# Ray transfer matrix method, I need explanations

Gold Member

## Homework Statement

It's very important to know how to construct the ray transfer matrix since it allows to solve many thick lenses problems.

In wikipedia: http://en.wikipedia.org/wiki/Ray_transfer_matrix_analysis#Definition_of_the_ray_transfer_matrix, I do not understand what they mean by $$A = {x_2 \over x_1 } \bigg|_{\theta_1 = 0}$$.
I don't understand the notation.
I can't even find the ray transfer matrix for a ray passing through a medium of refractive index=n (one of the most simple case I guess).

Can someone explain me what does the notation mean?

So that I can derive all the examples in the wikipedia page and solve a lot of exercises.
Thanks a lot... and sorry for being slow. I've also checked in Hecht's book, but I didn't understand most of it. As I said, I've absolutely NO IDEA about how to find the ray transfer matrix of the most simple cases.

In that case we can write $$\begin{bmatrix} x_2 \\ \theta _2 \end{bmatrix}= \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} x_1 \\ \theta _1 \end{bmatrix}$$.
Thus $$x_2=Ax_1+B \theta _1$$ and $$\theta _2=Cx_1 +D \theta _1$$.
Wikipedia's notation means that $$A=\frac{x_2- B \theta _1}{x_1}$$ evaluated in $$\theta _1=0$$ (because of the paraxial rays approximation) and we reach $$A=\frac{x_2}{x_1}$$.
We have that $$D=\frac{\theta _2 - Cx_2}{\theta _1}$$. According to wikipedia, if we set $$x_1 =0$$, we reach $$D=\frac{\theta _2}{\theta _1}$$. Why is it so? If they meant $$x_2=0$$ I would understand, but $$x_1$$?