Deriving the formula for double slit diffraction as in Young from Fraunhofer

1. Aug 24, 2008

NanakiXIII

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

I apologize for the blurriness in my title, I couldn't find anything better to fit within the length limit. The problem I'm stuck with, paraphrased, is to derive the formula for the diffraction pattern of a double slit, as found in the Young experiment, from the Fraunhofer formula for the same pattern. That is, you start with

$$I = 4 I_0 \frac{(\sin{\beta})^2}{\beta^2} (\cos{\alpha})^2$$

with

$$\alpha = \frac{k a}{2} \sin{\theta}; \beta = \frac{k a}{2} \sin{\theta}$$

where $$a$$ and $$b$$ are the distance between the slits and the width of the slits, respectively, and $$\theta$$ is the angle to the normal of the plane of the slits, and you have to prove that is identical to

$$I = 4 I_0 (\cos{\frac{y a \pi}{s \lambda}})^2$$

where $$y$$ is the distance from the point on the screen you're looking at to the center between the slits, projected onto the screen, and $$s$$ is the distance from the slits to the screen.

I hope that makes sense. I'm trying to be clear, but it's not so easy without images. Perhaps it would be best to hope these are the equations generally used for these situations.

2. Relevant equations

The relevant equation are really in the problem itself.

3. The attempt at a solution

I can in part see that the cosines in both equations are the same, because

$$\alpha = \frac{k a}{2} \sin{\theta} = \frac{2 \pi a}{2 \lambda} \sin{\theta} = \frac{2 \pi a}{2 \lambda} \frac{y}{\sqrt{y^2 + s^2}},$$

which, for small angles, approximates to

$$\alpha = \frac{\pi a y}{\lambda s}.$$

I'm guessing that's the way to handle the cosine. If it is, then that means that

$$\frac{(\sin{\beta})^2}{\beta^2} = 1.$$

I've tried filling in the expression for $$\beta$$ from above, and I also tried writing it just as I wrote the expression for $$\alpha$$ for the cosine, but it doesn't seem to get me anywhere. It makes me doubt, also, what I did with $$\alpha$$. Either I did that wrong, or I'm probably missing something small, but I'm stumped right now. Does anyone see the mistake I made or the direction in which I need to be pointed?

2. Aug 24, 2008

Staff: Mentor

You have a typo here. That second expression should have b instead of a.
Looks OK.

Right.
Hint: What happens when the slit width goes to zero?

3. Aug 24, 2008

NanakiXIII

Yes, that's a typo, sorry. I seem to have misplaced my Edit button, though.

When the slit width goes to zero, $$\beta$$ goes to zero and the Sinc function goes to one. Why would the slit width go to zero, though?

4. Aug 24, 2008

Staff: Mentor

In real life it doesn't (obviously). But to recover the Young diffraction pattern (which ignores single slit diffraction due to the width of the slits), one takes the limit of the Fraunhofer formula as slit width goes to zero. (Although your paraphrasing of the problem asks to show that the two formulas are identical--that doesn't make sense to me.)

5. Aug 24, 2008

NanakiXIII

Hmm, I see, I didn't know that. And indeed it wasn't clear that this was to be done from the question. If I am to translate the question more directly, it says

"Consider the Fraunhofer diffraction pattern for two slits with slit width b and distance a.

a) [Irrelevant]
b) The two-slit diffraction pattern, measured on a screen at distance s from the slits, is also described by the formula from the Young experiment:

[Formula from the Young experiment]

Derive this formula from the formula for the Fraunhofer diffraction pattern."

Apparently I was supposed to just know that you have to take the limit for b->0 if you want to get the Young formula, I just don't remember learning about that anywhere. Thanks a lot for pointing that out, that was the missing link.