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Young's double slit experiment (prob density)

  • Thread starter t_n_p
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1. Homework Statement

http://img141.imageshack.us/img141/1395/40941671kx8.jpg [Broken]
http://img141.imageshack.us/img141/982/82443157pt9.jpg [Broken]
2. Homework Equations

http://img85.imageshack.us/img85/4523/60192566ie7.jpg [Broken]

3. The Attempt at a Solution

I'm not sure where to start with this one. I've searched through textbooks and the internet and have not found anything that helps me remotely show the Pr density at D. Would appreciate it if someone could give me a start and help guide me through.

Thanks!
 
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Answers and Replies

Hootenanny
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One could start by finding the difference in path lengths between the 1D and 2D
 
595
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I can't see how. Both paths seem to travel the same length in the first two sections and differ in the third section. I can't see how you can quantitatively evaulate distance though..
 
595
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bump?
 
595
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up to the top.

I still need help on this one!
 
Hootenanny
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Hi t_n_p,

Sorry I completely missed your post, you should have PM'd me. After re-reading the question, it is a lot simpler than it first seems. Notice that the second wavefunction is given in terms of the phase difference, in other words, you are already given the difference in phase between the two waves so there no need to calculate the difference in path.

All you need to do is superimpose the two wavefunctions and then find the probability density.
 
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Sorry didn't want to bug you via PM!

You say superimpose, so I want to add wavefunction 1 to wavefunction 2? I don't really understand where to go from here on...
 
Hootenanny
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Sorry didn't want to bug you via PM!

You say superimpose, so I want to add wavefunction 1 to wavefunction 2? I don't really understand where to go from here on...
Correct, so

[tex]\psi_1+\psi_2 = A+Ae^{i\phi} = A\left(1+e^{i\phi}\right)[/tex]

And,

[tex]P = \left(\psi_1+\psi_2\right)\overline{\left(\psi_1+\psi_2\right)}[/tex]

(multiplication by the complex conjugate).
 
595
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Is that P, supposed to be "roh" the symbol for probabilty density?
 
Hootenanny
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what's with that bar over the second bracketed term?
Anyhow, where does the cos come in?

http://img148.imageshack.us/img148/5425/75948148ho4.jpg [Broken]
 
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Hootenanny
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what's with that bar over the second bracketed term?
The bar represents the complex conjugate of the bracket, as I said previously.
Anyhow, where does the cos come in?
You can write the solution in terms of real cosine as opposed to complex exponentials.
http://img148.imageshack.us/img148/5425/75948148ho4.jpg [Broken]
[/URL]
No, that isn't correct you multiply the original wavefunction by the complex conjugate.
 
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my bad, somehow missed that...

complex conjugate = A - Ae^(-iφ)?

then
http://img353.imageshack.us/img353/6237/29152244fb6.jpg [Broken]

my gut feeling tells me I'm wrong because I can't see how I can extract any form of those 2 relevant equations in the original post.
 
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Hootenanny
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complex conjugate = A - Ae^(-iφ)?
Not quite, complex conjugation means that you only reverse the sign of the imaginary part,

[tex]\overline{\psi_1+\psi_2} = A\left(1+e^{-i\varphi}\right)[/tex]
 
595
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Of course, I should have known.

So using that conjugate above and expanding gives me..
http://img353.imageshack.us/img353/2008/36482179rk6.jpg [Broken]

Now I can see how I can covert the middle term into a cos term to give me the equation below (using the relevant formula given in the original post), but I'm unsure how to proceed with the last term..
http://img353.imageshack.us/img353/1821/38229117kb9.jpg [Broken]
 
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Hootenanny
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Correct. It may be useful to note that,

[tex]e^{-a}\cdot e^a = e^{-a+a} = e^0 = 1 \hspace{1cm}\forall a[/tex]
 
595
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lol, sometimes I look past the easy things....

my only issue now, is that I have
http://img89.imageshack.us/img89/9929/60512908gp9.jpg [Broken]

and the formula states
http://img160.imageshack.us/img160/1535/94298372xw7.jpg [Broken]

the only difference having my cos term as cos (φ) and the formula stating cos (2φ). Is it possible to simply halve only the 2φ term?

Also I noticed that A is an absolute value, so should my values of A that appear throughout my working also appear with modulus signs?
 
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Hootenanny
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HINT:

[tex]\cos^2\theta = \frac{1}{2}\left(1+\cos\left(2\theta\right)\right) \Rightarrow \frac{1}{2}\left(1+\cos\left(\theta\right)\right) = \cos^2\left(\frac{\theta}{2}\right)[/tex]
 
595
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Got it! Thanks a WHOLE lot!

There's another follow on question..
Show that interference maxima is given by
http://img361.imageshack.us/img361/8347/72975778fp6.jpg [Broken]

Ignoring part c) for the time being, how exactly is pr density related to the interference maxima equation?
 
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595
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hmmm, still don't get it. :confused:
 
Hootenanny
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how the prob density equation just found is related to the interference equation.
 
Hootenanny
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how the prob density equation just found is related to the interference equation.
There's no need to relate the probability density to the interference pattern, the question simply asks you to derive the fringe separation, which can be done without using the probability density.
 
595
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hmm? It says "Using the results obtained in (a) [The pr density part], show that the interference maxima are given by..."
 

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