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Assume light travels from point a to b. Find the total time to get from a to b

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that light travels from point A to point B as shown in the figure/ Assume that the velocity of light above the boundary line is v1 and the velocity above the boundary line is v2. Find the total time T(x) to get from point A to point B, Write out the equation T'(x) = 0, replace the square roots using the sines of the angles in the figure and derive Snell's Law [tex]\frac{sinθ_1}{sinθ_2} = \frac{v_1}{v_2}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Here is the work I've done so far.

    [tex] T(x) = \frac{\sqrt{1+x^2}}{v_1} + (-\frac{\sqrt{1+(2-x)^2}}{v_2})[/tex]
    [tex] T'(x) = \frac{1}{v_1\sqrt{(1+x^2)}} - \frac{1(x-2)}{v_2(\sqrt{(1+(2-x)^2)}} = 0[/tex]
    [tex] T'(x) = \frac{1}{v_1\sqrt{(1+x)^2}} = \frac{x-2}{v_2\sqrt{1+(2-x)^2}}[/tex]

    And I know I have to get

    [tex]\frac{sinθ_1}{sinθ_2} = \frac{v_1}{v_2}[/tex]

    and I replaced the fractions with sines but the (x-2) on the right side of the equation messes up the formula, so now I'm confused.

    Also!!! Is my latex working for you guys? What did I do wrong? lol
     

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    Last edited: Dec 8, 2011
  2. jcsd
  3. Dec 8, 2011 #2

    Simon Bridge

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    You confused the parser by using bv code (all those and tags) inside the [tex] tag.
     
  4. Dec 8, 2011 #3

    Simon Bridge

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    I think you have forgotten to find x so that T(x) is a minima.
    (assuming x is the point the light hits the interface, A=(0,1) and B=(2,-1)?)
     
  5. Dec 8, 2011 #4
    Wouldn't making T'(x) = 0 be finding a minima?
     
  6. Dec 8, 2011 #5
    And what is the right way to type up this function then? haha
     
  7. Dec 8, 2011 #6
    Oh okay, so finding where the function = 0 would be finding the min. so before i use the sine functions I should solve for x?
     
  8. Dec 8, 2011 #7

    Simon Bridge

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    Sorry, misread .. you did do that step.
    You have a pic with two triangles, and the normal drawn in and the angles?

    Then, assuming A=(0,1) and B=(2,-1):
    [tex]\sin(\theta_1) = \frac{x}{\sqrt{1+x^2}}[/tex]

    you do it for [itex]\sin(\theta_2)[/itex]

    Aside:
    Sub and superscripts in LaTeX are like this: x_1^{23}would be [itex]x_1^{23}[/itex]
    For more than one character you have to enclose them in curly brackets.
     
    Last edited: Dec 8, 2011
  9. Dec 8, 2011 #8
    Also, I am soooooo stupid. There is a picture I forgot to attach. Wow, sorry anyone who was confused by this lol. The pic is up now.
     
  10. Dec 8, 2011 #9

    Simon Bridge

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    No worries, I stuffed up the sine in prev post.
    You should be able to find the other one now.

    In your last two equations, how come (x-2) instead of (2-x)? How come you get a 1 in the numerator of the first term in T' and not an x?

    In the last equation, as written, does not make sense ... presumably you mean:

    [tex]T'(x) = 0 \Rightarrow \frac{x}{v_1\sqrt{(1+x)^2}} = \frac{2-x}{v_2\sqrt{1+(2-x)^2}}[/tex]
     
    Last edited: Dec 8, 2011
  11. Dec 8, 2011 #10
    Why would

    [tex]sin(\theta_1) = \frac{x}{\sqrt{1+x^2}}[/tex]

    and not just

    [tex]sin(\theta_1) = \frac{1}{\sqrt{1+x^2}}[/tex] ?
     
  12. Dec 8, 2011 #11

    Simon Bridge

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    Look at the diagram.

    By definition: [itex]\sin \theta = \frac{O}{H}[/itex]
    The hypotenuse has length [itex]\sqrt{1+x^2}[/itex] because the opposite side has length [itex]x[/itex].

    (the angle is pointing the other way to how you are used to - [itex]1/\sqrt{1+x^2} = \cos \theta[/itex].)
     
  13. Dec 8, 2011 #12
    Ooooh ok I got it. So

    [tex]\sin(\theta_2) = \frac{2-x}{\sqrt{1+(2-x)^2}}[/tex]

    ? or would it be negative
     
  14. Dec 8, 2011 #13
    Ok no I got it it's [tex]\sin(\theta_2) = \frac{2-x}{\sqrt{1+(2-x)^2}}[/tex]

    so then I get [tex]\frac{\sin{\theta_1}}{v_1} = \frac{\sin{\theta_2}}{v_2}[/tex]
    which means we would get
    [tex]\frac{sin{\theta_1}}{sin{\theta_2}} = \frac{v_1}{v_2}[/tex]

    But now I am confused after reading it again and again...What am I exactly trying to find? The total time, right? So how do I use this formula it asked me to get to find the total time? Lol, I'm lost right now
     
  15. Dec 8, 2011 #14

    Simon Bridge

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    Weren't you supposed to derive Snell's Law?
    Isn't this what you have?

    Well done!

    An observation and a lesson:

    In general, light always takes the path of least time. This is just a specific example.

    Get the geometry right, everything else takes care of itself.
     
  16. Dec 8, 2011 #15
    Ok, so, how can I show that this will be the minimum time to travel from A to B?
     
  17. Dec 8, 2011 #16

    Simon Bridge

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    What did you find T'(x) for?
     
  18. Dec 9, 2011 #17

    HallsofIvy

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    a minimum. "Minima" is the plural of "minimum".
     
  19. Dec 9, 2011 #18
    Haha sorry, didn't mean to break the calculus rules...-.-
     
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