What Is the Correct Transmittance Function for Two Infinitely Narrow Slits?

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The discussion revolves around finding the transmittance function T(x,y) for two infinitely narrow slits of length L, separated by a distance 2d. The initial attempt defines T(x,y) as 1 within specific bounds and 0 otherwise, but the user seeks a more refined expression suitable for Fourier transforms. They successfully derived a function for slits with width but struggle to adapt it for infinitely narrow slits. The goal is to replace the rectangular function with a delta function while ensuring it maintains a finite value instead of approaching infinity. The challenge lies in accurately representing the two-slit scenario mathematically.
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Homework Statement


I have to find the transmittance function T(x,y) of two narrow slits of length L and separated by a distance 2d.

Homework Equations


[PLAIN]http://being.publicradio.org/programs/quarks/images/youngexperiment.gif
This is what I'm talking about. In my case the x-axis runs in between the slits from the top to bottom (or bottom to top), the slits have a length L and are a distance d from the x-axis.
<br /> \mathrm{rect}(\frac{t}{H}) = \begin{cases} 0 &amp; \mbox{if } |t| &gt; \frac{H}{2} \\ 1 &amp; \mbox{if } |t| &lt; \frac{H}{2}. \\ \end{cases} <br />

The Attempt at a Solution


<br /> T(x,y)=1 \mbox{ if } -\frac{L}{2} \leq x \leq \frac{L}{2} \mbox{ and } y=|d|<br />
<br /> \mbox{Otherwise } T=0<br />
I need to put this into a nicer form, one which will be useful for Fourier transforms. I can reproduce a single slit using step functions, but I'm not sure how to get the two slits at the same time.
 
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I figured out the case when the slit has some width: T(x,y)=rect\left(\frac{x}{L}\right) \left[rect\left(\frac{y+d}{t}\right)+rect\left(\frac{y-d}{t}\right)\right] where t is the thickness. However I'm looking for the case where t is "infinitely narrow"

What I want is to replace rect\left(\frac{y-d}{t}\right) with something like \delta(y-d), however I need it to equal 1 (or something finite) instead of infinity
 
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