E_{j} - E_{j}^{(0)})(\tfrac{x_{3}-2r} {c} +\xi _{0},x).}$$
By varying r ∈ (−, R), the electric field E can be obtained (to be more precise, its component in direction of the initial polarization) as a function of time at every detector position.

### Assumption 1.

The susceptibility χ is sufficiently small so that the Born approximation E(1)for the solution E of the Lippmann–Schwinger equation (14) can be applied.

### Assumption 2.

The detectors are sufficiently far away from the object so that one can use the far field asymptotics (20) for the measured field.

Under these assumptions, one can approximate the electric field by the far field expression of the Born approximation E(1) and plug in the expression in (20) to obtain the measurements Mr, j, j ∈ { 1, 2, 3}.

The above analysis, introducing appropriate operators, can then be formulated as an operator equation. The integral equation (20) can be formally written as
$$\displaystyle{(\hat{E}^{(1)} -\hat{ E}^{(0)})(\omega,x) = (\mathcal{K}_{ 0}\hat{\chi })(\omega,x),}$$
for a given $$\hat{E}^{(0)}$$ where the operator $$\mathcal{K}_{0}:\hat{\chi } \mapsto \hat{E}^{(1)} -\hat{ E}^{(0)}$$ is given by
$$\displaystyle\begin{array}{rcl} (\mathcal{K}_{0}v)(\omega,\rho \vartheta )& =& -\frac{\omega ^{2}{\mathrm{e}}^{{\mathrm{i}} \frac{\omega } { c}\rho }} {4\pi \rho c^{2}} \int _{\mathbb{R}^{3}}\vartheta \times \big (\vartheta \times (v(\omega,y)\hat{E}^{(0)}(\omega,y))\big){\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}\left <\vartheta,y\right >}{\mathrm{d}}y, {}\\ & & \rho > 0,\vartheta \in S^{2}. {}\\ \end{array}$$
It is emphasized that $$\hat{\chi }: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{C}^{3\times 3}$$ and $$\hat{E}^{(1)} -\hat{ E}^{(0)}: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{C}^{3},$$ that is $$\mathcal{K}_{0}v$$ is a function from $$\mathbb{R} \times \mathbb{R}^{3}$$ into $$\mathbb{C}^{3}.$$ Equivalently, considering the Eq. (13b), one has
$$\displaystyle{M(r,x) = \left (M_{r,j}(x)\right )_{j=1}^{3} =\big (\mathcal{M}(\hat{E}^{(1)} -\hat{ E}^{(0)})\big)(r,x),}$$
where the operator $$\mathcal{M}$$ is defined by
$$\displaystyle{(\mathcal{M}v)(r,x) = \left (\int _{-\infty }^{\infty }v_{ j}(\omega,x)\overline{(\hat{E}_{r,j} -\hat{ E}_{j}^{(0)})(\omega,x)}{\mathrm{d}}\omega \right )_{ j=1}^{3},\quad x \in \mathcal{D}.}$$
Here, $$\mathcal{M}v$$ is a function from $$\mathbb{R} \times \mathcal{D}$$ to $$\mathbb{R}^{3}.$$ Thus, combining the operators $$\mathcal{K}_{0}$$ and $$\mathcal{M}$$, the forward operator $$\mathcal{F}:\hat{\chi } \mapsto M,$$$$\mathcal{F} = \mathcal{M}\mathcal{K}_{0}$$ models the direct problem. The inverse problem of OCT is then formulated as an operator equation
$$\displaystyle{ \mathcal{F}\hat{\chi } = M. }$$
(25)
For the specific illumination (21), one has
$$\displaystyle{ \hat{E}^{(0)}(\omega,x) = \left (\int _{ -\infty }^{\infty }f(t + \tfrac{x_{3}} {c} ){\mathrm{e}}^{{\mathrm{i}}\omega t}{\mathrm{d}}t\right )p =\hat{ f}(\omega ){\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}x_{3} }p. }$$
(26)
Then, the operators $$\mathcal{K}_{0}$$ and $$\mathcal{M}$$ simplify to
$$\displaystyle{ (\mathcal{K}_{0}v)(\omega,\rho \vartheta ) = -\frac{\omega ^{2}{\mathrm{e}}^{{\mathrm{i}} \frac{\omega } { c}\rho }} {4\pi \rho c^{2}} \hat{f}(\omega )\int _{\mathbb{R}^{3}}\vartheta \times \big (\vartheta \times (v(\omega,y)\,p)\big){\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}\left <\vartheta +e_{3},y\right >}{\mathrm{d}}y }$$
(27)
and, recalling that Mr, 3 = 0 since p3 = 0 (the polarization in the incident direction is zero),
$$\displaystyle{ (\mathcal{M}v)(r,x) = \left (-\frac{p_{j}} {2\pi } \int _{-\infty }^{\infty }v_{ j}(\omega,x)\hat{f}(-\omega ){\mathrm{e}}^{{\mathrm{i}} \frac{\omega } {c} (2r-x_{3})}{\mathrm{d}}\omega \right )_{ j=1}^{2}. }$$
(28)
The operator $$\mathcal{K}_{0}$$ is derived from the Born approximation taking into account the far field approximation for the solution of the Lippmann–Schwinger equation (14). But, one could also neglect 1 and 2 and use the operator $$\mathcal{K}$$ corresponding to Eq. (14), that is,
$$\displaystyle{(\mathcal{K}v)(\omega,x) = \left ( \frac{\omega ^{2}} {c^{2}}\mathbb{1} + {\mathrm{grad}}_{x}{\mathrm{div}}_{x}\right )\int _{\mathbb{R}^{3}}G(\omega,x - y)v(\omega,y)\hat{E}(\omega,y){\mathrm{d}}y,}$$
and considering the nonlinear forward operator $$\mathcal{F} = \mathcal{M}\mathcal{K}.$$

The next section focuses on the solution of (25), considering the operators $$\mathcal{K}_{0}$$ and $$\mathcal{M},$$ given by (27) and (28), respectively. The inversion of $$\mathcal{F}$$ is performed in two steps, first $$\mathcal{M}$$ is inverted and then $$\mathcal{K}_{0}.$$

## 5 The Inverse Scattering Problem

In optical coherence tomography, the susceptibility χ of the sample is imaged from the measurements Mr(x), r ∈ (−, R), $$x \in \mathcal{D}$$. In a first step, it is shown that the measurements allow us to reconstruct the scattered field on the detector $$\mathcal{D},$$ that is inverting the operator (28).

The vertical distribution $$f: \mathbb{R} \rightarrow \mathbb{R}$$ should additionally satisfy (see Fig. 2)
$$\displaystyle{ {\mathrm{supp}}f \subset (\tfrac{R} {c}, \tfrac{R} {c} + \tfrac{2\delta } {c}) \subset (\tfrac{R} {c}, \tfrac{d} {c})\quad \text{for some}\quad \delta > 0. }$$
(29)
This guarantees that the initial field E(0) (and also the magnetic field H(0)) not interact with the sample or the mirror for t ≤ 0 and neither contribute to the measurement at the detectors for t ≥ 0 as required by (8) and (10).

The condition that the length of the support of E(0) is at most 2δ (the assumption that the support starts at $$\frac{R} {c}$$ is only made to simplify the notation) is required for 8. It ensures that the formula (13a) for the measurement data Mr(x), $$x \in \mathcal{D}$$, vanishes for values r ≥ R so that the integral on the right hand side of (30) is only over the interval (−, R) where measurement data are obtained (recall that measurements are only performed for positions r < R of the mirror).

### Proposition 8.

Let E(0)be an initial illumination of the form (21) satisfying (29) . Then, the measurements Mrfrom7imply for the electric field E:
$$\displaystyle{ (\hat{E}_{j} -\hat{ E}_{j}^{(0)})(\omega,x)\overline{\hat{f}(\omega )}p_{ j} = -\frac{2} {c}\int _{-\infty }^{R}M_{ r,j}(x){\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}(2r-x_{3})}{\mathrm{d}}r }$$
(30)
for all j ∈{ 1,2,3}, $$\omega \in \mathbb{R}$$, and$$x \in \mathcal{D}$$.

