Ray tracing with transfer matrix method

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The discussion revolves around simulating ray propagation through a simple optical system consisting of two thin lenses with specified focal lengths and separation. The user successfully calculated the final image distance using the transfer matrix method, arriving at a distance of 20.37 mm. However, they are struggling to interpret the output rays generated by their simulation, particularly in understanding the relationship between input and output parameters. The user provides specific input values for height and angle, seeking clarification on how these translate through the system's transfer matrix. Visualizing the system with a diagram of input and output rays is suggested as a potential aid in comprehension.
mester1025
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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:
\begin{equation} 55-2.7x = 0\end{equation}
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{equation}
\begin{bmatrix} y \\ θ \end{bmatrix}
=
\begin{bmatrix} 0.0002 \\ -0.54 \end{bmatrix}

\end{equation}
where y is the input height and θ is the input angle in radian.
\begin{equation}
\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}
\end{equation}

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

I apologize for the wall of text :)
 
Mathematics news on Phys.org
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.
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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