Young's double slit experiment (prob density)

[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 definition 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.