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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 $e^{ikz}$ givesHere $I_{\mathrm{inc}}$ 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 toThis 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 $z=0.0200\mathrm{m}$, $N_D\simeq 12.6382$, $N_a\simeq 3.15956$ and the edge-to-centre wave phase change is $9.92604$ radians. At $z=2.00\mathrm{m}$, $N_D\simeq 0.126382$, $N_a\simeq 0.0315956$ 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.

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 $1.14812I_{\mathrm{inc}}$, with brighter off-axis maxima. At 2 m it is about $0.126272I_{\mathrm{inc}}$ and is the peak. The dashed Fraunhofer diffraction curve has its own central normalization; its first zeros are at $x=\pm \lambda z/D=\pm 3.165\mathrm{mm}$. 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 givesThis 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.

For the free parabolic wave equation $E_x=i{E}_{zz}/(2k)$, input $E(0,z)=e^{-A{z}^2}$ with $\mathrm{Re}A>0$ propagates asThe branch is continuous from $q(0{)}^{-1/2}=1$. One-dimensional transverse Fresnel propagation or two Gaussian integrals prove the formula. With $A=D^{-2}+ik/(2F)$, the squared envelope is $|q{|}^{-1}\exp [-2z^2/(D^2|q{|}^2)]$. The amplitude power is $-1/2$ because there is only one transverse coordinate. A two-transverse-coordinate Gaussian beam instead has power $-1$ when both coordinates share the same $q$.

PutIn free space, the parabolic wave equation is $E_x=i{E}_{zz}/(2k)$. Under the Fourier transform convention $\hat{E}(p)=\int E(z)e^{-ipz}dz$, it becomesThe Gaussian integral is legitimate because $\mathrm{Re}A=1/D^2>0$. Invert the transform after multiplying by $e^{-i{p}^{2}x/(2k)}$. A second Gaussian integral, or equivalently one-dimensional transverse Fresnel propagation, givesChoose the square-root branch continuously from $q(0{)}^{-1/2}=1$; for real $x\ge 0$ there is no zero of $q$. This gives the correct incident field at $x=0$, and direct differentiation verifies the free parabolic wave equation.

For clarity, the squared envelope magnitude isThe Gaussian beam with one transverse coordinate remains Gaussian, with one-transverse-coordinate amplitude factor $q^{-1/2}$, not the $q^{-1}$ of a beam with two transverse coordinates. The negative initial quadratic phase produces focusing for $F>0$; diffraction prevents a singularity at $x=F$. These expressions describe the paraxial approximation to free propagation, rather than an exact unrestricted Helmholtz equation beam.

Use the time convention $e^{-i\omega t}$. The free-space Helmholtz equation, with $\psi =e^{ikx}E$, gives the exact reduced equationFor a forward plane wave at angle $\alpha$, $E_x=O(k{\alpha }^{2}E)$ and $E_{zz}=O(k^2{\alpha }^{2}E)$, whereas $E_{xx}=O(k^2{\alpha }^{4}E)$. Thus the paraxial approximation neglects $E_{xx}$ relative to $2ik{E}_x$. The reduced field satisfies the parabolic wave equationWith the Fourier transform convention used below and its inverse, this becomes an ordinary differential equation for each transverse wavenumber:Consequently the initial-value solution isEquivalently, evaluating the oscillatory Gaussian integral gives the one-dimensional transverse Fresnel propagation formulaThe square-root branch has $\sqrt{1/i}=e^{-i\pi /4}$. The formula is an oscillatory integral for general data, or the Fresnel propagator acting on square-integrable functions. It approaches the initial field as $x\downarrow 0$. A plane wave has transverse wavenumber $\nu =k\sin \alpha$; the reduced axial phase $-{\nu }^{2}x/(2k)$ agrees with $k(\cos \alpha -1)x$ through order ${\alpha }^2$. This also checks the sign. The paraxial approximation applies to $|\nu |\ll k$.