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) $L -> \infty$, $y \approx n \lambda L$