How to use ray transfer matrices for practical purposes?

Question

I am familiar with the concept of a ray transfer matrix (ABCD matrix) and how it describes the transformation of a ray with height and angle $\pmatrix{x_0 \ \theta_0}$ into a ray with height and angle $\pmatrix{x_1 \ \theta_1}$. However, I am having trouble using this to answer practical questions about an optical system.

For instance, if I have a complex optical setup described by the matrix $$\textbf{M} = \pmatrix{A && B\\ C && D}$$ and I need to determine the optimal position for placing an image sensor to capture the output, how should I proceed?

I understand the relationship $$\pmatrix{x_1 \\ \theta_1} = \pmatrix{A && B \\ C && D} \pmatrix{x_0 \\ \theta_0}$$ but I'm not sure how to use this information to find the correct position for the image sensor. Could someone explain the steps or provide guidance on how to approach this problem?

Answer 1

In the example you are discussing, consider the following scenario. To take an image of an object you need to bring all the rays emitted from a single point to a single point on your camera with your optical system. It means that you want all the rays of type $(x, \theta_1)$, $(x, \theta_2)$, ... after traveling distance $d_1$ to your system described by an ABCD matrix to converge after another distance $d_2$ to $(x', \theta_1')$, $(x', \theta_2')$, ... . Let's now do the math for a single ray.

$$\begin{bmatrix}1&d_2\\0&1\end{bmatrix} \begin{bmatrix}A&B\\C&D\end{bmatrix} \begin{bmatrix}1&d_1\\0&1\end{bmatrix} \begin{bmatrix}x\\\theta\end{bmatrix} = \begin{bmatrix}A(x+d_1\theta) + B\theta+d_2(C(x+d_1\theta)+D\theta)\\...\end{bmatrix}. $$

The equation for the angular part is omitted, as it's not important for our reasoning. After all, it's not important with what angle your ray hits your sensor. From the demand that the height $x'$ have to be common between the rays you can state: $$ A(x+d_1\theta_1) + B\theta_1+d_2(C(x+d_1\theta_1)+D\theta_1) =A(x+d_1\theta_2) + B\theta_2+d_2(C(x+d_1\theta_2)+D\theta_2) \\ (Ad_1+B+Cd_2d_1+Dd_2)\theta_1 = (Ad_1+B+Cd_2d_1+Dd_2)\theta_2 .$$

And that equation has to be true for any pair of $\theta$s, so:

$$ Ad_1+B+Cd_2d_1+Dd_2 = 0.$$ And finally: $$ d_2 = -\frac{Ad_1 +B}{Cd_1 + D},$$

which is the position where your image is created, and where you should put your sensor. You can easily check that for the ABCD matrix of a thin lens you recreate the lens equation: $1/d_1 + 1/d_2 = 1/f$.