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Optics question using ray matrices

  1. Nov 30, 2006 #1
    1. The problem statement, all variables and given/known data
    The problem is: The left end of a long glass rod 10.0 cm in diameter, with an index of refraction 1.5, is ground and polished to a convex hemispherical surface with a radius of 5.0cm. An object in the form of an arrow 2.00mm tall, at right angles to the axis of the rod, is located on the axis 25.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface.

    From this I know that R= 5cm, do = 25cm, ho=.2cm, n1 = 1, n2 = 1.5

    2. Relevant equations

    n1/n2
    -(nlarge - nsmall)/n2R



    3. The attempt at a solution (and what an attempt it is.)
    The teacher wants us to use ray matrices to figure out the answer. I figure there are 3 matrices that will need to be formed.

    Here is what I have:
    [tex]m1 = \left(\begin{array}{cc}
    1 & 25 \\
    0 & 1\\
    \end{array}
    \right)[/tex]

    [tex]m2 = \left(\begin{array}{cc}
    1 & 0 \\
    -0.067 & .67\\
    \end{array}
    \right)[/tex]

    [tex]m3 = \left(\begin{array}{cc}
    1 & di \\
    0 & 1\\
    \end{array}
    \right)[/tex]

    When I multiply the 3 together I get my final result as:

    [tex]mfinal = \left(\begin{array}{cc}
    2.675 & 2.675di + 16.75 \\
    -.067 & -.067di +.67\\
    \end{array}
    \right)[/tex]

    For some reason I am not getting the right answers in my final matrix to lign up with the image matrix. Any help would be appreciated.
     
    Last edited: Nov 30, 2006
  2. jcsd
  3. Nov 30, 2006 #2

    OlderDan

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    What order did you use for the matrices when you multiplied them?
     
  4. Nov 30, 2006 #3
    I multiplied [m1] [m2] I then took the result and multiplied it by [m3].
     
  5. Nov 30, 2006 #4

    OlderDan

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    Try [m3][m2][m1] and think about why you might want it that way
     
  6. Nov 30, 2006 #5
    I did it that way and that is how I got mFinal. I don't know if my math is not right or if something else is not right.
     
  7. Nov 30, 2006 #6

    OlderDan

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    I think you need to check it. Do you know that [m2][m1] is not the same as [m1][m2]?? I see no way you can have di in the bottom row of the final matrix. I also see no way to get any negatives.
     
  8. Nov 30, 2006 #7
    I switched the result of [m2][m1] with where [m3] was suppose to be. Thanks for you help, so here is what I have for my mFinal:
    [tex]mfinal = \left(\begin{array}{cc}
    2.675 + di & 16.75 -.67di\\
    -.067 & .67\\
    \end{array} \right)[/tex]

    the negetive I have from my m2 matrix, I posted it wrong orignaly.
     
    Last edited: Nov 30, 2006
  9. Nov 30, 2006 #8

    OlderDan

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    The only way I can see you getting a 16.75 anywhere is from multiplying [m1][m2] in that order instead of [m2][m1]. Then [m3][m1][m2] gives your new bottom row but not your top row.

    You still need to compute [m3][m2][m1]. You can do it as

    {[m3][m2]}[m1] or [m3]{[m2][m1]}

    Matrix multiplication is associative, but it is not commutative.

    I think you are correct about that sign in [m2]. I assume you changed that back in your original post.
     
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