# Homework Help: Interpreting a system matrix (optics)

1. Dec 3, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! While doing some homework for school, I realised that I still struggle to get what are the elements of an optical system matrix referring to. Here is the problem:

An optical tube with length $L=50$cm has at one end a convex lens ($D=2$) and at the other end a concave lens ($D=-2$). A mirror is placed a distance $x$ from the concave lens (outside the tube) perpendicular to the optical axis. An object is placed at a distance $d=100$cm from the convex lens (see attached picture).

For which distance $x$ can the real image of the object lay in the object plane? How big is the magnification? Is the image straight or inverted?

2. Relevant equations

I used the following convention for the matrices:

1. Translation matrix: $\begin{bmatrix} 1 & 0 \\ d & 1 \end{bmatrix}$
2. Thin lens matrix: $\begin{bmatrix} 1 & -1/f \\ 0 & 1 \end{bmatrix}$
3. Flat mirror matrix: $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

3. The attempt at a solution

So here is my attempt at solving the problem. First I determined the system matrix:

$\mathcal{M} = \begin{bmatrix} 1 & 0 \\ d & 1 \end{bmatrix} \begin{bmatrix} 1 & -1/f_1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ L & 1 \end{bmatrix} \begin{bmatrix} 1 & -1/f_2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ x & 1 \end{bmatrix} \begin{bmatrix} 1 & -1/f_2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ L & 1 \end{bmatrix} \begin{bmatrix} 1 & -1/f_1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ d & 1 \end{bmatrix}$
$= \begin{bmatrix} -1 & 8x \\ 0 & -1 \end{bmatrix}$

(I hope there is no mistake in the matrix. I've checked it many times already so it should be fine. The only doubts I have would concern the signs in the matrices when the ray comes back towards the object plane after hitting the mirror.)

Then I tried simply multiplying the matrix with an arbitrary height $h$ and angle $phi_0$ and got:

$\begin{bmatrix} \phi_f \\ h_f \end{bmatrix} = \mathcal{M} \begin{bmatrix} \phi_0 \\ h \end{bmatrix} = \begin{bmatrix} -\phi_0 + 8xh \\ -h \end{bmatrix}$

I can see that the resulting height is independent of the starting angle of the ray and moreover it is equal to $-h$. Thus my interpretation is that the magnification factor is $1$ and that the image is inverted (because of the negative sign). Is that correct?

Then I have trouble interpreting the question about $x$. Any idea I have always ends up into having $x=0$ which seems unlikely. Did I make a mistake calculating the system matrix or am I not seeing something? As I said in the introduction I am unsure on how to interprete the matrix component $M_{12} = 8x$. Is it the focal length of the whole system? Maybe it should be equal to something but I can't see what right now. The inverse of the total length times 2?

Julien.

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2. Dec 3, 2016

### JulienB

$8x = \frac{1}{f_{\mathcal{M}}} = \frac{1}{2(L + d + x)}$
$\equiv x^2 + x(L+d) - \frac{1}{4} = 0$
$\implies x= \frac{(L+d)^2}{2} = \frac{9}{8} = 112.5$cm