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Help Formulating Eqation

  1. Jul 4, 2009 #1
    I got this from physics' double slit experiment. The way it is done on the physics book is by approximation. Since i am interested in the mathematics behind it, i have been trying to formulate an equation which gives me the exact value.

    http://img183.imageshack.us/img183/60/figure.png [Broken]

    Like the double slit experiment i want to find the difference r1 - r2.
    The approximation given on the physics books is:
    [itex]r_1 - r_2 = n \lambda = \frac{d Y}{L}[/itex]
    Therefore [itex]y = \frac{n L \lambda}{d}[/itex]

    Here are some relevant formulas I came up using the geometry of the problem:
    01) [itex]r_1 sin \theta_1 = Y - \frac{d}{2}[/itex]
    02) [itex]r_2 sin \theta_2 = Y + \frac{d}{2}[/itex]
    03) [itex]r_1 cos \theta_1 = L[/itex]
    04) [itex]r_2 cos \theta_2 = L[/itex]
    05) [itex]L tan \theta_1 = Y - \frac{d}{2}[/itex]
    06) [itex]L tan \theta_2 = Y + \frac{d}{2}[/itex]
    07) [itex]r_1^2 = L^2 + (Y - \frac{d}{2})^2[/itex]
    08) [itex]r_2^2 = L^2 + (Y + \frac{d}{2})^2[/itex]
    09) [itex]y(x=L) = aL + \frac{d}{2}[/itex]
    10) [itex]y(x=L) = bL - \frac{d}{2}[/itex]
    Note: [itex]\theta_1[/itex] is the angle between r1 and L. Similarly, [itex]\theta_2[/itex] is the angle between r2 and L

    I have been trying to combine these formulas and get something nice and simple but i never was able to get rid of the variables or at least reduce it into one single variable.

    Here is my best attempt so far:
    -Subtracting equation 8 and 7
    [itex](r_2 - r_1)(r_2 + r_1) = 2dY[/itex]
    -Solving for Y and let [itex]r_1 - r_2 = n \lambda[/itex]
    [itex]Y = \frac{n \lambda}{2d}(r_2 + r_1)[/itex]
    -Substitute r_1 and r_2 using equation 3 and 4
    [itex]Y = \frac{n \lambda}{2d}(\frac{L}{cos \theta_2} + \frac{L}{cos \theta_1})[/itex]
    [itex]Y = \frac{n L \lambda}{2d}(\frac{1}{cos \theta_2} + \frac{1}{cos \theta_1})[/itex]
    Which still has 2 variables.

    I know that the hint lies on something like the parabola. This problem, just like the parabola, has 2 fixed points, and 2 lines that goes along a curve.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 30, 2009 #2
    I got the answer. The way i derived was using only 3 of the equations:
    [itex]r_1^2 = L^2 + (y - \frac{d}{2})^2 => r_1^2 = L^2 + y^2 - dy + d^2/4[/itex] (1)
    [itex]r_2^2 = L^2 + (y + \frac{d}{2})^2 => r_1^2 = L^2 + y^2 + dy + d^2/4[/itex] (2)
    [itex]r_2 - r_1 = n \lambda[/itex] (3)

    Subtracting equation (2) and (1)
    [itex](r_2 - r_1)(r_2 + r_1) = 2dy[/itex] (4)

    Substitute equation (3) into (4)
    [itex]n \lambda(r_2 + r_1) = 2dy[/itex] (5)

    But according to equation (3)
    [itex]r_2 = n \lambda + r_1[/itex] (6)

    Substitute equation (6) into (5)
    [itex]n \lambda(n \lambda + 2r_1) = 2dy[/itex] (7)

    Solve for [itex]r_1[/itex] in equation (7)
    [itex]n \lambda + 2r_1 = (2dy)/(n \lambda)[/itex]
    [itex]2r_1 = (2dy)(n \lambda) - n \lambda[/itex]
    [itex]r_1 = (2dy - n^2 \lambda^2)/(4n \lambda)[/itex]
    Square both sides
    [itex]r_1^2 = (4d^2y^2 - 4dyn^2 \lambda^2 + n^4 \lambda^4)/(4n^2 \lambda^2)[/itex]
    [itex]r_1^2 = (d^2y^2)/(n^2 \lambda^2) - dy + (n^2 \lambda^2)/(4)[/itex] (8)

    Substitute equation (1) into (8)
    [itex]L^2 + y^2 - dy + (d^2)/4 = (d^2y^2)/(n^2 \lambda^2) - dy + (n^2 \lambda^2)/4[/itex]
    Solve for y:
    [itex]L^2 + (d^2)/4 - (n^2 \lambda^2)/4 = (d^2y^2)/(n^2 \lambda^2) - y^2[/itex]
    [itex]L^2 + (d^2 - n^2 \lambda^2)/4 = y^2[(d^2)/(n^2 \lambda^2) - 1][/itex]
    [itex](4L^2 + d^2 - n^2 \lambda^2)/4 = y^2 (d^2 - n^2 \lambda^2)/(n^2 \lambda^2)[/itex]
    [itex]y = \pm \sqrt {[(4L^2 + d^2 - n^2 \lambda^2)*(n^2 \lambda^2)] / [4 * d^2 - n^2 \lambda^2]}[/itex]
    [itex]y = \pm \frac{n \lambda}{2}\sqrt {(4L^2 + d^2 - n^2 \lambda^2) / (d^2 - n^2 \lambda^2)}[/itex] !!!

    Analyzing the formula, we can say that:
    (a) [itex]L -> 0[/itex], [itex]y \approx \frac{n \lambda}{2}[/itex]

    (b) [itex]L -> \infty[/itex], [itex]y \approx n \lambda L[/itex]
     
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