Kirchhoff's diffraction formula

Point source

Consider a monochromatic point source at P₀, which illuminates an aperture A₁ in a screen as shown in the right figure. The intensity of the wave emitted by a point source falls off as the inverse square of the distance traveled, so the amplitude falls off as the inverse of the distance. The complex amplitude of the disturbance at a distance ${\displaystyle r}$ is given by

$${\displaystyle U(r)={\frac {ae^{ikr}}{r}},}$$ where ${\displaystyle a}$ represents the magnitude of the disturbance at the point source.

The disturbance at a spatial position P can be found by applying the Kirchhoff's integral theorem to the closed surface formed by the screen, consisted of A₁ and A₂, and a unblocked part of a sphere of radius R centered at P, A₃. The integration is performed over the areas A₁, A₂ and A₃, giving $${\displaystyle U(P)={\frac {1}{4\pi }}\left[\int _{A_{1}}+\int _{A_{2}}+\int _{A_{3}}\right]\left(U{\frac {\partial }{\partial n}}\left({\frac {e^{iks}}{s}}\right)-{\frac {e^{iks}}{s}}{\frac {\partial U}{\partial n}}\right)dS.}$$

To solve the equation, it is assumed that the values of ${\displaystyle U}$ and ${\displaystyle {\frac {\partial U}{\partial n}}}$ in the aperture area A₁ are the same as when the screen is not present (1st assumption of the Kirchhoff's diffraction formula, one of two Kirchhoff's boundary conditions), so at the position Q on the aperture, $${\displaystyle U_{A_{1}}={\frac {ae^{ikr}}{r}},}$$ $${\displaystyle {\frac {\partial U_{A_{1}}}{\partial n}}=\nabla U_{A_{1}}\cdot n={\frac {ae^{ikr}}{r}}\left[ik-{\frac {1}{r}}\right]\cos(n,r),}$$ ^{[note 2]}where ${\displaystyle r}$ is the length of the straight line P₀Q, and ${\displaystyle (n,r)}$ is the angle between a straightly extended version of P₀Q (a forward extension of the line of ${\displaystyle r}$) and the (inward) normal ${\displaystyle n}$ to the aperture. Note that ${\displaystyle 0<(n,r)<{\frac {\pi }{2}}}$ so ${\displaystyle \cos(n,r)}$ is a positive real number on A₁.

At Q, we also have $${\displaystyle {\frac {\partial }{\partial n}}\left({\frac {e^{iks}}{s}}\right)={\frac {e^{iks}}{s}}\left[ik-{\frac {1}{s}}\right]\cos(n,s),}$$ where ${\displaystyle s}$ is the length of the straight line PQ, and ${\displaystyle (n,s)}$ is the angle between a straightly extended version of PQ (a backward extension of the line of ${\displaystyle s}$) and the (inward) normal ${\displaystyle n}$ to the aperture. Note that ${\displaystyle {\frac {\pi }{2}}<(n,s)<{\frac {3\pi }{2}}}$ so ${\displaystyle \cos(n,s)}$ is a negative real number on A₁.

Two more following assumptions are made.

• (2nd assumption of the Kirchhoff's diffraction formula) In the above normal derivatives, the terms ${\displaystyle {\frac {1}{r}}}$ and ${\displaystyle {\frac {1}{s}}}$ in the both square brackets are assumed to be negligible compared with the wavenumber ${\displaystyle k={\frac {2\pi }{\lambda }}}$, means ${\displaystyle r}$ and ${\displaystyle s}$ are much greater than the wavelength ${\displaystyle \lambda }$.
• (3rd assumption of the Kirchhoff's diffraction formula, one of the two Kirchhoff's boundary conditions) Kirchhoff assumes that the values of ${\displaystyle U}$ and ${\displaystyle {\frac {\partial U}{\partial n}}}$ on the opaque areas marked by A₂ are zero. This implies that ${\displaystyle U}$ and ${\displaystyle {\frac {\partial U}{\partial n}}}$ are discontinuous at the edge of the aperture A₁. Of course this is generally not the case.^[4][5]

The contribution from A₃ to the integral is expected to be zero, and it can be justified by one of the following reasons.

