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Geometric optics and Fermat's principle

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A ray travels as shown in the image attached below. In this case, Fermat's principle may be written as

    ##A =\frac{n(1+ay)}{\sqrt{1+(y')^2}}##

    Where y' is dy/dx, n is the index of refraction and A is a real constant.

    The trajectory of a ray of light is given by

    ##y = -\frac{1}{a}## ## + \frac{A}{na}## ##\cosh{\frac{na}{A}(x-x0)}##

    When the ray is observed from a height y above the x axis, it seems to come from the ground at horizontal distance d from the observer (see figure). Determine d if a(x−x0) is small. Write your answer in terms of y and a.

    2. Relevant equations


    3. The attempt at a solution
    I've tried to work out the specific trajectory for the ray that grazes the x axis at xg. For x=xg, dy/dx is 0 and y=0, so at that point A = n. Although there is a sqrt involved, take positive n because n=c/v is always positive as far as I know.

    Sub that into the second equation for y, taking y=0 again, then I got xg = x0.

    So the specific trajectory of this ray would then be

    ##y = -\frac{1}{a}## ## + \frac{1}{a}## ##\cosh{a(x-xg)}##

    Really have no clue what to do from there! I did try finding the tangent to the curve to maybe find an expression for the length of the hypotenuse, but that didn't work. So I don't know.
     

    Attached Files:

  2. jcsd
  3. Oct 11, 2015 #2

    andrewkirk

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    If the eye is at coordinates (x,y) and y' represents the gradient at that point then what expression can you write for d in terms of y and y'?

    If you can do that, you just need to get rid of y'. Could the Fermat formula be of any help?
     
  4. Oct 12, 2015 #3
    Oh yeah! An expression such as y'2= d2+y2? Rearranged to make d the subject, then rearrange the fermat formula to make y' the subject and sub in to the d = expression, along with the given value of y!
     
  5. Oct 12, 2015 #4

    andrewkirk

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    Yes, but not that particular expression. ##y'## is the gradient, which is the tangent of the angle of the hypotenuse to the horizontal. What is the relation of that to ##d## and ##y##?
     
  6. Oct 13, 2015 #5
    Oh, ok... so tan(θ) = ##\frac{y}{d}##=##y'##? So in fact, d = y/y' then?

    I don't understand why not the first expression isn't right though. Because the gradient is dy/dx, and dx at the point we're looking at is d, isn't it? I thought the hypotenuse would be represented, when squared, by ##d^2 + y^2##.
    Although I do know why dy/dx is the tan of the angle of the hypotenuse to the horizontal.
     
    Last edited: Oct 13, 2015
  7. Oct 13, 2015 #6
    But then, if I use

    ##y = -\frac{1}{a}## ## + \frac{1}{a}## ##\cosh{a(x-xg)}##

    Then since I've been given that a(x-xg) is small, cosh(a(x-xg)) = 1. So I'd get that y=0 and therefore d = 0, wouldn't I?
     
  8. Oct 13, 2015 #7

    andrewkirk

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    I wouldn't use that, because it has x in it and you are looking for a formula for d in terms of only y and a. Try instead using the first formula in the OP:
    $$
    A =\frac{n(1+ay)}{\sqrt{1+(y')^2}}
    $$
     
  9. Oct 14, 2015 #8
    It does have x in it, but x disappears because I've been given that a(x-xg) is small. I know that y isn't zero! So why would I get that result?
     
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