(adsbygoogle = window.adsbygoogle || []).push({}); 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

[tex]

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

[/tex]

with

[tex]

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

[/tex]

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

[tex]

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

[/tex]

where [tex]y[/tex] is the distance from the point on the screen you're looking at to the center between the slits, projected onto the screen, and [tex]s[/tex] 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

[tex]

\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}},

[/tex]

which, for small angles, approximates to

[tex]

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

[/tex]

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

[tex]\frac{(\sin{\beta})^2}{\beta^2} = 1.[/tex]

I've tried filling in the expression for [tex]\beta[/tex] from above, and I also tried writing it just as I wrote the expression for [tex]\alpha[/tex] for the cosine, but it doesn't seem to get me anywhere. It makes me doubt, also, what I did with [tex]\alpha[/tex]. 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?

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# Homework Help: Deriving the formula for double slit diffraction as in Young from Fraunhofer

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