1. Make the assumption that the source starts to radiate at a particular time, and make R large enough, so that when the disturbance at P is being considered, no contributions from A₃ will have arrived there.^[1] Such a wave is no longer monochromatic, since a monochromatic wave must exist at all times. This assumption is not necessary, and a more formal argument avoiding it has been derived.^[6]
2. A wave emanated from the aperture A₁ is expected to evolve toward a spherical wave as it propagates (Water wave examples of this can be found in many pictures showing a water wave passing through a relatively narrow opening.). So, if R is large enough, then the integral on A₃ becomes $${\displaystyle U(P)\sim -{\frac {ia}{2\lambda }}\int _{A_{3}}{\frac {e^{2ik(r')}}{r'^{2}}}[\cos(n,r')-\cos(n,r')]d{A_{3}}=-{\frac {ia}{2\lambda }}\int _{A_{3}}e^{2ik(r')}[\cos(n,r')-\cos(n,r')]d{\Omega }=0}$$where ${\displaystyle r'}$ and ${\displaystyle d\Omega }$ are the distance from the center of the aperture A₁ to an integral surface element and the differential solid angle in the spherical coordinate system respectively.

As a result, finally, the integral above, which represents the complex amplitude at P, becomes $${\displaystyle U(P)=-{\frac {ia}{2\lambda }}\int _{A_{1}}{\frac {e^{ik(r+s)}}{rs}}[\cos(n,r)-\cos(n,s)]d{A_{1}}.}$$

This is the Kirchhoff or Fresnel–Kirchhoff diffraction formula.

This formula can also be derived by using a different geometry in which non-physical (no so wave blocking) arbitrary closed surface, that is apart from the point wave source P₀ and surrounds the observation point P, is the integral surface to which the Kirchhoff's integral theorem is applied.^[7] In this geometry, among three assumptions made above, only the 2nd assumption - ${\displaystyle r}$ (The distance between P₀ and an element of the integral surface) and ${\displaystyle s}$ (The distance between the integral surface element and P) are much greater than the wavelength ${\displaystyle \lambda }$ - is required. The result is the same integral form of the Fresnel–Kirchhoff diffraction formula above, except that the integral surface is the chosen closed surface.

Equivalence to Huygens–Fresnel principle

The Huygens–Fresnel principle can be derived by integrating over a different closed surface (the boundary of some volume having an observation point P). In the right diagram, the aperture area A₁ in the diagram of the above section is replaced by (1) a part of a wavefront (emitted from a P₀) at r₀, which is the closest to the aperture, and (2) a portion of a cone with a vertex at P₀, labelled A₄, that is straight lines perpendicular to the wavefront and between the wavefront and the edges of the aperture. If the wavefront is positioned such that the wavefront is very close to the edges of the aperture, then the contribution from A₄ can be neglected (an additional assumption on the top of three Kirchhoff's diffraction formula assumptions above). On this new A₁, the inward (toward the volume enclosed by the closed integral surface, so toward the right side in the diagram) normal ${\displaystyle n}$ to A₁ is along the radial direction from P₀, i.e., the direction perpendicular to the wavefront. As a result, the angle ${\displaystyle (n,r)=0}$ and the angle ${\displaystyle (n,s)}$ is related with the angle ${\displaystyle \chi }$ (the angle as defined in Huygens–Fresnel principle) as $${\displaystyle (n,s)=\pi -\chi .}$$

The complex amplitude of the wavefront at r₀ is given by $${\displaystyle U(r_{0})={\frac {ae^{ikr_{0}}}{r_{0}}}.}$$

So, the diffraction formula in the above section becomes $${\displaystyle U(P)=-{\frac {i}{2\lambda }}{\frac {ae^{ikr_{0}}}{r_{0}}}\int _{S}{\frac {e^{iks}}{s}}(1+\cos \chi )dS,}$$ where the integral is done over the part of the wavefront at r₀ which is the closest to the aperture in the diagram. This integral leads to the Huygens–Fresnel principle (with the obliquity factor ${\textstyle {\frac {1+\cos \chi }{2}}}$).

