= Fresnel slit propagation
{c}
{title2=$u(x,z)=e^{-i\pi/4}(\lambda z)^{-1/2}\int_{-D/2}^{D/2}e^{i\pi(x-\xi)^2/(\lambda z)}d\xi$}
Consider a uniform <plane wave> of unit <complex amplitude> transmitted by an ideal <optical slit> of full width $D$. Treat the field as invariant along the <optical slit> height: this is <one-dimensional transverse Fresnel propagation>, so the prefactor is $1/\sqrt{i\lambda z}$, rather than the two-dimensional prefactor $1/(i\lambda z)$. Removing the carrier $e^{ikz}$ gives
$$
u(x,z)=\frac{e^{-i\pi/4}}{\sqrt{\lambda z}}\int_{-D/2}^{D/2}\exp\!\left[\frac{i\pi(x-\xi)^2}{\lambda z}\right]d\xi,\qquad\frac{I(x,z)}{I_{\rm inc}}=|u(x,z)|^2.
$$
Here $I_{\rm inc}$ is incident <irradiance>. This infinitely long slit model describes a transverse cut away from the ends of a sufficiently long physical slit; its finite conserved quantity is transmitted <power> per unit <optical slit> height.
Put $t=\xi/D$ and $q=xD/(\lambda z)$. Substitution reduces the <integral> to
$$
A(q,N_D)=\int_{-1/2}^{1/2}e^{i\pi N_Dt^2-2\pi iqt}dt,\qquad\frac{I(x,z)}{I_{\rm inc}}=N_D|A(q,N_D)|^2.
$$
This form separates the pattern's shape from its physical scale. Reflection $t\mapsto-t$ proves $A(-q,N_D)=A(q,N_D)$, so the <light intensity> is an <even function> of $x$. At $N_D=0$, direct integration gives $A(q,0)=\sin(\pi q)/(\pi q)$, with <limit> one at $q=0$. Hence the <Fraunhofer diffraction> comparison is the squared <sinc function>. In this paragraph the unnormalized <sinc function> takes argument $\pi q$.
For an original example, choose <wavelength> $\lambda=633\,\mathrm{nm}$ and full width $D=0.400\,\mathrm{mm}$. At $z=0.0200\,\mathrm m$, $N_D\simeq12.6382$, $N_a\simeq3.15956$ and the edge-to-centre <wave phase> change is $9.92604$ <radians>. At $z=2.00\,\mathrm m$, $N_D\simeq0.126382$, $N_a\simeq0.0315956$ and that <wave phase> change is only $0.0992604$ <radians>. Even the nearer screen is fifty slit widths away; that geometric comparison alone misses its strong <Fresnel diffraction>.
\Image[/fresnel-slit-near-far-profiles.png]
{title=Original calculated <Fresnel diffraction> profiles of a 0.400 mm slit at 633 nm, compared with the <Fraunhofer diffraction> shape}
These are numerical predictions from the displayed scalar <integral>, using the <composite trapezoidal rule>, rather than measured profiles. Each curve is divided by its own peak <light intensity>; the two horizontal scales differ. At 2 cm the computed central <irradiance> is about $1.14812I_{\rm inc}$, with brighter off-axis maxima. At 2 m it is about $0.126272I_{\rm inc}$ and is the peak. The dashed <Fraunhofer diffraction> curve has its own central normalization; its first zeros are at $x=\pm\lambda z/D=\pm3.165\,\mathrm{mm}$. The finite-distance <Fresnel diffraction> curve remains slightly nonzero there.
Normalization hides the brightness change, but not the change in shape. The <Fourier transform> propagation multiplier has unit <complex modulus>, so <Plancherel theorem> gives
$$
\int_{\mathbb R}\frac{I(x,z)}{I_{\rm inc}}dx=\int_{-D/2}^{D/2}1\,d\xi=D.
$$
This is conservation of transmitted <power> per unit height in the ideal scalar model. An <integral> over only the displayed screen <interval> misses the tails; a peak-normalized profile is not a unit-<integral> <probability density>. Spreading and decreasing brightness together preserve the transmitted <power>.
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