### Proof.

Remark that the formula (24a) can be extended to all $$r \in \mathbb{R}$$ by setting Mr, j(x) = 0 for r ≥ R. Indeed, from (29) it follows that E(t, ⋅ ) = E(0)(t, ⋅ ) for all $$t < \frac{\delta } {c}$$. Since E is a solution of the linear wave equation with constant wave speed c on the half space given by x3 > Rδ, the difference between E and E(0) caused by the sample needs at least time $$\tfrac{d-R+\delta } {c}$$ to travel from the point at x3 = Rδ to the detector at x3 = d, so:
$$\displaystyle{E(t,x) = E^{(0)}(t,x)\quad \text{for all}\quad x \in \mathbb{R}^{3}\;\text{with}\;x_{ 3} = d\;\text{and}\;ct < 2\delta + d - R.}$$
This means that the integrand vanishes for $$t < \frac{2\delta +d-R} {c}$$. In the case of $$t \geq \frac{2\delta +d-R} {c},$$ it holds for r ≥ R that
$$\displaystyle{ct + 2r - d \geq 2\delta + d - R + 2R - d = R + 2\delta,}$$
so that $$f(t + \frac{2r-d} {c} ) = 0$$ by the assumption (29) on the support of f. Therefore, for r ≥ R, always one of the factors in the integrand in (24a) is zero which implies that Mr(x) = 0 for r ≥ R and $$x \in \mathcal{D}$$.
Thus, Eq. (24b) holds for all $$r \in \mathbb{R}$$ and applying the inverse Fourier transform with respect to r, using that $$\hat{f}(-\omega ) = \overline{\hat{f}(\omega )}$$ because f is real valued, yields
$$\displaystyle{\frac{2} {c}\int _{-\infty }^{\infty }M_{ r,j}(x){\mathrm{e}}^{-{\mathrm{i}}\frac{2\omega r} {c} }{\mathrm{d}}r = -p_{j}(\hat{E}_{j} -\hat{ E}_{j}^{(0)})(\omega,x)\overline{\hat{f}(\omega )}{\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } { c}x_{3}},}$$
which can equivalently be written as (30). □

This means that one can calculate from the Fourier transform of the measurements rMr(x) at some frequency ω the Fourier transform of the electric field at ω as long as the Fourier transform of the initial wave E(0) does not vanish at ω, that is for $$\hat{f}(\omega )\neq 0$$. Thus, under the 1 and 2, Eq. (30) can be solved for the electric field $$\hat{E}$$. 8 thus provides the inverse of the operator $$\mathcal{M}$$ defined by (28). Now, the inversion of the operator $$\mathcal{K}_{0}$$ given by (27) is performed considering the optical properties of the sample.

### Proposition 9.

Let E(0)(t,x) be given by the form (21) with p3= 0 and the additional assumption (29) . Then, for every$$\omega \in \mathbb{R}\setminus \{0\}$$with$$\hat{f}(\omega )\neq 0$$, the formula
$$\displaystyle{ p_{j}\big[\vartheta \times (\vartheta \times \tilde{\chi }(\omega, \tfrac{\omega } {c}(\vartheta +e_{3}))p)\big]_{j} \simeq \frac{8\pi \rho c} {\omega ^{2}\vert \hat{f}(\omega )\vert ^{2}}\int _{-\infty }^{R}M_{ r,j}(\rho \vartheta ){\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}(2r-\rho (\vartheta _{3}-1))}{\mathrm{d}}r }$$
(31)
holds for all j ∈{ 1,2}, $$\vartheta \in S_{+}^{2}:=\{\eta \in S^{2}\mid \eta _{3} > 0\}$$, and$$\rho = \frac{d} {\vartheta _{3}}$$(asymptotically for χ → 0 and ρ →∞).
Here$$\tilde{\chi }$$denotes the Fourier transform of χ with respect to time and space, that is
$$\displaystyle{ \tilde{\chi }(\omega,k) =\int _{ -\infty }^{\infty }\int _{ \mathbb{R}^{3}}\chi (t,x){\mathrm{e}}^{-{\mathrm{i}}\left <k,x\right >}{\mathrm{e}}^{{\mathrm{i}}\omega t}{\mathrm{d}}x{\mathrm{d}}t =\int _{ \mathbb{R}^{3}}\hat{\chi }(\omega,x){\mathrm{e}}^{-{\mathrm{i}}\left <k,x\right >}{\mathrm{d}}x. }$$
(32)

### Proof.

Because of (26), the Fourier transform of the electric field EE(0) can be approximated, using (20) with E ≃ E(1) (by 1 and 2), by
$$\displaystyle\begin{array}{rcl} (\hat{E} -\hat{ E}^{(0)})(\omega,\rho \vartheta ) \simeq -\frac{\omega ^{2}\hat{f}(\omega ){\mathrm{e}}^{{\mathrm{i}} \frac{\omega } {c}\rho }} {4\pi \rho c^{2}} \int _{\mathbb{R}^{3}}{\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}(\left <y,\vartheta \right >+y_{3})}\vartheta \times (\vartheta \times \hat{\chi }(\omega,y)p){\mathrm{d}}y.& & {}\\ \end{array}$$
Then, applying (32), one obtains
$$\displaystyle{ (\hat{E} -\hat{ E}^{(0)})(\omega,\rho \vartheta ) \simeq -\frac{\omega ^{2}\hat{f}(\omega ){\mathrm{e}}^{{\mathrm{i}} \frac{\omega } {c}\rho }} {4\pi \rho c^{2}} \vartheta \times (\vartheta \times \tilde{\chi }(\omega, \tfrac{\omega } {c}(\vartheta +e_{3}))p). }$$
(33)
From (30), it is known that, for $$\hat{f}(\omega )\neq 0$$ and pj ≠ 0,
$$\displaystyle{(\hat{E}_{j} -\hat{ E}_{j}^{(0)})(\omega,\rho \vartheta ) = - \frac{2} {p_{j}c\overline{\hat{f}(\omega )}}\int _{-\infty }^{R}M_{ r,j}(\rho \vartheta ){\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}(2r-\rho \vartheta _{3})}{\mathrm{d}}r.}$$
This identity together with (33), asymptotically for ω ≠ 0, yields the statement (31). □
To derive reconstruction formulas, 9 is used, which states that from the measurements Mr (under the 1 and 2) the expression
$$\displaystyle{ p_{j}\big[\vartheta \times (\vartheta \times \tilde{\chi }(\omega, \tfrac{\omega } {c}(\vartheta +e_{3}))p)\big]_{j},\quad j = 1,2, }$$
(34)
can be calculated. Here, $$p \in \mathbb{R}^{2} \times \{ 0\}$$ denotes the polarization of the initial illumination E(0), see (21), and ϑ ∈ S+2 is the direction from the origin (where the sample is located) to a detector.

### The Isotropic Case

This section analyzes the special case of an isotropic medium, meaning that the susceptibility matrix χ is just a multiple of the unit matrix, so in the following χ is identified with a scalar.

Then, from the sum of the measurements Mr, 1 and Mr, 2, using the formula (31), one obtains the expression
$$\displaystyle{\tilde{\chi }(\omega, \tfrac{\omega } {c}(\vartheta +e_{3}))\left < p,\vartheta \times (\vartheta \times p)\right > =\tilde{\chi } (\omega, \tfrac{\omega } {c}(\vartheta +e_{3}))(\left < \vartheta,p\right >^{2} -\vert p\vert ^{2}).}$$
Since $$\left < \vartheta,p\right >^{2} < \vert p\vert ^{2}$$ for every combination of $$p \in \mathbb{R}^{2} \times \{ 0\}$$ and $$\vartheta \in S_{+}^{2}$$, one has direct access to the spatial and temporal Fourier transform
$$\displaystyle{ \tilde{\chi }(\omega, \tfrac{\omega } {c}(\vartheta +e_{3})),\quad \omega \in \mathbb{R}\setminus \{0\},\;\vartheta \in S_{+}^{2}, }$$
(35)
of χ in a subset of $$\mathbb{R} \times \mathbb{R}^{3}.$$
However, it remains the problem of reconstructing the four-dimensional susceptibility data χ from the three-dimensional measurement data (35). In the following, some different additional assumptions are discussed to compensate the lack of dimension, see Table 1.
Table 1

Different assumptions about the susceptibility and the corresponding reconstruction formulas

