Formulating the Exact Equation for the Double Slit Experiment in Physics"

[cshum00]

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

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

Here are some relevant formulas I came up using the geometry of the problem:
01) r_1 sin \theta_1 = Y - \frac{d}{2}
02) r_2 sin \theta_2 = Y + \frac{d}{2}
03) r_1 cos \theta_1 = L
04) r_2 cos \theta_2 = L
05) L tan \theta_1 = Y - \frac{d}{2}
06) L tan \theta_2 = Y + \frac{d}{2}
07) r_1^2 = L^2 + (Y - \frac{d}{2})^2
08) r_2^2 = L^2 + (Y + \frac{d}{2})^2
09) y(x=L) = aL + \frac{d}{2}
10) y(x=L) = bL - \frac{d}{2}
Note: \theta_1 is the angle between r₁ and L. Similarly, \theta_2 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
(r_2 - r_1)(r_2 + r_1) = 2dY
-Solving for Y and let r_1 - r_2 = n \lambda
Y = \frac{n \lambda}{2d}(r_2 + r_1)
-Substitute r_1 and r_2 using equation 3 and 4
Y = \frac{n \lambda}{2d}(\frac{L}{cos \theta_2} + \frac{L}{cos \theta_1})
Y = \frac{n L \lambda}{2d}(\frac{1}{cos \theta_2} + \frac{1}{cos \theta_1})
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.

 

[cshum00]

I got the answer. The way i derived was using only 3 of the equations:
r_1^2 = L^2 + (y - \frac{d}{2})^2 => r_1^2 = L^2 + y^2 - dy + d^2/4 (1)
r_2^2 = L^2 + (y + \frac{d}{2})^2 => r_1^2 = L^2 + y^2 + dy + d^2/4 (2)
r_2 - r_1 = n \lambda (3)

Subtracting equation (2) and (1)
(r_2 - r_1)(r_2 + r_1) = 2dy (4)

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

But according to equation (3)
r_2 = n \lambda + r_1 (6)

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

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

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

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

(b) L -> \infty, y \approx n \lambda L