IAM physical model notes

Infinite internal reflections between parallel planes

This is related to PR 1616. The following sections all refer to the sketch lower down.

first boundary (12)

At the 12 boundary, the reflectance is $\rho_{12}$ and therefore transmittance is $\tau_{12} = \left(1-\rho_{12}\right)$.

second boundary (23)

At the 23 boundary, the transmitted ray is reflected internally by $\rho_{23}$ , and therefore the first transmittance from the 23 boundary is as follows:

$$ \tau_{23} = \tau_{12}\left(1-\rho_{23}\right) = \left(1-\rho_{12}\right)\left(1-\rho_{23}\right) $$

first boundary (12) again, first internal reflection

The internally reflected ray from the 23 boundary, returns to the 12 boundary at the same angle since the planes are parallel and Snell's law. The internally reflected ray is $\tau_{12}\rho_{23}$ part of which is retransmitted back to the medium on the 1 side of the first boundary as $\tau_{12}\rho_{23}\left(1-\rho_{12}\right)$ and part of which is reflected internally again back into the medium on the 2 side between the parallel boundaries as $\tau_{12}\rho_{23}\rho_{12}$.

Note that the reflectance $\rho_{12}$ is the same regardless of which side of boundary 12 it is incident on. This is because Fresnel's equations are squared, therefore always positive, regardless of the order of indices. For example:

$$ \rho_0 = \left(\frac{n_1 - n_2}{n_1 + n_2}\right)^2 = \left(\frac{n_2 - n_1}{n_2 + n_1}\right)^2 $$

second boundary (23) again, first internal reflection

Therefore the retransmitted ray from the 23 boundary after the first internal reflection (shown as exponent 😜) is as follows:

$$ \tau^1_{23} = \tau_{12}\rho_{23}\rho_{12}\left(1-\rho_{23}\right) = \left(1-\rho_{12}\right)\rho_{23}\rho_{12}\left(1-\rho_{23}\right) $$

One can imagine that each consecutive internal reflection will always result in multiplication by $\rho_{12}\rho_{23}$ as the ray bounces up to the 12 boundary and back down to the 23 boundary. For example, the 2nd internal reflection would be $\tau^2_{23} = \tau_{12}\rho^2_{23}\rho^2_{12}\left(1-\rho_{23}\right)$. Therefore, with infinite internal reflections the total transmittance from the 3 side of the second boundary would be as follows:

$$ \sum\tau = \tau_{12}\left(1-\rho_{23}\right)\left(1 + \rho_{23}\rho_{12} + \rho^2_{23}\rho^2_{12} + \cdots \right) = \left(1-\rho_{12}\right)\left(1-\rho_{23}\right)\left(1 + \rho_{23}\rho_{12} + \rho^2_{23}\rho^2_{12} + \cdots \right) $$

first boundary (12) again, 2nd internal reflection

Similar to the second boundary, each time the ray returns to the first boundary it is further reduced by $\rho_{12}\rho_{23}$. Therefore, the transmittance from the 1 side of the first boundary on the 2nd internal reflection is $\tau_{12}\rho_{23}\rho_{12}\rho_{23}\left(1-\rho_{12}\right)$.

zero absorption assumption, energy conservation

If we assume as the number internal reflections approaches infinity that zero energy is absorbed between the boundaries then all of the incident light will be retransmitted from either side of the boundaries. Let's call the transmittance from the 1 side of the first boundary $\sum\rho$ for convenience. then $\sum\rho + \sum\tau = 1$ where the total transmittance from the 3 side of the second boundary was already given. Similarly to transmittance from the bottom boundary, each consecutive internal reflection will always result in multiplication by $\rho_{12}\rho_{23}$ of the total transmittance from the top boundary. Therefore, its total transmittance, remember we're calling it $\sum\rho$ for convenience, is as follows:

$$\begin{split} \sum\rho &= \rho_{12} + \tau_{12}\rho_{23}\left(1-\rho_{12}\right) + \tau_{12}\rho_{23}\rho_{12}\rho_{23}\left(1-\rho_{12}\right) + \tau_{12}\rho_{23}\rho^2_{12}\rho^2_{23}\left(1-\rho_{12}\right) + \cdots \\ &= \rho_{12} + \tau_{12}\rho_{23}\left(1-\rho_{12}\right) \left(1 + \rho_{23}\rho_{12} + \rho^2_{23}\rho^2_{12} + \cdots \right) \\ &= \rho_{12} + \rho_{23}\left(1-\rho_{12}\right)^2 \left(1 + \rho_{23}\rho_{12} + \rho^2_{23}\rho^2_{12} + \cdots \right) \end{split}$$

algebra

We can simplify the energy conservation expression given above if we let $\mathit{HOT} = \left(1 + \rho_{23}\rho_{12} + \rho^2_{23}\rho^2_{12} + \cdots \right)$ as follows:

$$\begin{split} 1 &= \sum\rho + \sum\tau \\ 1 &= \left(1-\rho_{12}\right)\left(1-\rho_{23}\right)\mathit{HOT} + \rho_{12} + \rho_{23}\left(1-\rho_{12}\right)^2\mathit{HOT} \\ 1-\rho_{12} &= \left(1-\rho_{12}\right)\left(1-\rho_{23}\right)\mathit{HOT} + \rho_{23}\left(1-\rho_{12}\right)^2\mathit{HOT} \\ \frac{1-\rho_{12}}{\mathit{HOT}} &= \left(1-\rho_{12}\right)\left(1-\rho_{23}\right) + \rho_{23}\left(1-\rho_{12}\right)^2 \\ \frac{1}{\mathit{HOT}} &= \left(1-\rho_{23}\right) + \rho_{23}\left(1-\rho_{12}\right) \\ \frac{1}{\mathit{HOT}} &= 1-\rho_{23} + \rho_{23}-\rho_{23}\rho_{12} \\ \frac{1}{\mathit{HOT}} &= 1-\rho_{23}\rho_{12} \\ \mathit{HOT} &= \frac{1}{1-\rho_{23}\rho_{12}} \\ \end{split}$$

This leaves us with a very convenient expression for the total transmittance from the bottom of the second boundary as follows:

$$ \sum\tau = \tau_{12}\left(1-\rho_{23}\right)\mathit{HOT} = \frac{\tau_{12}\left(1-\rho_{23}\right)}{1-\rho_{23}\rho_{12}} $$

QED

sketch of two planes with boundaries 12 and 23