 Assumptions Reconstruction method Section $$\tilde{\chi }(\omega,k) =\tilde{\chi } (k)$$ Reconstruction from partial (three dimensional) Fourier data:$$\tilde{\chi }(k)$$, $$k \in \mathbb{R}^{3}$$, $$\measuredangle (k,e_{3}) \in (-\tfrac{\pi }{4}, \tfrac{\pi } {4})$$ Non-dispersive Medium in Full Field OCT $$\tilde{\chi }(\omega,k) =\tilde{\chi } (k_{3})$$ Reconstruction from full (one dimensional) Fourier data:$$\tilde{\chi }(k_{3})$$, $$k_{3} \in \mathbb{R}\setminus \{0\}$$ Non-dispersive Medium with Focused Illumination supp $$\chi (\cdot,x) \subset [0,T]$$$$\mathcal{R}(\chi (\tau,\cdot ))(\cdot,\varphi )$$ is piecewise constant Recursive formula to get limited angle Radon data$$\mathcal{R}(\chi (\tau,\cdot ))(\sigma,\varphi )$$, $$\sigma \in \mathbb{R}$$, φ ∈ S2with $$\measuredangle (\varphi,e_{3}) \in (-\frac{\pi }{4}, \frac{\pi } {4})$$ Dispersive Medium $$\chi (\tau,x) =\delta (x_{1})\delta (x_{2})\chi (\tau,x_{3})$$,supp χ(⋅ , x) ⊂ [0, T], andχ(τ, ⋅ ) is piecewise constant Recursive formula toreconstruct χ Dispersive Layered Medium with Focused Illumination

Here, $$(\mathcal{R}g)(\sigma,\varphi ) =\int _{\{x\in \mathbb{R}^{3}\mid \left <x,\varphi \right >=\sigma \}}g(y){\mathrm{d}}s(y)$$,$$\sigma \in \mathbb{R}$$,φ ∈ S2, denotes the Radontransform of a function $$g: \mathbb{R}^{3} \rightarrow \mathbb{R}$$.

#### Non-dispersive Medium in Full Field OCT

The model is simplified by assuming an immediate reaction of the sample to the exterior electric field in (2a). This means that χ can be considered as a delta distribution in time so that its temporal Fourier transform $$\hat{\chi }$$ does not depend on frequency, that is $$\hat{\chi }(\omega,x) =\hat{\chi } (x).$$ Thus, the reconstruction reduces to the problem of finding $$\hat{\chi }$$ from its partial (spatial) Fourier data
$$\displaystyle\begin{array}{rcl} & & \tilde{\chi }(k)\quad \text{for}\quad k \in \{ \tfrac{\omega } {c}(\vartheta +e_{3}) \in \mathbb{R}^{3}\mid \vartheta \in S_{ +}^{2},\;\omega \in \mathbb{R}\setminus \{0\}\} {}\\ & & \qquad \qquad \qquad \qquad =\{\kappa \in \mathbb{R}^{3}\setminus \{0\}\mid \arccos (\langle \tfrac{\kappa } {\vert \kappa \vert },e_{3}\rangle ) \in (-\tfrac{\pi } {4}, \tfrac{\pi } {4})\}. {}\\ \end{array}$$
Thus, only the Fourier data of χ in the right circular cone $$\mathcal{C}$$ with axis along e3 and aperture $$\frac{\pi }{2}$$ are observed (see Fig. 3). In practice, these data are usually only available for a small range of frequencies ω.

Inverse scattering for full field OCT, under the Born approximation, has been considered by Marks et al. [28, 29] where algorithms to recover the scalar susceptibility were proposed.

#### Non-dispersive Medium with Focused Illumination

In standard OCT, the illumination is focused to a small region inside the object so that the function χ can be assumed to be constant in the directions e1 and e2 (locally the illumination is still assumed to be properly described by a plane wave). Then, the problem can be reduced by two dimensions assuming that the illumination is described by a delta distribution in these two directions. As before, χ is assumed to be frequency independent, so that
$$\displaystyle{\hat{\chi }(\omega,x) =\delta (x_{1})\delta (x_{2})\hat{\chi }(x_{3}),}$$
this means that the spatial and temporal Fourier transform (32) fulfils $$\tilde{\chi }(\omega,k) =\tilde{\chi } (k_{3}).$$ In this case, the two-dimensional detector array can be replaced by a single point detector located at de3.
Then, the measurement data (35) in direction ϑ = e3 provide the Fourier transform
$$\displaystyle{\tilde{\chi }(\tfrac{2\omega } {c})\quad \text{for all}\quad \omega \in \mathbb{R}\setminus \{0\}.}$$

Therefore, the reconstruction of the (one dimensional) susceptibility $$x_{3}\mapsto \hat{\chi }(x_{3})$$ can be simply obtained by an inverse Fourier transform.

This one-dimensional analysis, has been used initially by Fercher et al. [18], reviewed in [13] and by Hellmuth [22] to describe time domain OCT. Ralston et al. [34, 35] described the OCT system using a single backscattering model. The solution was given through numerical simulation using regularized least squares methods.

#### Dispersive Medium

However, in the case of a dispersive medium, that is frequency-dependent $$\hat{\chi },$$ the difficulty is to reconstruct the four-dimensional function $$\chi: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{C}$$ from the three-dimensional data
$$\displaystyle{ \hat{m}: \mathbb{R} \times S_{+}^{2} \rightarrow \mathbb{C},\quad \hat{m}(\omega,\vartheta ) =\tilde{\chi } (\omega, \tfrac{\omega } {c}(\vartheta +e_{3})). }$$
(36)

### Lemma 1.

Let$$\hat{m}$$be given by (36) . Then its inverse Fourier transform$$m: \mathbb{R} \times S_{+}^{2} \rightarrow \mathbb{C}$$with respect to the first variable is given by
$$\displaystyle{ m(t,\vartheta ) = \frac{c} {\sqrt{2(1 +\vartheta _{3 } )}}\int _{-\infty }^{\infty }\bar{\chi }(\tau;\tau -t,\vartheta ){\mathrm{d}}\tau,\quad t \in \mathbb{R},\;\vartheta \in S_{ +}^{2}, }$$
(37)
where
$$\displaystyle{\bar{\chi }(\tau;\sigma,\vartheta ) =\int _{E_{\sigma,\vartheta }}\chi (\tau,y){\mathrm{d}}s(y),\quad \tau,\sigma \in \mathbb{R},\,\vartheta \in S_{+}^{2},}$$
and$$E_{\sigma,\vartheta }$$denotes the plane
$$\displaystyle{ E_{\sigma,\vartheta } =\{ y \in \mathbb{R}^{3}\mid \left < \vartheta +e_{ 3},y\right > = c\sigma \},\quad \sigma \in \mathbb{R},\;\vartheta \in S_{+}^{2}. }$$
(38)

