# Ray tracing with transfer matrix method

1. Sep 24, 2015

### mester1025

Hi,
I'm new in physics and optics so I need a little help. I've a simple optical system from 2 thin lenses.
The first thin lens has a focal distance of 50 [mm] , and the second one has 25 [mm]. The 2 lenses are separated by 40 [mm] and the object is placed 75 [mm] before the first lens.

I've to find the position of the final image and simulate howto propagate the rays across the optical system.

So, I made a little program to determine the ray transfer matrix of the system.
This is the transfer matrix of the system:
\begin{bmatrix} 0.20-0.028x & 55-2.7x \\ -0.0028 & -2.7 \end{bmatrix}

where the x variable is the final image distance from the last lens.

If I do the following:
$$55-2.7x = 0$$
Then x = 20.37 [mm] , this is the image distance, exactly what I was looking for.
So the system matrix is:
\begin{bmatrix} -0.37036 & 0.001 \\ -0.0028 & -2.7 \end{bmatrix}

My problem is:

I have lots of input rays to simulate the ray propagation, but I can't understand the output rays.

For example :

\begin{bmatrix} y \\ θ \end{bmatrix}
=
\begin{bmatrix} 0.0002 \\ -0.54 \end{bmatrix}

where y is the input height and θ is the input angle in radian.

\begin{bmatrix} 0.0002 \\ -0.54 \end{bmatrix}
=
\begin{bmatrix} -0.37036 & 0.001 \\ -0.0028 & -2.7 \end{bmatrix}

\begin{bmatrix} 0 \\ 0.2 \end{bmatrix}

Could you help me explain the output ray of the system?

I apologize for the wall of text :)

2. Sep 25, 2015

### Simon Bridge

A ray input at height 0 and grad 0.2 will output at height 0.0002 and gradient -0.54 ... draw a box for your system of lenses, draw the input and output lines at the appropriate ends of the box.