add_same_slides

iadpython.combine.add_same_slides(sample, R01, R10, T01, T10, R, T)[source]

Find matrix when slab is sandwiched between identical slides.

This routine is optimized for a slab with equal boundaries on each side. It is assumed that the slab is homogeneous and therefore the ‘R’ and ‘T’ matrices are identical for upward or downward light directions.

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: R_{01}=R_{32}, R_{10}=R_{23}, T_{01}=T_{32}, and T_{10}=T_{23}.

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 R_{30} and T_{03} respectively. These are given by

\[A_{XX} = T_{12}(E-R_{10}R_{12})^{-1}\]

\[R_{20} = A_{XX} R_{10}T_{21} + R_{21}\]

\[B_{XX} = T_{10}(E-R_{20}R_{10})^{-1}\]

\[T_{03} = B_{XX} A_{XX} T_{01}\]

\[R_{30} = B_{XX} R_{20} T_{01} + R_{01}/(2 𝜈 w)^2\]

Parameters:

R01 – R, T for slide assuming 0=air and 1=slab
R10, T01, T10 – R, T for slide assuming 0=air and 1=slab
T01, T10 – R, T for slide assuming 0=air and 1=slab
T10 – R, T for slide assuming 0=air and 1=slab
R, T – R12=R21, T12=T21 for homogeneous slab

Returns:

T30, T03 – R, T for all 3 with top = bottom boundary