### Proof.

Taking the inverse temporal Fourier transform of $$\hat{m}$$ and using (32), it follows that
$$\displaystyle\begin{array}{rcl} m(t,\vartheta )& =& \frac{1} {2\pi }\int _{-\infty }^{\infty }\tilde{\chi }(\omega, \tfrac{\omega } {c}(\vartheta +e_{3})){\mathrm{e}}^{-{\mathrm{i}}\omega t}{\mathrm{d}}\omega {}\\ & =& \frac{1} {2\pi }\int _{-\infty }^{\infty }\int _{ \mathbb{R}^{3}}\hat{\chi }(\omega,x){\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}\left <\vartheta +e_{3},x\right >}{\mathrm{e}}^{-{\mathrm{i}}\omega t}{\mathrm{d}}x{\mathrm{d}}\omega. {}\\ \end{array}$$
Interchanging the order of integration, the integral over ω is again described by an inverse Fourier transform and the previous equation becomes
$$\displaystyle{m(t,\vartheta ) =\int _{\mathbb{R}^{3}}\chi (t + \tfrac{1} {c}\left < \vartheta +e_{3},x\right >,x){\mathrm{d}}x.}$$
Substituting then the variable x3 by $$\tau = t + \frac{1} {c}\left < \vartheta +e_{3},x\right >$$, this can be written as
$$\displaystyle{ m(t,\vartheta ) = \frac{c} {1 +\vartheta _{3}}\int _{-\infty }^{\infty }\int _{ -\infty }^{\infty }\int _{ -\infty }^{\infty }\chi (\tau,\psi _{\tau -t,\vartheta }(x_{1},x_{2})){\mathrm{d}}x_{1}{\mathrm{d}}x_{2}{\mathrm{d}}\tau }$$
(39)
with the function
$$\displaystyle{\psi _{\sigma,\vartheta }: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3},\quad \psi _{\sigma,\vartheta }(x_{1},x_{2}) = \left (x_{1},x_{2}, \frac{c\sigma } {1 +\vartheta _{3}} -\frac{\vartheta _{1}x_{1} +\vartheta _{2}x_{2}} {1 +\vartheta _{3}} \right ).}$$
Now, $$\psi _{\sigma,\vartheta }$$ is seen to be the parametrization of the plane
$$\displaystyle{E_{\sigma,\vartheta } =\{ y \in \mathbb{R}^{3}\mid \left < v_{\vartheta },y\right > = a_{\sigma,\vartheta }\}\quad \text{with}\quad v_{\vartheta } = \left (\begin{array}{*{10}c} \frac{\vartheta _{1}} {1+\vartheta _{3}} \\ \frac{\vartheta _{2}} {1+\vartheta _{3}} \\ 1 \end{array} \right )\quad \text{and}\quad a_{\sigma,\vartheta } = \frac{c\sigma } {1 +\vartheta _{3}},}$$
see Fig. 4. The square root of the Gram determinant of the parametrization $$\psi _{\sigma,\vartheta }$$ is now given by the length of the vector $$v_{\vartheta } = \frac{\partial \psi _{\sigma,\vartheta }} {\partial x_{1}} \times \frac{\partial \psi _{\sigma,\vartheta }} {\partial x_{2}}$$, which implies that
$$\displaystyle{\int _{-\infty }^{\infty }\int _{ -\infty }^{\infty }\chi (\tau,\psi _{\tau -t,\vartheta }(x_{1},x_{2})){\mathrm{d}}x_{1}{\mathrm{d}}x_{2} = \sqrt{\frac{1 +\vartheta _{3 } } {2}} \int _{E_{\tau -t,\vartheta }}\chi (\tau,y){\mathrm{d}}s(y).}$$
Plugging this into (39) yields the claim. □
Thus, the measurements give the combination (37) of values $$\bar{\chi }$$ of the Radon transform of the function χ(τ, ⋅ ). It seems, however, impossible to recover the values $$\bar{\chi }(\tau;\sigma,\vartheta )$$ from this combination
$$\displaystyle{m(t,\vartheta ) = \frac{c} {\sqrt{2(1 +\vartheta _{3 } )}}\int _{-\infty }^{\infty }\bar{\chi }(\tau;\tau -t,\vartheta ){\mathrm{d}}\tau,}$$
since (for every fixed angle ϑ ∈ S+2) one would have to reconstruct a function on $$\mathbb{R}^{2}$$ from one dimensional data.

To overcome this problem, the function $$\bar{\chi }(\tau;\cdot,\vartheta )$$ is going to be discretized for every $$\tau \in \mathbb{R}$$ and $$\vartheta \in S_{+}^{2}$$, where the step size will depend on the size of the support of $$\chi (\cdot,x)$$.

Let us therefore consider the following assumption.

### Assumption 3.

The support of χ in the time variable is contained in a small interval [0,T] for some T > 0:
$$\displaystyle{ {\mathrm{supp}}\chi (\cdot,x) \subset [0,T]\quad \text{for all}\quad x \in \mathbb{R}^{3}. }$$
Then, the following discretization
$$\displaystyle{\bar{\chi }_{n}(\tau,\vartheta ) =\int _{E_{nT,\vartheta }}\chi (\tau,y){\mathrm{d}}s(y),\quad n \in \mathbb{Z},\;\tau \in (0,T),\;\vartheta \in S_{+}^{2},}$$
of the Radon transform of the functions χ(τ, ⋅ ) is considered, where Eσ, ϑ denotes the plane defined in (38).

### Assumption 4.

The value$$\bar{\chi }_{n}(\tau,\vartheta )$$is a good approximation for the integral of the function χ(τ,⋅) over the planes$$E_{nT+\varepsilon,\vartheta }$$for all$$\varepsilon \in [-\frac{T} {2}, \frac{T} {2} )$$(see Fig. 4), that is
$$\displaystyle{ \bar{\chi }_{n}(\tau,\vartheta ) \approx \int _{E_{nT+\varepsilon,\vartheta }}\chi (\tau,y){\mathrm{d}}s(y),\quad \varepsilon \in [-\tfrac{T} {2}, \tfrac{T} {2} ),\;n \in \mathbb{Z},\;\tau \in (0,T),\;\vartheta \in S_{+}^{2}. }$$
Under the 4, Eq. (37) can be rewritten in the form
$$\displaystyle{m(t,\vartheta ) \approx \frac{c} {\sqrt{2(1 +\vartheta _{3 } )}}\int _{0}^{T}\bar{\chi }_{ N(\tau -t)}(\tau,\vartheta ){\mathrm{d}}\tau,}$$
where $$N(\sigma ) = \left \lfloor \frac{\sigma }{T} + \frac{1} {2}\right \rfloor$$ denotes the integer closest to $$\frac{\sigma }{T}$$. This (approximate) identity can now be iteratively solved for $$\bar{\chi }$$.

### Proposition 10.

Let
$$\displaystyle{\bar{m}(t,\vartheta ) = \frac{c} {\sqrt{2(1 +\vartheta _{3 } )}}\int _{0}^{T}\bar{\chi }_{ N(\tau -t)}(\tau,\vartheta ){\mathrm{d}}\tau,\quad \vartheta \in S_{+}^{2},\;t \in \mathbb{R},}$$
for some constant T > 0 with the integer-valued function$$N(\sigma ) = \left \lfloor \frac{\sigma }{T} + \frac{1} {2}\right \rfloor$$.
Then, $$\bar{\chi }$$fulfils the recursion relation
$$\displaystyle\begin{array}{rcl} & & \bar{\chi }_{n}(\tau,\vartheta ) =\bar{\chi } _{n+1}(\tau,\vartheta ) + \frac{\sqrt{2(1 +\vartheta _{3 } )}} {c} \frac{\partial \bar{m}} {\partial t} (\tau -(n + \tfrac{1} {2})T,\vartheta ), \\ & & \quad n \in \mathbb{Z},\;\tau \in (0,T),\;\vartheta \in S_{+}^{2}. {}\end{array}$$
(40)

### Proof.

Let $$t = -nT+\varepsilon$$ with $$\varepsilon \in [-\frac{T} {2}, \frac{T} {2} )$$ and $$n \in \mathbb{Z}$$, then
$$\displaystyle{N(\tau -t) = \left \{\begin{array}{@{}l@{\quad }l@{}} n \quad &\text{if}\;\tau \in (0, \tfrac{T} {2} +\varepsilon ), \\ n + 1\quad &\text{if}\;\tau \in [\tfrac{T} {2} +\varepsilon,T). \end{array} \right.}$$
Taking an arbitrary $$\varepsilon \in [-\frac{T} {2}, \frac{T} {2} )$$ and $$n \in \mathbb{Z},$$$$\bar{m}$$ can be formulated as
$$\displaystyle{\bar{m}(-nT+\varepsilon,\vartheta ) = \frac{c} {\sqrt{2(1 +\vartheta _{3 } )}}\left (\int _{0}^{\frac{T} {2} +\varepsilon }\bar{\chi }_{n}(\tau,\vartheta ){\mathrm{d}}\tau +\int _{ \frac{T} {2} +\varepsilon }^{T}\bar{\chi }_{ n+1}(\tau,\vartheta ){\mathrm{d}}\tau \right ).}$$
Differentiating this equation with respect to $$\varepsilon$$, it follows that
$$\displaystyle{\frac{\partial \bar{m}} {\partial t} (-nT+\varepsilon,\vartheta ) = \frac{c} {\sqrt{2(1 +\vartheta _{3 } )}}\left (\bar{\chi }_{n}(\tfrac{T} {2} +\varepsilon,\vartheta ) -\bar{\chi }_{n+1}(\tfrac{T} {2} +\varepsilon,\vartheta )\right ),}$$
which (with $$\tau = \frac{T} {2} +\varepsilon$$) is equivalent to (40). □
Thus, given that $$\bar{\chi }_{n}(t,\vartheta ) = 0$$ for sufficiently large $$n \in \mathbb{Z}$$ (recall that supp $$\chi (\tau,\cdot ) \subset \Omega$$ for all τ ∈ (0, T)), one can recursively reconstruct $$\bar{\chi }$$, to obtain the data
$$\displaystyle{\int _{E_{\sigma,\vartheta }}\chi (\tau,y){\mathrm{d}}s(y)\quad \text{for all}\quad \tau \in [0,T),\;\sigma \in \mathbb{R},\;\vartheta \in S_{+}^{2}}$$
for the Radon transform of χ(τ, ⋅ ).

