Double slit diffraction derivation The task is to derive the intensity pattern of a double slit diffraction, where light travels along the $z$-axis, the point of interest lies on the $x$-$z$ plane, the slits are wide (in $z$-direction) and infinite in extent in the $y$-direction.
But how can this be derived by Fraunhofer diffraction, when one of the assumptions of Fraunhofer is, that the aperture is much smaller in size than the distance of aperture-POI?
Because when using the Fraunhofer formula:
$$U(P)\propto e^{ik(s_0+d_0)} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} t(x,y) e^{-ik(Xx+Yy)}\mathrm dx\mathrm dy$$
where $d_0, s_0$ are the distances of the source (here infinity) to the aperture and the distance of aperture - POI, and $X = x_s/s_0 + x_p/d_0$ and $Y = y_s/s_0 + y_p/d_0$. The problem comes from $y_p = 0, y_s=0 \implies Y=0$, so an integration over $y$ from $-\infty$ to $+\infty$ would diverge. How can I deal with this?
 A: Indeed, an infinite slit would give infinite illumination.
To see what happen, we can give the slit a finite height $H$ and integrate from $-H / 2$ to $H / 2$:
$s=\int\limits_{-H/2}^{H/2}{{{e}^{jkYy}}dy=}H\text{sinc}\left( \frac{kAY}{2} \right)$ with the sinc function defined by $\text{sinc}(u)=\frac{\sin (u)}{u}$
https://en.wikipedia.org/wiki/Sinc_function
The factor $H$ must be grouped in the intensity in the center of the figure: $I=cst{{s}^{2}}=\underbrace{cst{{H}^{2}}}_{I\text{center}}\text{sin}{{\text{c}}^{\text{2}}}\left( \frac{kHY}{2} \right)={{I}_{\text{center}}}\text{sin}{{\text{c}}^{\text{2}}}\left( \frac{kHY}{2} \right)$
You then compare the illumination in the center and the illumination at any point on the screen. We have the first zero minimum:  $\frac{kHY}{2}=\pm \pi $ give $Y=\pm \frac{2\pi }{kH}=\pm \frac{\lambda }{H}$ decrease when $H$ increase.
Mathematically, at the limit of $H$ infinite, we have a Dirac: infinite amplitude over a null width : $\underset{H\to +\infty }{\mathop{\lim }}\,H\text{sinc}(Hx)=\delta (x)$
Sorry for my English !
A: T(x, y) is t(x) for an infinite slit in the y direction. You can drop the integration over y since the problem is independent of y.
