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Ray transfer matrix method, I need explanations

  1. Apr 5, 2010 #1


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    1. The problem statement, all variables and given/known data
    I can't understand my class notes nor wikipedia about this.
    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 [tex]A = {x_2 \over x_1 } \bigg|_{\theta_1 = 0} [/tex].
    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.
  2. jcsd
  3. Apr 6, 2010 #2


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    Ok, I now understand a bit better. We have to assume that the rays entering an optic system can be described by a linear approximation (that is, if the rays are close enough to the optical axis).
    In that case we can write [tex]\begin{bmatrix} x_2 \\ \theta _2 \end{bmatrix}= \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} x_1 \\ \theta _1 \end{bmatrix}[/tex].
    Thus [tex]x_2=Ax_1+B \theta _1[/tex] and [tex]\theta _2=Cx_1 +D \theta _1[/tex].
    Wikipedia's notation means that [tex]A=\frac{x_2- B \theta _1}{x_1}[/tex] evaluated in [tex]\theta _1=0[/tex] (because of the paraxial rays approximation) and we reach [tex]A=\frac{x_2}{x_1}[/tex].
    I have no problem understanding how to get A, B and C. But D really is an obstacle.

    We have that [tex]D=\frac{\theta _2 - Cx_2}{\theta _1}[/tex]. According to wikipedia, if we set [tex]x_1 =0[/tex], we reach [tex]D=\frac{\theta _2}{\theta _1}[/tex]. Why is it so? If they meant [tex]x_2=0[/tex] I would understand, but [tex]x_1[/tex]?
    I would appreciate an explanation about how to reach D.
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