However, since the plane Eσ, ϑ is by its Definition (38) orthogonal to the vector ϑ + e3 for ϑ ∈ S+2, this provides only the values of the Radon transform corresponding to planes which are orthogonal to a vector in the cone $$\mathcal{C}$$, see Fig. 3. For the reconstruction, one therefore still has to invert a limited angle Radon transform.

#### Dispersive Layered Medium with Focused Illumination

Except from ophthalmology, OCT is also widely used for investigation of skin deceases, such as cancer. From the mathematical point of view, this simplifies the main model since the human skin can be described as a multilayer structure with different optical properties and varying thicknesses in each layer.

Here the incident field is considered to propagate with normal incidence to the interface x3 = L and the detector array is replaced by a single point detector located at de3. The susceptibility is simplified as
$$\displaystyle{\chi (t,x) =\delta (x_{1})\delta (x_{2})\chi (t,x_{3}),}$$
and therefore the measurements provide the data, see (37) with $$\tilde{\chi }(\omega,k) =\tilde{\chi } (\omega,k_{3}),$$
$$\displaystyle{ \hat{m}(\omega ) =\tilde{\chi } (\omega, \tfrac{\omega } {c}2e_{3}),\quad \omega \in \mathbb{R}\setminus \{0\}. }$$
Considering the special structure of a layered medium, the susceptibility is described by a piecewise constant function in x3. This means explicitly that χ has the form
$$\displaystyle{ \chi (t,x_{3}) = \left \{\begin{array}{ll} \chi _{0}:= 0,&x_{3}\notin [0,L] \\ \chi _{n}(t), &x_{3} \in [L_{n},L_{n+1}) \end{array} \right.,\quad n = 1,\ldots,N }$$
(41)
with (unknown) parameters L = L1 > L2 >  > LN+1 = 0 characterizing the thicknesses of the N layers and (unknown) functions χn.
1, for ϑ = e3, gives
$$\displaystyle{m(t) = \frac{c} {2}\int _{-\infty }^{\infty }\bar{\chi }(\tau;\tau -t){\mathrm{d}}\tau,\quad \mbox{ where}\quad \bar{\chi }(\tau;\sigma ) =\chi (\tau, \tfrac{c\sigma } {2}).}$$

Remarking that $$\bar{\chi }$$ is piecewise constant (41) and additionally assuming that χ(⋅ , x3) has compact support, see 3, with $$T < \tfrac{2} {c}\min _{n}(L_{n} - L_{n+1})$$10 can be applied for ϑ = e3 to iteratively reconstruct χ starting from χ0 = 0.

##### Modified Born Approximation

In the proposed iteration scheme, 10, the traveling of the incident field through the sample before reaching a “specific” layer, where the susceptibility is to be reconstructed, is not considered. To do so, a modified iteration method is presented describing the traveling of the light through the different layers using Frensel’s equations.

The main idea is to consider, for example, in the second step of the recursive formula, given χ1 to find χ2, as incident the field $$\hat{E}^{(0)},$$ given by (26), traveled also through the first layer. This process can be continued to the next steps.

Let us first introduce some notations which will be used in the following. The fields $$\hat{E}_{n}^{(r)}$$ and $$\hat{E}_{n}^{(t)}$$ denote the reflected and the transmitted fields, with respect to the boundary Ln, respectively. The transmitted field $$\hat{E}_{n}^{(t)}$$ after traveling through the n-th layer is incident on the Ln+1 boundary and is denoted by $$\hat{E}_{n+1}^{(0)}.$$ The reflected field by the Ln+1 boundary back to the Ln boundary will be denoted by $$\hat{E}_{n+1}^{(r)}$$ and by $$\hat{E}_{n}^{(r')}$$ after traveling through the n-th layer (see Fig. 5). To simplify this model, multiple reflections are not included and the electric fields are taken to be tangential to the interface planes.

### Lemma 2.

Let the sample have susceptibility given by (41) and let ρnand τndenote the reflection and the corresponding transmission coefficients for the Lnboundary, respectively. Then, the field incident on the n-th layer with respect to the initial incident field$$\hat{E}^{(0)}:=\hat{ E}_{1}^{(0)}$$is given by
$$\displaystyle{ \left (\begin{array}{*{10}c} \hat{E}_{n}^{(0)} \\ 0 \end{array} \right ) = \left (\mathcal{M}_{1} \cdot \mathcal{M}_{2} \cdot \ldots \cdot \mathcal{M}_{n-1}\right )^{-1}\left (\begin{array}{*{10}c} \hat{E}^{(0)} \\ \hat{E}_{1}^{(r)} \end{array} \right )\quad \mbox{ for}\quad n = N-1,\ldots,2 }$$
assuming no backward field in the n-th layer, where
$$\displaystyle{\mathcal{M}_{n} = \frac{1} {\tau _{n}}\left (\begin{array}{*{10}c} {\mathrm{e}}^{{\mathrm{i}} \tfrac{\omega } {c}\sqrt{\chi _{n } +1}(L_{n}-L_{n+1})} & \rho _{ n}{\mathrm{e}}^{-{\mathrm{i}} \tfrac{\omega } {c}\sqrt{\chi _{n } +1}(L_{n}-L_{n+1})} \\ \rho _{n}{\mathrm{e}}^{{\mathrm{i}} \tfrac{\omega } {c}\sqrt{\chi _{n } +1}(L_{n}-L_{n+1})} & {\mathrm{e}}^{-{\mathrm{i}} \tfrac{\omega } {c}\sqrt{\chi _{n } +1}(L_{n}-L_{n+1})} \end{array} \right ).}$$

### Proof.

Because of the assumptions (normal incidence, $$\hat{E}^{(0)}$$ tangential to the boundary) the boundary conditions require the continuity of the total (upward and downward) electric and magnetic fields. Then, the reflection ρn and the corresponding transmission τn coefficients for the Ln boundary in terms of the susceptibility are given by [21]
$$\displaystyle{ \rho _{n} = \frac{\sqrt{\chi _{n-1 } + 1} -\sqrt{\chi _{n } + 1}} {\sqrt{\chi _{n-1 } + 1} + \sqrt{\chi _{n } + 1}},\quad \tau _{n} = 1 +\rho _{n}. }$$
To determine the propagation equations for the electric fields, the transfer matrices formulation is applied [30]. In particular, the fields at the top of the n-th layer can be computed with respect to the fields at the top of the (n + 1)th using
$$\displaystyle{ \left (\begin{array}{*{10}c} \hat{E}_{n}^{(0)} \\ \hat{E}_{n}^{(r)} \end{array} \right ) = \mathcal{M}_{n}\left (\begin{array}{*{10}c} \hat{E}_{n+1}^{(0)} \\ \hat{E}_{n+1}^{(r)} \end{array} \right )\quad \mbox{ for}\quad n = N-1,\ldots,1. }$$
and with respect to the incident field,
$$\displaystyle{ \left (\begin{array}{*{10}c} \hat{E}_{1}^{(0)} \\ \hat{E}_{1}^{(r)} \end{array} \right ) = \mathcal{M}_{1}\cdot \ldots \cdot \mathcal{M}_{n-1}\left (\begin{array}{*{10}c} \hat{E}_{n}^{(0)} \\ \hat{E}_{n}^{(r)} \end{array} \right )\quad \mbox{ for}\quad n = N-1,\ldots,2. }$$
□
From the previous result, given χn (by the recursion relation of 10), the matrix $$\mathcal{M}_{n+1}$$ is computed to obtain the update $$\hat{E}_{n+1}^{(0)}$$ which is then incident to the rest part of the sample. This means that $$\hat{E}^{(0)}$$ is replaced by $$\hat{E}_{n+1}^{(0)}$$ in the derivation of the measurements and the recursion relation (1 and 10) for computing χn+1. For example, in the second step to reconstruct χ2, the incident field is simply given by
$$\displaystyle{\hat{E}_{2}^{(0)} =\tau _{ 1}{\mathrm{e}}^{-{\mathrm{i}} \tfrac{\omega } {c}\sqrt{\chi _{1 } +1}(L_{1}-L_{2})}\hat{E}^{(0)}.}$$

The only unknown in this representation is the boundary L2 which can be approximated considering the point where change in the value of the measured function $$\bar{m}$$ is observed. The following analysis can be also extended for anisotropic media, but in a more complicated context since the displacement D and the electric field E are not always parallel.

