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The left end of a long glass rod 6.00cm in diameter has a convex

  1. Jun 30, 2012 #1
    The left end of a long glass rod 6.00cm in diameter has a convex hemispherical surface 3.00cm in radius. The refractive index of the glass is 1.60. Distances are measured from the vertex of the hemispherical surface.

    A.)Determine the position of the image if an object is placed in air on the axis of the rod at the infinitely far distance to the left of the vertex of the curved end.

    I did not know where to start here so I proceeded to B.

    B.)Determine the position of the image if an object is placed in air on the axis of the rod at the distance of 12.00cm to the left of the vertex of the curved end.

    Here I took the equation (n1/s)+(n2/s')=(n2-n1)/R and solved for s'=n2Rs/((n2-n1)-R) and plugged in 12.00cm in for s and solved to get 8.00cm. I then accidentally hit that answer for A.) and it was right. The correct answer for this part turns out to be 13.7cm, but I do not know why either is that way.

    Thanks, Stephen





    3. The attempt at a solution
     
  2. jcsd
  3. Jun 30, 2012 #2

    Simon Bridge

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    Re: Optics:Lenses

    What is it about the answer you don't understand? Can you think of any reason it shouldn't be that way?
    BTW: I'd have used the matrix form for this.
     
  4. Jun 30, 2012 #3
    Re: Optics:Lenses

    I can't think of any reason that part B.) shouldn't be 12, but it is 13.7cm, and I have no idea how A.) is 12. I do not know the matrix form
     
  5. Jun 30, 2012 #4

    Simon Bridge

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    Re: Optics:Lenses

    What you are doing is called Par-axial geometric optics ... look through that handout for "ray transfer matrix".

    At x=0 you have a spherical interface between n=1 (x<0) and n=1.6 (x>0) with R=3.0cm ... we don't care about the situation for |y|>3cm... in fact, we are only using the bit of the system close to the axis (par-axial).

    For passage through a spherical surface, the transfer matrix is:
    [tex]M=\left (
    \begin{array}{cc}
    1 & 0\\
    \frac{n_1-n_2}{R} & 1
    \end{array}\right )[/tex]

    I'm not sure there is a special advantage to this representation for what you need to get. I think you aught to be able to figure it from a ray diagram. The matrix approach is invaluable for systems that are tricky to think about.


    Part A wants to know about the situation for parallel rays... what would you expect to happen for this situation?

    (aside: I take it you have been given answers of 12cm and 13.7cm respectively?)
     
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