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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 r

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 r

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.

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

Like the double slit experiment i want to find the difference r

_{1}- r_{2}.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 r

_{1}and L. Similarly, [itex]\theta_2[/itex] is the angle between r_{2}and LI 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.

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