A simplification usually made here is to consider the sample field as the sum of all the discrete reflections and neglect dispersion. This mathematical model was adopted by Bruno and Chaubell [7] for solving the inverse scattering problem of determining the refractive index and the width of each layer from the output data. The solution was obtained using the Gauss–Newton method and the effect of the initial guesses was also considered.

In conclusion, the traveling of the scattered field from the n-th layer through the sample could also be considered. Since the spherical waves can be represented as a superposition of plane waves by using similar techniques, in a more complicated form, one can obtain the transmitted scattered field.

### The Anisotropic Case

In the anisotropic case, the susceptibility χ cannot be considered a multiple of the identity. Therefore, the problem is to reconstruct from the expressions
$$\displaystyle{p_{j}\big[\vartheta \times (\vartheta \times \tilde{\chi }(\omega, \tfrac{\omega } {c}(\vartheta +e_{3}))p)\big]_{j},\quad j = 1,2,}$$
see (34), the matrix-valued function $$\chi: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3\times 3}$$, where it is assumed that measurements for every polarization $$p \in \mathbb{R}^{2} \times \{ 0\}$$ of the initial field E(0) are available.
Introducing in analogy to (36) the function
$$\displaystyle{\hat{m}_{p,j}: \mathbb{R} \times S_{+}^{2} \rightarrow \mathbb{C},\quad \hat{m}_{ p,j}(\omega,\vartheta ) =\tilde{\chi } _{\vartheta,p,j}(\omega, \tfrac{\omega } {c}(\vartheta +e_{3})),}$$
where $$\tilde{\chi }_{\vartheta,p,j}$$ is for every ϑ ∈ S+2, $$p \in \mathbb{R}^{2} \times \{ 0\}$$, and j ∈ { 1, 2} the (spatial and temporal) Fourier transform of
$$\displaystyle{\chi _{\vartheta,p,j}: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R},\quad \chi _{\vartheta,p,j}(t,x) = p_{j}\big[\vartheta \times (\vartheta \times \chi (t,x)p)\big]_{j},}$$
1 (with m replaced by mp, j and χ replaced by χϑ, p, j) can be applied to find that the inverse Fourier transform of $$\hat{m}_{p,j}$$ with respect to its first variable fulfils
$$\displaystyle{m_{p,j}(t,\vartheta ) = \frac{c} {\sqrt{2(1 +\vartheta _{3 } )}}\int _{-\infty }^{\infty }\int _{ E_{\tau -t,\vartheta }}\chi _{\vartheta,p,j}(\tau,y){\mathrm{d}}s(y){\mathrm{d}}\tau.}$$

Now, the same assumptions as in the isotropic case are considered, namely 3 and similar to 4:

### Assumption 5.

The approximation
$$\displaystyle{\int _{E_{nT,\vartheta }}\chi _{\vartheta,p,j}(\tau,y){\mathrm{d}}s(y) \approx \int _{E_{nT+\varepsilon,\vartheta }}\chi _{\vartheta,p,j}(\tau,y){\mathrm{d}}s(y)\quad \text{for all}\quad \varepsilon \in [-\tfrac{T} {2}, \tfrac{T} {2} )}$$
is for every$$\tau \in \mathbb{R}$$, ϑ ∈ S+2, $$n \in \mathbb{Z}$$, $$p \in \mathbb{R}^{2} \times \{ 0\}$$, and j ∈{ 1,2} justified.
Then, 10 provides an approximate reconstruction formula for the functions
$$\displaystyle{ \bar{\chi }_{p,j}(\tau;\sigma,\vartheta ) =\int _{E_{\sigma,\vartheta }}\chi _{\vartheta,p,j}(\tau,y){\mathrm{d}}s(y) = p_{j}\left [\vartheta \times \left (\vartheta \times \bar{\chi }(\tau;\sigma,\vartheta )p\right )\right ]_{j} }$$
(42)
for all $$p \in \mathbb{R}^{2} \times \{ 0\}$$, $$\tau \in \mathbb{R}$$, $$\sigma \in \mathbb{R}$$, ϑ ∈ S+2, and j ∈ { 1, 2}, where
$$\displaystyle{ \bar{\chi }(\tau;\sigma,\vartheta ) =\int _{E_{\sigma,\vartheta }}\chi (\tau,y){\mathrm{d}}s(y) }$$
(43)
denotes the two-dimensional Radon transform data of the function χ(τ, ⋅ ).

### Proposition 11.

Let ϑ ∈ S+2be fixed and ap,j, $$p \in \mathbb{R}^{2} \times \{ 0\}$$, j = 1,2, be such that the equations
$$\displaystyle{ p_{j}[\vartheta \times (\vartheta \times Xp)]_{j} = a_{p,j}\quad \text{for all}\quad p \in \mathbb{R}^{2} \times \{ 0\},\;j \in \{ 1,2\}, }$$
(44)
for the matrix$$X \in \mathbb{R}^{3\times 3}$$have a solution.
Then$$X \in \mathbb{R}^{3\times 3}$$is a solution of (44) if and only if
$$\displaystyle{ (P_{\vartheta }X)_{k\ell} = B_{k\ell},\quad B = \left (\begin{array}{*{10}c} -a_{e_{1},1} & a_{e_{1},1} - a_{e_{1}+e_{2},1} \\ a_{e_{2},2} - a_{e_{1}+e_{2},2} & -a_{e_{2},2} \end{array} \right ),\quad k,\ell\in \{ 1,2\}, }$$
(45)
where$$P_{\vartheta } \in \mathbb{R}^{3\times 3}$$denotes the orthogonal projection in direction ϑ.

### Proof.

First, remark that the equation system (44) is equivalent to the four equations
$$\displaystyle{ \begin{array}{llll} a_{e_{1},1} & = [\vartheta \times (\vartheta \times Xe_{1})]_{1},\qquad \qquad a_{e_{1}+e_{2},1} & = a_{e_{1},1} + [\vartheta \times (\vartheta \times Xe_{2})]_{1}, \\ a_{e_{2},2} & = [\vartheta \times (\vartheta \times Xe_{2})]_{2},\qquad \qquad a_{e_{1}+e_{2},2} & = a_{e_{2},2} + [\vartheta \times (\vartheta \times Xe_{1})]_{2},\end{array} }$$
(46)
which correspond to the Eq. (44) for (p, j) ∈ { (e1, 1), (e2, 2), (e1 + e2, 1), (e1 + e2, 2)}. Indeed, for arbitrary polarization p = p1e1 + p2e2, the expression pj[ϑ × (ϑ × Xp)]j can be written as a linear combination of the four expressions $$[\vartheta \times (\vartheta \times Xe_{i})]_{k}$$, i, k = 1, 2:
$$\displaystyle\begin{array}{rcl} p_{1}[\vartheta \times (\vartheta \times Xp)]_{1}& =& p_{1}^{2}[\vartheta \times (\vartheta \times Xe_{ 1})]_{1} + p_{1}p_{2}[\vartheta \times (\vartheta \times Xe_{2})]_{1}, {}\\ p_{2}[\vartheta \times (\vartheta \times Xp)]_{2}& =& p_{1}p_{2}[\vartheta \times (\vartheta \times Xe_{1})]_{2} + p_{2}^{2}[\vartheta \times (\vartheta \times Xe_{ 2})]_{2}, {}\\ \end{array}$$
and is thus determined by (46).
Now, the equation system (46) written in matrix form reads
$$\displaystyle{ [\vartheta \times (\vartheta \times Xp)]_{k} = -\left [B\left (\begin{array}{*{10}c} p_{1} \\ p_{2}\end{array} \right )\right ]_{k},\quad k \in \{ 1,2\}, }$$
(47)
for all $$p \in \mathbb{R}^{2} \times \{ 0\}$$ with B defined by (45).
Decomposing $$Xp = \left < \vartheta,Xp\right >\vartheta + P_{\vartheta }Xp$$, where $$P_{\vartheta } \in \mathbb{R}^{3\times 3}$$ denotes the orthogonal projection in direction ϑ, and using that
$$\displaystyle{\vartheta \times (\vartheta \times Xp) =\vartheta \times (\vartheta \times P_{\vartheta }Xp) = \left < \vartheta,P_{\vartheta }Xp\right >\vartheta - P_{\vartheta }Xp = -P_{\vartheta }Xp,}$$
the Eq. (47) can be written in the form (45). □
11 applied to the Eq. (42) for the matrix $$X =\bar{\chi } (\tau;\sigma,\vartheta )$$ for some fixed values $$\tau,\sigma \in \mathbb{R}$$ and $$\vartheta \in S_{+}^{2}$$ shows that the data $$a_{p,j} = p_{j}[\vartheta \times (\vartheta \times \bar{\chi }(\tau;\sigma,\vartheta ))]_{j}$$ for j = 1, 2 and the three different polarization vectors p = e1, p = e2, and p = e1 + e2 uniquely determine with Eq. (45) the projection
$$\displaystyle{(P_{\vartheta }\bar{\chi }(\tau;\sigma,\vartheta ))_{k,\ell} =\int _{E_{\sigma,\vartheta }}(P_{\vartheta }\chi (\tau,y))_{k,\ell}\,{\mathrm{d}}s(y)\quad \text{for}\quad k,\ell\in \{ 1,2\}.}$$
Moreover, measurements for additional polarizations p do not provide any further information so that at every detector point, corresponding to a direction ϑ ∈ S+2, only the four elements (Pϑχ)k, , $$k,\ell= 1,2$$, of the projection $$P_{\vartheta }\chi$$ influence the measurements.

