5 Slit Experimen: Probability Distribution Function

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The discussion revolves around deriving the probability distribution function for a five-slit experiment, expressed as P(y) = A | ∑(from n=-2 to 2) exp(ikR(n)) |^2. The goal is to show that this leads to the angle-dependent probability distribution P(theta) = |A|^2 |2 cos(2b[theta]) + 2 cos(b[theta]) + 1|^2, with b needing clarification. The participants are working through the equations by substituting values for k and R(n) and expanding the terms, resulting in a complex mix of cosine and sine functions. There is an indication that a solution may be forthcoming as they continue to manipulate the equations. The discussion highlights the complexities involved in deriving the probability distribution from the initial function.
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1. Homework Statement [/b]

given the following probability distribution function from 5 slits (y is verticle on screen)

P(y) = A | \sum(from n=-2 to 2) exp(ikR(n)) |^2

where k = 2\pi/\lambda

and R(n) ~ \approx R-nyD/L

Show that the probability distribution as a function of angle from normal is:

P(theta) = |A|^2 |2 cos(2b[theta]) + 2 cos(b[theta]) + 1|^2

What is b??

2. Relevant equation

[theta] = y/L

Complex numers, Euler's equation

3. The Attempt at a Solution

I expanded out the equation with k and R_{n} plugged in - got a bunch of cos's and i*sin's. Too much to type...
 
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A wrote out the exp(ikR(n)) 's, and crossed multiplied 5 terms by 5 terms...

may get an answer soon?
 
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