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Interpreting a system matrix (optics)

  1. Dec 3, 2016 #1
    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?


    Thank you very much in advance for your suggestions.


    Julien.
     

    Attached Files:

  2. jcsd
  3. Dec 3, 2016 #2
    What about that:

    ##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

    Does that make sense?
     
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