To obtain additional data which make a full reconstruction of χ possible, one can carry out extra measurements after slight rotations of the sample.

So, let $$R \in \mathrm{ SO}(3)$$ describe the rotation of the sample. Then the transformed susceptibility χR is given by
$$\displaystyle{ \chi _{R}(t,y) = R\chi (t,R^{{\mathrm{T}}}y)R^{{\mathrm{T}}}. }$$
(48)

### Lemma 3.

Let$$\chi: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3\times 3}$$be the susceptibility of the sample and ϑ ∈ S+2be given. Furthermore, let R ∈ SO (3) be such that there exists a constant αR> 0 and a direction ϑR∈ S+2with
$$\displaystyle{ \vartheta _{R} + e_{3} =\alpha _{R}R(\vartheta +e_{3}) }$$
(49)
and define the susceptibility χRof the rotated sample by (48) .
Then, the data
$$\displaystyle{ \bar{\chi }_{R,p,j}(\tau;\sigma,\vartheta _{R}) = p_{j}\left [\vartheta _{R} \times \left (\vartheta _{R} \times \int _{E_{\sigma,\vartheta _{ R}}}\chi _{R}(\tau,y){\mathrm{d}}s(y)\right )\right ]_{j}, }$$
(50)
corresponding to the measurements of the rotated sample at the detector in direction ϑR, see (42), fulfil that
$$\displaystyle{ \bar{\chi }_{R,p,j}(\tau;\alpha _{R}\sigma,\vartheta _{R}) = p_{j}[\vartheta _{R} \times (\vartheta _{R} \times R\bar{\chi }(\tau;\sigma,\vartheta )R^{{\mathrm{T}}}]_{ j} }$$
(51)
for all$$\tau,\sigma \in \mathbb{R}$$, $$p \in \mathbb{R}^{2} \times \{ 0\}$$, j = 1,2, where$$\bar{\chi }$$is given by (43) .

### Proof.

Inserting Definition (48) and substituting z = RTy, formula (50) becomes
$$\displaystyle{\bar{\chi }_{R,p,j}(\tau;\sigma,\vartheta _{R}) = p_{j}\left [\vartheta _{R} \times \left (\vartheta _{R} \times \int _{R^{{\mathrm{T}}}E_{\sigma,\vartheta _{ R}}}R\chi (\tau,z)R^{{\mathrm{T}}}{\mathrm{d}}s(z)\right )\right ]_{ j}.}$$
Since now, by the Definition (38) of the plane Eσ, ϑ and by the Definition (49) of ϑR,
$$\displaystyle\begin{array}{rcl} R^{{\mathrm{T}}}E_{\sigma,\vartheta _{R}}& =& \{R^{{\mathrm{T}}}y \in \mathbb{R}^{3}\mid \left < \vartheta _{ R} + e_{3},y\right > = c\sigma \} {}\\ & =& \{z \in \mathbb{R}^{3}\mid \left < R^{{\mathrm{T}}}(\vartheta _{ R} + e_{3}),z\right > = c\sigma \} {}\\ & =& \{z \in \mathbb{R}^{3}\mid \alpha _{ R}\left < \vartheta +e_{3},z\right > = c\sigma \} = E_{ \frac{\sigma }{\alpha _{ R}},\vartheta }, {}\\ \end{array}$$
it follows (51). □

This means that the data $$\bar{\chi }_{R,p,j}(\tau;\alpha _{R}\sigma,\vartheta _{R})$$ obtained from a detector placed in the direction $$\vartheta _{R}$$, defined by (49), depends only on the Radon transform data $$\bar{\chi }(\tau;\sigma,\vartheta )$$. However, it still remains the algebraic problem of solving the Eq. (51) for different rotations R to obtain the matrix $$\bar{\chi }(\tau;\sigma,\vartheta ) \in \mathbb{R}^{3\times 3}$$.

### Proposition 12.

Let$$A \in \mathbb{R}^{3\times 3}$$and$$\vartheta \in S_{+}^{2}$$be given. Moreover, let$$R_{0},R_{1},R_{2} \in \mathrm{ SO}(3)$$be rotations so that every proper subset of$$\{R_{0}^{{\mathrm{T}}}e_{3},R_{1}^{{\mathrm{T}}}e_{3},R_{2}^{{\mathrm{T}}}e_{3},\vartheta +e_{3}\}$$is linearly independent and such that there exist for every R ∈{ R0, R1, R2} constants αR> 0 and$$\vartheta _{R} \in S_{+}^{2}$$fulfilling (49) .

Let further$$P \in \mathbb{R}^{2\times 3}$$be the orthogonal projection in direction e3, $$P_{\theta } \in \mathbb{R}^{3\times 3}$$the orthogonal projection in direction$$\theta \in \mathbb{R}^{3}$$, and
$$\displaystyle\begin{array}{rcl} B_{R}& =& \left (\begin{array}{*{10}c} -a_{R,e_{1},1} & a_{R,e_{1},1} - a_{R,e_{1}+e_{2},1} \\ a_{R,e_{2},2} - a_{R,e_{1}+e_{2},2} & -a_{R,e_{2},2} \end{array} \right ), {}\\ & & \qquad a_{R,p,j} = p_{j}[\vartheta _{R} \times (\vartheta _{R} \times RAR^{{\mathrm{T}}}p)]_{ j}, {}\\ \end{array}$$
for every$$R \in \{ R_{0},R_{1},R_{2}\}$$.
Then, the equations
$$\displaystyle{ PP_{\vartheta _{R}}RXR^{{\mathrm{T}}}P^{{\mathrm{T}}} = B_{ R},\quad R \in \{ R_{0},R_{1},R_{2}\} }$$
(52)
have the unique solution X = A.

