Light Transport in Tissue

Index mismatch, no incident diffuse light, both media scattering

Since the radiance over the square of the index of refraction is constant across an interface (Section B.1) then as long as the radiance is divided by the square of the index of refraction of the medium it is in, the same methods used previously will suffice. Consequently, for light travelling upwards

$\displaystyle \int_{2\pi\mu\,\,\le0}{L^{(1)}(\mathbf{r},\hat\mathbf{s})\over n_1^2}(-\hat\mathbf{z}\cdot\hat\mathbf{s})\,d\omega$	=	$\displaystyle \int_{2\pi\mu\,\,\ge0}{L^{(1)}(\mathbf{r},\hat\mathbf{s})\over n_... ...(\hat\mathbf{z}\cdot\hat\mathbf{s})(\hat\mathbf{z}\cdot\hat\mathbf{s})\,d\omega$
	+	$\displaystyle \int_{2\pi\mu\,\,\le0}{L^{(2)}(\mathbf{r},\hat\mathbf{s})\over n_... ...\hat\mathbf{z}\cdot\hat\mathbf{s})(-\hat\mathbf{z}\cdot\hat\mathbf{s})\,d\omega$	(4.64)

This means that the total amount of light travelling upwards in layer 1 from the boundary equals the light reflected back into layer 1 plus that transmitted from the lower layer. Similarly, for light travelling downwards

$\displaystyle \int_{2\pi\mu\,\,\ge0}{L^{(2)}(\mathbf{r},\hat\mathbf{s})\over n_2^2}(\hat\mathbf{z}\cdot\hat\mathbf{s})\,d\omega$	=	$\displaystyle \int_{2\pi\mu\,\,\le0}{L^{(2)}(\mathbf{r},\hat\mathbf{s})\over n_... ...\hat\mathbf{z}\cdot\hat\mathbf{s})(-\hat\mathbf{z}\cdot\hat\mathbf{s})\,d\omega$
	+	$\displaystyle \int_{2\pi\mu\,\,\ge0}{L^{(1)}(\mathbf{r},\hat\mathbf{s})\over n_... ...(\hat\mathbf{z}\cdot\hat\mathbf{s})(\hat\mathbf{z}\cdot\hat\mathbf{s})\,d\omega$	(4.65)

The superscripts on the reflection r and the transmission t indicate the medium from which light is incident on the boundary

For upwards travelling light, Equation (4.64) can be simplified to

$\displaystyle {1\over n_2^2}\Big[\left(1-R_1^{21}\right)\varphi_d^{(2)}(\mathbf{r})$	-	$\displaystyle 2\left(1-R_2^{21}\right)\mathbf{F}_d^{(2)}(\mathbf{r})\cdot\hat\mathbf{z}\Big]$	(4.66)
	=	$\displaystyle {1\over n_1^2}\Big[\left(1-R_1^{12}\right)\varphi_d^{(1)}(\mathbf... ...-2\left(1+R_2^{12}\right)\mathbf{F}_d^{(1)}(\mathbf{r})\cdot\hat\mathbf{z}\Big]$

For downwards travelling light, Equation (4.65) becomes

$\displaystyle {1\over n_2^2}\Big[\left(1-R_1^{21}\right)\varphi_d^{(2)}(\mathbf{r})$	+	$\displaystyle 2\left(1+R_2^{21}\right)\mathbf{F}_d^{(2)}(\mathbf{r})\cdot\hat\mathbf{z}\Big]$	(4.67)
	=	$\displaystyle {1\over n_1^2}\Big[\left(1-R_1^{12}\right)\varphi_d^{(1)}(\mathbf... ...+2\left(1-R_2^{12}\right)\mathbf{F}_d^{(1)}(\mathbf{r})\cdot\hat\mathbf{z}\Big]$

Subtracting yields

$\begin{displaymath} \mathbf{F}_d^{(1)}(\mathbf{r})\cdot\hat\mathbf{z}=\left({n_1... ..._2}\right)^2 \mathbf{F}_d^{(2)}(\mathbf{r})\cdot\hat\mathbf{z} \end{displaymath}$

(4.68)

Thus the flux across a boundary behaves the same way that the radiance does. Equation (4.68) may be rewritten

$\begin{displaymath} {1\over n_1^2\mu_{tr}^{(1)}} {d\varphi^{(1)}(\mathbf{r})\ove... ...over n_2^2\mu_{tr}^{(2)}} {d\varphi^{(2)}(\mathbf{r})\over dz} \end{displaymath}$

(4.69)

Adding (4.67) and (4.68) yields

$\displaystyle {1\over n_1^2}\Big[\left(1-R_1^{12}\right)\varphi_d^{(1)}(\mathbf{r})$	-	$\displaystyle 2 R_2^{12}\mathbf{F}_d^{(1)}(\mathbf{r})\cdot\hat\mathbf{z}\Big]$	(4.70)
	=	$\displaystyle -{1\over n_2^2}\Big[\left(1-R_1^{21}\right)\varphi_d^{(2)}(\mathbf{r}) + 2 R_2^{21}\mathbf{F}_d^{(2)}(\mathbf{r})\cdot\hat\mathbf{z}\Big]$

where R₂²¹ is the second moment of the Fresnel reflection R₂ for light passing from n₁ to n₂. Substituting for F_d(r)

$\displaystyle {1\over n_1^2}\Big[\left(1-R_1^{12}\right)\varphi_d^{(1)}(\mathbf{r})$	-	$\displaystyle R_2^{12}\left({1\over\mu_{tr}^{(1)}} {\partial\varphi^{(1)}(\mathbf{r})\over \partial z}-Q'(\mathbf{r}) \right)\Big]$	(4.71)
	=	$\displaystyle -{1\over n_2^2}\Big[\left(1-R_1^{21}\right)\varphi_d^{(2)}(\mathb... ...{\partial\varphi^{(2)}(\mathbf{r})\over \partial z}-Q'(\mathbf{r}) \right)\Big]$

This equation with Equation (4.68) provides the two boundary conditions at an interface. If n₁=n₂ then R₂¹²=R₂²¹=0.