Young's double slit experiment (prob density)

[t_n_p]

Homework Statement

http://img141.imageshack.us/img141/1395/40941671kx8.jpg
http://img141.imageshack.us/img141/982/82443157pt9.jpg

Homework Equations

http://img85.imageshack.us/img85/4523/60192566ie7.jpg

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!

 

[Hootenanny]

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

 

[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..

 

[t_n_p]

bump?

 

[t_n_p]

up to the top.

I still need help on this one!

 

[Hootenanny]

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.

 

[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...

 

[Hootenanny]

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

\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).

 

[t_n_p]

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

 

[Hootenanny]

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

Yes, P is the probability density.

 

[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

 

[Hootenanny]

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

[/URL]
No, that isn't correct you multiply the original wavefunction by the complex conjugate.

 

[t_n_p]

my bad, somehow missed that...

complex conjugate = A - Ae^(-iφ)?

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

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.

 

[Hootenanny]

complex conjugate = A - Ae^(-iφ)?

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)

 

[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

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

 

[Hootenanny]

Correct. It may be useful to note that,

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

 

[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

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

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?

 

[Hootenanny]

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)

 

[t_n_p]

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

Ignoring part c) for the time being, how exactly is pr density related to the interference maxima equation?

 

[Hootenanny]

Check out the slit interference pages at hyperphysics: http://hyperphysics.phy-astr.gsu.edu/Hbase/phyopt/slits.html

 

[t_n_p]

hmmm, still don't get it.

 

[Hootenanny]

hmmm, still don't get it.

What specifically don't you understand?

 

[t_n_p]

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

 

[Hootenanny]

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.

 

[t_n_p]

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

 

[Hootenanny]

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

Okay, how does the maxima relate to \rho? What is a maxima?

 

[t_n_p]

Okay, how does the maxima relate to \rho? What is a maxima?

So derive \rho in terms of \psi and set to zero?
What about A (imaginary number?)

 

[t_n_p]

Also, where would lambda come from?

 

[Hootenanny]

So derive \rho in terms of \psi and set to zero?

Who would one set \rho to zero? Wouldn't a maxima occur when \rho is greatest?

What about A (imaginary number?)

What is the maximum value of \rho?

 

[t_n_p]

Who would one set \rho to zero? Wouldn't a maxima occur when \rho is greatest?

What is the maximum value of \rho?

The maximum value of \rho is 1, or at least that is what I think. But where to from there?

 

[Hootenanny]

The maximum value of \rho is 1, or at least that is what I think.

No it isn't

 

[t_n_p]

No it isn't

infinity?

 

[Hootenanny]

What is the maximum value of \cos^2\theta?

 

[t_n_p]

What is the maximum value of \cos^2\theta?

1!

 

[Hootenanny]

1!

Correct, therefore the maximum value of \rho is...?

 

[t_n_p]

4?

 

[Hootenanny]

4?

You're getting closer, look at the equation.

 

[t_n_p]

4|a|²?

Don't know how it helps though..

 

[Hootenanny]

4|a|²?

Don't know how it helps though..

Correct ! So a maxima occurs when,

\rho = 4|A^2| = 4|A^2|\cos^2\left(\frac{\varphi}{2}\right)

 

[t_n_p]

But how do I get anything remotely looking like dsin(θ) = mλ?

 

[Hootenanny]

But how do I get anything remotely looking like dsin(θ) = mλ?

How do you think your going to do it? What is the defintion of \varphi? How does that relate to the wavelength? How does it relate to the maximum value of \rho?

Come on t_n_p, I'm not going to walk you through the whole question, you're going to have to think for yourself at some point.

 

[t_n_p]

\varphi is the wavefunction and relates to the wavelength in some way I don't know. Probability density, roh is the wavefunction multiplied by its complex conjugate.

if \rho max is 4|A|², max of wavefunction is 16|A|^4?

 

[Hootenanny]

\varphi is the wavefunction and relates to the wavelength in some way I don't know. Probability density, roh is the wavefunction multiplied by its complex conjugate.

No, \varphi is not the wavefunction. Read the question.

 

[t_n_p]

ok, phase difference = mλ where where m = 0, 1, 2, 3, 4...

can't find anything about the phase difference/prob density relationship though..

 

[Hootenanny]

Yes, \varphi is the phase difference. To add a little more explanation; as I said previously, a maxima occurs when \rho is maximal, or when,

\cos\left(\frac{\varphi}{2}\right) = 1 \Rightarrow \frac{\varphi}{2} = n\pi \hspace{1cm}n\in\mathhbb{Z}

So a maxima occurs when the phase difference between the two waves is,

\varphi = 2n\pi \hspace{1cm}n\in\mathbb{Z}

which corresponds to a complete wavelength, which is the expression you stated above. Now, this is the RHS of the equation we are attempting to derive, so we're half way there. Next, you need to work out how the relate the angle of the maxima to the wavelength. To do this, try working out the path difference between the two waves, notice that their paths are parallel until they interfere.