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Math Help - light traveling in minimal time

  1. #1
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    light traveling in minimal time

    Iíve been working on this task for a while but i donít know where to start.
    Anyone that can give me any tips on how to solve it?

    Light travels in such a way that it requires the minimum possible time to get from one point to another. A ray of light from C reflects of a plane mirror AB at X and then passes through D. (see figure) Show that the rays CX and XD make equal angles with the normal to AB at X. (Remark: you may wish to give a proof based on elementary geometry without using any calculus, or you can minimize the travel time on CXD.)

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  2. #2
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    i have started something, dont know if its right.

    I know that

    sin(θ1) = L1/CX
    CX = L1/sinθ

    sin(θ2) = L2/XD
    XD = L2/sin(θ2)

    CXD = (L1/sin(θ1)) + (L2/sin(θ2))

    hmm and now im stuck again
    Last edited by 93mickor; October 12th 2006 at 04:12 PM.
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  3. #3
    MHF Contributor Quick's Avatar
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    A simple approach

    Warning: I'm just learning geometry, so some of this may be wrong (that's for someone else to decide)...

    Make a point on line AB so that it is directly under point C, label it E

    Now continue ray CX past line AB

    Now place a point on ray CX so that it is directly under point D, label it F

    Make a point on line AB so that it is directly above point F, label it G

    Now angles AXC and BXF are vertical, thus congruent.

    This means that triangles CXE and FXG are similar

    Now when the line hits the mirror, it's path is reflected across the line AB

    Thus, reflect triangle FXG across the line (you'll get triangle DGX)

    Triangles DGX and CXE are still similar

    Thus angles BXD and AXC are congruent.
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  4. #4
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    Thank you for a fast answer.

    Your solution seems to be correct.
    Now I only need to formulate an answer, to show that the relation between the triangels proves that the shortest way between CXD is when angles BXD and AXC are congruent.

    Shouldnít bee to hard though if I assume that the fastest way between two points is the straight way. This is the case if the hypotenuse of triangel FXG and GXD is the same.

    If anyone has something further to add Iím glad to hear.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by 93mickor View Post
    Thank you for a fast answer.

    Your solution seems to be correct.
    Now I only need to formulate an answer, to show that the relation between the triangels proves that the shortest way between CXD is when angles BXD and AXC are congruent.

    Shouldnít bee to hard though if I assume that the fastest way between two points is the straight way. This is the case if the hypotenuse of triangel FXG and GXD is the same.

    If anyone has something further to add Iím glad to hear.
    The path taken is a stationary point of the time for a photon to travel
    from the source to the detector. The direct path and the reflected path
    are both stationary in the sense that the travel time along these paths
    is less than that along any other path close to them.

    Of course the direct path in this case takes less time and is shorter than
    the reflected path. But they are both local minima for travel time.

    RonL
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  6. #6
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    Quote Originally Posted by 93mickor View Post
    Iíve been working on this task for a while but i donít know where to start.
    ... you may wish to give a proof based on elementary geometry without using any calculus, ...
    Hi,

    if you only have the points C and D and the plane AB then you find the point X by reflecting D at AB to D'. Connect the points C and D'. The intersection of CD' and AB gives point X. (I've attached your modified diagram to show what I mean).

    You now are dealing with isoscele triangles so you can easily proof the values of the angles.

    EB
    Attached Thumbnails Attached Thumbnails light traveling in minimal time-spiegl_licht.gif  
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