# Homework Help: Young's double slit experiment (prob density)

1. Apr 4, 2008

### t_n_p

1. The problem statement, all variables and given/known data

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

Last edited by a moderator: May 3, 2017
2. Apr 4, 2008

### Hootenanny

Staff Emeritus
One could start by finding the difference in path lengths between the 1D and 2D

3. Apr 4, 2008

### t_n_p

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..

4. Apr 5, 2008

### t_n_p

bump?

5. Apr 11, 2008

### t_n_p

up to the top.

I still need help on this one!

6. Apr 11, 2008

### Hootenanny

Staff Emeritus
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.

Last edited: Apr 11, 2008
7. Apr 13, 2008

### t_n_p

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...

8. Apr 13, 2008

### Hootenanny

Staff Emeritus
Correct, so

$$\psi_1+\psi_2 = A+Ae^{i\phi} = A\left(1+e^{i\phi}\right)$$

And,

$$P = \left(\psi_1+\psi_2\right)\overline{\left(\psi_1+\psi_2\right)}$$

(multiplication by the complex conjugate).

9. Apr 13, 2008

### t_n_p

Is that P, supposed to be "roh" the symbol for probabilty density?

10. Apr 13, 2008

### Hootenanny

Staff Emeritus
Yes, P is the probability density.

11. Apr 13, 2008

### t_n_p

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]

Last edited by a moderator: May 3, 2017
12. Apr 13, 2008

### Hootenanny

Staff Emeritus
The bar represents the complex conjugate of the bracket, as I said previously.
You can write the solution in terms of real cosine as opposed to complex exponentials.
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No, that isn't correct you multiply the original wavefunction by the complex conjugate.

Last edited by a moderator: May 3, 2017
13. Apr 13, 2008

### t_n_p

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.

Last edited by a moderator: May 3, 2017
14. Apr 13, 2008

### Hootenanny

Staff Emeritus
Not quite, complex conjugation means that you only reverse the sign of the imaginary part,

$$\overline{\psi_1+\psi_2} = A\left(1+e^{-i\varphi}\right)$$

15. Apr 13, 2008

### t_n_p

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]

Last edited by a moderator: May 3, 2017
16. Apr 13, 2008

### Hootenanny

Staff Emeritus
Correct. It may be useful to note that,

$$e^{-a}\cdot e^a = e^{-a+a} = e^0 = 1 \hspace{1cm}\forall a$$

17. Apr 13, 2008

### t_n_p

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?

Last edited by a moderator: May 3, 2017
18. Apr 13, 2008

### Hootenanny

Staff Emeritus
HINT:

$$\cos^2\theta = \frac{1}{2}\left(1+\cos\left(2\theta\right)\right) \Rightarrow \frac{1}{2}\left(1+\cos\left(\theta\right)\right) = \cos^2\left(\frac{\theta}{2}\right)$$

19. Apr 13, 2008

### t_n_p

Got it! Thanks a WHOLE lot!

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?

Last edited by a moderator: May 3, 2017
20. Apr 13, 2008

### Hootenanny

Staff Emeritus
21. Apr 13, 2008

### t_n_p

hmmm, still don't get it.

22. Apr 13, 2008

### Hootenanny

Staff Emeritus
What specifically don't you understand?

Last edited: Apr 13, 2008
23. Apr 13, 2008

### t_n_p

how the prob density equation just found is related to the interference equation.

24. Apr 13, 2008

### Hootenanny

Staff Emeritus
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.

25. Apr 13, 2008

### t_n_p

hmm? It says "Using the results obtained in (a) [The pr density part], show that the interference maxima are given by..."