Light Transport in Tissue

Boundary conditions

Boundary conditions are implemented in the adding-doubling method by creating a layer which mimics the reflection and transmission at a boundary. This layer is added to a slab to find the reflection and transmission for a slab including the boundary conditions. If $r(\mu)$ is the Fresnel reflection for light incident from a medium with index of refraction n0 on a medium with index of refraction n₁ then

$\begin{displaymath} R_{01}(\mu_i,\mu_j) = {r(\mu_i)\over2\mu_i}\delta_{ij} \end{displaymath}$

(3.35)

$\begin{displaymath} T_{01}(\mu_i,\mu_j) = {1-r(\mu_i)\over2\mu_i} \left({n_1\over n_0}\right)^2 \delta_{ij} \end{displaymath}$

(3.36)

The square of the ratio of the indices of refraction is due to the n²-Law of radiance (Appendix B) which accounts for the difference in radiances across across an index of refraction mismatch. The factor of $2\mu$ is included to compensate for star multiplication. If a glass slide is present then Equation (B.21) should be used to calculate $r(\mu_i)$ with $n_i=n_{\rm tissue}$ , $n_g=n_{\rm glass}$ and $n_t=n_{\rm outside}$ . Both operators are diagonal because light is specularly reflected and the angle of incidence equals the angle of reflection. The reflection and transmission operators for light travelling from medium 1 into medium 0 are

$\begin{displaymath} R_{10}(\mu_i,\mu_j) = R_{01}(\mu_i,\mu_j) \end{displaymath}$

(3.37)

$\begin{displaymath} T_{10}(\mu_i,\mu_j) = T_{01}(\mu_i,\mu_j) \left({n_0\over n_1}\right)^4 \delta_{ij} \end{displaymath}$

(3.38)

Since light is refracted at the boundary, care must be taken to ensure that the incident and reflected fluxes are identified with the proper angles. For example, if eleven Gaussian quadrature angles are used, all but three are totally internally reflected at an interface with n₁=1.5 and n₂=1.0. This is caused by the uneven distribution of quadrature points over the integration interval--in the Gaussian quadrature scheme oblique angles are emphasized over angles near normal and, therefore, most of the quadrature angles undergo total internal reflection.

If equal boundary conditions exist on both sides of the slab then, by symmetry, the transmission and reflection operator for light travelling from the top to the bottom are equal to those for light propagating from the bottom to the top. Consequently only one set need be calculated. This leads to a faster method for calculating the reflection and transmission for a slab with equal boundary conditions on each side. Let the top boundary be layer 01, the medium layer 12, and the bottom layer 23. The boundary conditions on each side are equal: R₀₁=R₃₂, R₁₀=R₂₃, T₀₁=T₃₂, and T₁₀=T₂₃. For example the light reflected from layer 01 (travelling from boundary 0 to boundary 1) will equal the amount of light reflected from layer 32, since there is no physical difference between the two cases. The switch in the numbering arises from the fact that light passes from the medium to the outside at the top surface by going from 1 to 0, and from 2 to 3 on the bottom surface. The reflection and transmission for the slab with boundary conditions are R₃₀ and T₀₃ respectively. These are given by

T₀₂ = T₁₂(E-R₁₀R₁₂)^-1T₀₁

(3.39)

R₂₀ = T₁₂(E-R₁₀R₁₂)^-1 R₁₀T₂₁+R₂₁

(3.40)

and

T₀₃ = T₁₀(E-R₂₀R₁₀)^-1T₀₂

(3.41)

R₃₀ = T₁₀(E-R₂₀R₁₀)^-1 R₂₀T₀₁+R₀₁

(3.42)

Further increases in efficiency may be made by exploiting the diagonal nature of the reflection and transmission operators for an interface, since most matrix/matrix multiplications above become vector/matrix multiplications.

S. A. Prahl."Light Transport in Tissue," PhD thesis, University of Texas at Austin, 1988.