In the derivation of this integral, instead of the geometry depicted in the diagram of this section, two closed surfaces (no physical surfaces so they do not block waves), one is a sphere of radius r₀ centered at the point wave source P₀, and the other is an arbitrary closed surface that contains this inner sphere and is infinitely far from it (e.g., another sphere of infinite radius), can be used.^[7] In this geometry, the observation point P is located in the volume enclosed by these closed surfaces, so the Fresnel-Kirchhoff diffraction formula is applied on the two spheres. (The surface normal on these integral surfaces are, say again, toward the enclosed volume.) In the formula application to this geometry, the additional assumption of neglecting the integral on A4, used in the above diagram of this section, is not required. The integral on the outer surface is zero as long as it is infinitely far from P, so as a result, the Huygens–Fresnel principle in the same integral from above is derived, except that the integral surface is the inner sphere.

Extended source

Assume that an aperture is illuminated by waves from an extended monochromatic wave source.^[8] The complex amplitude at the aperture is given by U₀(r). (In a point wave source case above, it is ${\displaystyle {\frac {ae^{ikr}}{r}}}$.)

It is assumed, as before, that the values of ${\displaystyle U}$ and ${\displaystyle {\frac {\partial U}{\partial n}}}$ in the area A₁ are the same as when a screen having the aperture is not present, that the values of ${\displaystyle U}$ and ${\displaystyle {\frac {\partial U}{\partial n}}}$ in A₂ are zero (Kirchhoff's boundary conditions), and that 1/s is negligible compared with a wavenumber k (s is much greater than the wavelength ${\displaystyle \lambda }$). By the same logic as before, the integral over A₃ is zero. By applying the Kirchhoff's integral theorem to A₁, we then have $${\displaystyle U(P)={\frac {1}{4\pi }}\int _{A_{1}}{\frac {e^{iks}}{s}}\left[ikU_{0}(r)\cos(n,s)-{\frac {\partial U_{0}(r)}{\partial n}}\right]\,dS.}$$

This is the most general form of the Kirchhoff diffraction formula. To solve this equation for an extended source, an additional integration would be required to sum the contributions made by the individual points in the source. If, however, we assume that the light from the source at each point in the aperture has a well-defined direction, which is the case if the distance between the source and the aperture is significantly greater than the wavelength, then we can write $${\displaystyle U_{0}(r)\approx a(r)e^{ikr},}$$ where a(r) is the magnitude of the disturbance at the point r in the aperture. We then have $${\displaystyle {\frac {\partial {U_{0}(r)}}{\partial n}}=ika(r)e^{ikr}\cos(n,r)}$$ and thus $${\displaystyle U(P)=-{\frac {i}{2\lambda }}\int _{S}a(r){\frac {e^{ik(s+r)}}{s}}[\cos \chi +\cos(n,r)]\,dS.}$$

Fraunhofer diffraction

If all the terms in f(x', y') can be neglected except for the terms in x' and y', we have the Fraunhofer diffraction equation. If the direction cosines of P₀Q and PQ are $${\displaystyle {\begin{aligned}l_{0}&=-x_{0}/r',\\m_{0}&=-y_{0}/r',\\l&=x/s',\\m&=y/s'.\end{aligned}}}$$

The Fraunhofer diffraction equation is then $${\displaystyle U(P)=C\int _{S}e^{ik[(l_{0}-l)x'+(m_{0}-m)y']}\,dx'dy',}$$ where C is a constant. This can also be written in the form $${\displaystyle U(P)=C\int _{S}e^{i(\mathbf {k} _{0}-\mathbf {k} )\cdot \mathbf {r} '}\,dr',}$$ where k₀ and k are the wave vectors of the waves traveling from P₀ to the aperture and from the aperture to P respectively, and r' is a point in the aperture.

If the point source is replaced by an extended source whose complex amplitude at the aperture is given by U₀(r' ), then the Fraunhofer diffraction equation is: $${\displaystyle U(P)\propto \int _{S}a_{0}(\mathbf {r} ')e^{i(\mathbf {k} _{0}-\mathbf {k} )\cdot \mathbf {r} '}\,dr',}$$ where a₀(r') is, as before, the magnitude of the disturbance at the aperture.

In addition to the approximations made in deriving the Kirchhoff equation, it is assumed that

• r and s are significantly greater than the size of the aperture,
• second- and higher-order terms in the expression f(x', y') can be neglected.

Fresnel diffraction

When the quadratic terms cannot be neglected but all higher order terms can, the equation becomes the Fresnel diffraction equation. The approximations for the Kirchhoff equation are used, and additional assumptions are:

• r and s are significantly greater than the size of the aperture,
• third- and higher-order terms in the expression f(x', y') can be neglected.