### Proof.

Using that $$\vartheta _{R} =\alpha _{R}R(\vartheta +e_{3}) - e_{3}$$, see (49), it follows with Pe3 = 0 that
$$\displaystyle{PP_{\vartheta _{R}} = P(\mathbb{1} -\vartheta _{R}\vartheta _{R}^{{\mathrm{T}}}) = P -\alpha _{ R}PR(\vartheta +e_{3})\vartheta _{R}^{{\mathrm{T}}}.}$$
With this identity, the Eq. (52) can be written in the form
$$\displaystyle{ PR(X -\alpha _{R}(\vartheta +e_{3})\vartheta _{R}^{{\mathrm{T}}}RX)R^{{\mathrm{T}}}P^{{\mathrm{T}}} = B_{ R}. }$$
(53)
Let now $$\eta _{R} \in \mathbb{R}^{3}$$ denote a unit vector orthogonal to ϑ + e3 and orthogonal to RTe3. Then R ηR is orthogonal to e3 and therefore, with $$P^{{\mathrm{T}}}P = P_{e_{3}}$$,
$$\displaystyle{(PR\eta _{R})^{{\mathrm{T}}}PR = (P_{ e_{3}}R\eta _{R})^{{\mathrm{T}}}R = (R\eta _{ R})^{{\mathrm{T}}}R =\eta _{ R}^{{\mathrm{T}}}.}$$
Thus, multiplying the Eq. (53) from the left with (PR ηR)T, it follows that
$$\displaystyle{\eta _{R}^{{\mathrm{T}}}(X -\alpha _{ R}(\vartheta +e_{3})\vartheta _{R}^{{\mathrm{T}}}RX)R^{{\mathrm{T}}}P^{{\mathrm{T}}} = (PR\eta _{ R})^{{\mathrm{T}}}B_{ R}.}$$
Since now ηR is orthogonal to ϑ + e3, this simplifies to
$$\displaystyle{ \eta _{R}^{{\mathrm{T}}}XR^{{\mathrm{T}}}P^{{\mathrm{T}}} = (PR\eta _{ R})^{{\mathrm{T}}}B_{ R}. }$$
(54)
Remarking that the orthogonal projection onto the line $$\mathbb{R}\eta _{R}$$ can be written as the composition of two orthogonal projections onto orthogonal planes with intersection $$\mathbb{R}\eta _{R}$$:
$$\displaystyle{\eta _{R}\eta _{R}^{{\mathrm{T}}} = P_{\nu _{ R}}P_{\vartheta +e_{3}},}$$
where νR is a unit vector orthogonal to ηR and ϑ + e3, and multiplying Eq. (54) from the left with ηR, one finds that
$$\displaystyle{P_{\nu _{R}}(P_{\vartheta +e_{3}}X)R^{{\mathrm{T}}}P^{{\mathrm{T}}} =\eta _{ R}(PR\eta _{R})^{{\mathrm{T}}}B_{ R}.}$$
Applying then the projection P from the right and using that $$R^{{\mathrm{T}}}P^{{\mathrm{T}}}P = R^{{\mathrm{T}}}P_{e_{3}} = P_{R^{{\mathrm{T}}}e_{3}}R^{{\mathrm{T}}}$$, this can be written in the form
$$\displaystyle{ P_{\nu _{R}}(P_{\vartheta +e_{3}}X)P_{R^{{\mathrm{T}}}e_{3}} =\eta _{R}(PR\eta _{R})^{{\mathrm{T}}}B_{ R}PR. }$$
(55)
Evaluating this equation now for R = R0, R1, R2 and remarking that $$\{\nu _{R_{0}},\nu _{R_{1}},\nu _{R_{2}}\}$$ and $$\{R_{0}^{{\mathrm{T}}}e_{3},R_{1}^{{\mathrm{T}}}e_{3},R_{2}^{{\mathrm{T}}}e_{3}\}$$ are linearly independent, one concludes that the 3 × 3 matrix $$P_{\vartheta +e_{3}}X$$ is uniquely determined by (55).
However, it is also possible to calculate X (and not only its projection $$P_{\vartheta +e_{3}}X)$$ from Eq. (53). Because
$$\displaystyle{X = P_{\vartheta +e_{3}}X + \frac{1} {\vert \vartheta + e_{3}\vert ^{2}}(\vartheta +e_{3})(\vartheta +e_{3})^{{\mathrm{T}}}X,}$$
it follows from (53) that
$$\displaystyle\begin{array}{rcl} & & PR\left ( \frac{\vartheta +e_{3}} {\vert \vartheta + e_{3}\vert ^{2}}(1 -\alpha _{R}\vartheta _{R}^{{\mathrm{T}}}R(\vartheta +e_{ 3}))(\vartheta +e_{3})^{{\mathrm{T}}}X\right )R^{{\mathrm{T}}}P^{{\mathrm{T}}} \\ & & \qquad \qquad \qquad \ \qquad = B_{R} - PR(\mathbb{1} -\alpha _{R}(\vartheta +e_{3})\vartheta _{R}^{{\mathrm{T}}}R)P_{\vartheta +e_{3}}XR^{{\mathrm{T}}}P^{{\mathrm{T}}}.{}\end{array}$$
(56)
Plugging in the identity
$$\displaystyle{\alpha _{R}\vartheta _{R}^{{\mathrm{T}}}R(\vartheta +e_{ 3}) =\vartheta _{ R}^{{\mathrm{T}}}(\vartheta _{ R} + e_{3}) = 1 +\vartheta _{R,3},}$$
which follows from the Definition (49) of ϑR, applying PT from the left and P from the right, and using as before $$R^{{\mathrm{T}}}P^{{\mathrm{T}}}P = P_{R^{{\mathrm{T}}}e_{3}}R^{{\mathrm{T}}}$$, the Eq. (56) yields
$$\displaystyle\begin{array}{rcl} & & -\vartheta _{R,3}P_{R^{{\mathrm{T}}}e_{3}}\left ( \frac{\vartheta +e_{3}} {\vert \vartheta + e_{3}\vert ^{2}}(\vartheta +e_{3})^{{\mathrm{T}}}X\right )P_{ R^{{\mathrm{T}}}e_{3}} \\ & & \qquad \quad \qquad = R^{{\mathrm{T}}}P^{{\mathrm{T}}}B_{ R}PR - P_{R^{{\mathrm{T}}}e_{3}}(\mathbb{1} -\alpha _{R}(\vartheta +e_{3})\vartheta _{R}^{{\mathrm{T}}})P_{\vartheta +e_{3}}XP_{R^{{\mathrm{T}}}e_{3}}.{}\end{array}$$
(57)
Since the right hand side is already known (it depends only on $$P_{\vartheta +e_{3}}X$$), the equation system (57) for R = R0, R1, R2 can be uniquely solved for
$$\displaystyle{ \frac{\vartheta +e_{3}} {\vert \vartheta + e_{3}\vert ^{2}}(\vartheta +e_{3})^{{\mathrm{T}}}X.}$$

Therefore, the Eq. (52) uniquely determine X and because A is by construction a solution of the equations, this implies that X = A. □

Thus, applying 12 to the matrix $$A =\bar{\chi } (\tau;\sigma,\vartheta )$$ shows that the measurements aR, p, j obtained at the detectors ϑR for the polarizations p = e1, e2, e1 + e2 and rotations R = R0, R1, R2, fulfilling the assumptions of 12 provide sufficient information to reconstruct the Radon data $$\bar{\chi }(\tau;\sigma,\vartheta )$$. Calculating these two-dimensional Radon data for all directions ϑ in some subset of S+2 (by considering some additional rotations so that for every direction ϑ, there exist three rotations fulfilling the assumptions of 12), it is possible via an inversion of a limited angle Radon transform to finally recover the susceptibility χ.

## 6 Conclusion

In this chapter, a general mathematical model of OCT based on Maxwell’s equations has been presented. As a consequence of this modeling, OCT was formulated as an inverse scattering problem for the susceptibility χ. It was shown that without additional assumptions about the optical properties of the medium, in general, χ cannot be reconstructed due to lack of measurements. Some reasonable physical assumptions were presented, under which the medium can, in fact, be reconstructed. For instance, if the medium is isotropic, iterative schemes to reconstruct the susceptibility were developed. Dispersion and focus illumination are also considered. For an anisotropic medium, it follows that different incident fields, with respect to direction (rotating the sample) and polarization, should be considered to completely recover χ.

## Notes

### Acknowledgements

The authors would like to thank Wolfgang Drexler and Boris Hermann from the Medical University Vienna for their valuable comments and stimulating discussions. This work has been supported by the Austrian Science Fund (FWF) within the national research network Photoacoustic Imaging in Biology and Medicine, projects S10501-N20 and S10505-N20.

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## Authors and Affiliations

• Peter Elbau
• 1
• Leonidas Mindrinos
• 1
• Otmar Scherzer
• 2
• 3
1. 1.Computational Science CenterUniversity of ViennaViennaAustria
2. 2.Computational Science CenterUniversity of ViennaViennaAustria