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The diffraction integral makes it possible to calculate the diffraction of light in optics through an aperture of any shape. In particular, the intensity of the light incident at a point on the observation screen is calculated on the basis of an incident elementary wave and the diaphragm function , which describes the light transmittance of the diaphragm.
Two borderline cases of the diffraction integral are the approximations for the far field ( Fraunhofer diffraction ) and for the near field ( Fresnel diffraction ). See the relevant subsections.
Experimental setup for diffraction of light at a diaphragm
The sketch opposite shows the experimental arrangement, consisting of a light source , a diaphragm , on which the incident light is diffracted, and an observation screen on which the incident light intensity is examined. The shape and properties of the diaphragm determine how the intensity distribution looks on the observation screen.
${\ displaystyle Q}$${\ displaystyle S}$${\ displaystyle P}$
Has the aperture z. B. the shape of a double slit , the known interference pattern results as the intensity distribution . Further applications of the diffraction integral are e.g. B. Diffraction disks and Klotoids .
Kirchhoff's diffraction integral
Sketch for the Fraunhofer / Fresnel approximation of the diffraction integral
The Kirchhoff diffraction integral , also called FresnelKirchhoff diffraction integral , reads
 ${\ displaystyle \ psi _ {P} = {\ frac {a_ {Q} \, k_ {0}} {2 \ pi \, \ mathrm {i}}} \ int _ {\ text {aperture}} \ mathrm {d} S \, f_ {S} \, {\ frac {e ^ {\ mathrm {i} \, k_ {0} (d + d_ {1})}} {d \ cdot d_ {1}}} \ left [{\ frac {\ cos \ theta + \ cos {\ theta _ {1}}} {2}} \ right].}$
Designate

${\ displaystyle a_ {Q}}$the amplitude of the source,

${\ displaystyle k_ {0} = 2 \ pi / \ lambda}$the magnitude of the wave vector ,

${\ displaystyle \ lambda}$the wavelength of light,

${\ displaystyle \ mathrm {d} S}$ an infinitesimal surface element of the diaphragm,

${\ displaystyle f_ {S}}$the aperture function ,

${\ displaystyle (\ cos \ theta + \ cos \ theta _ {1}) / 2}$the slope factor and finally

${\ displaystyle \ psi _ {P}}$the amplitude at the point on the observation screen.${\ displaystyle P}$
Since the distances and in most applications are sufficiently perpendicular to the diaphragm, the inclination factor can be set equal to one in these cases. In this case, or are the angles between the lines marked with and and a perpendicular to the diaphragm plane at the intersection of the lines.
${\ displaystyle d_ {1}}$${\ displaystyle d}$${\ displaystyle \ theta _ {1}}$${\ displaystyle \ theta}$${\ displaystyle d_ {1}}$${\ displaystyle d}$
The intensity at the point results from the square of the absolute value of${\ displaystyle P}$${\ displaystyle \ psi _ {P}}$
 ${\ displaystyle I (P) =  \ psi _ {P}  ^ {2} = {\ frac {a_ {Q} ^ {2} \, k_ {0} ^ {2}} {4 \ pi ^ { 2}}} \ left  \ int _ {\ text {aperture}} \ mathrm {d} S \, f_ {S} \, {e ^ {\ mathrm {i} \, k_ {0} (d + d_ {1})} \ over d \ cdot d_ {1}} \ left [{\ frac {\ cos \ theta + \ cos {\ theta _ {1}}} {2}} \ right] \ right  ^ { 2}.}$
Fraunhofer and Fresnel diffraction
The principle of Fresnel diffraction explained using a slit diaphragm
The principle of Fraunhofer diffraction explained using a slit diaphragm
The principle of Fraunhofer diffraction explained using a lens system and a slit diaphragm
The geometric relationships apply to the light paths and (see sketch)
${\ displaystyle d}$${\ displaystyle d_ {1}}$

${\ displaystyle d = {\ sqrt {L ^ {2} +  {\ vec {s}} + ( {\ vec {p}})  ^ {2}}}}$ and

${\ displaystyle d_ {1} = {\ sqrt {L_ {1} ^ {2} +  {\ vec {s}}  ^ {2}}}}$.
Under the assumptions and , the roots can be approximated by a Taylor expansion.
${\ displaystyle L_ {1} \ gg  {\ vec {s}}  = {\ sqrt {x ^ {2} + y ^ {2}}}}$${\ displaystyle L \ gg  {\ vec {p}}  = {\ sqrt {x '^ {2} + y' ^ {2}}}}$
This approximation corresponds precisely to the case that , i. That is, for these considerations, the inclination factor can be set approximately equal to 1. The diffraction integral is thus
${\ displaystyle \ theta \ approx \ theta _ {1} \ approx 0}$
 ${\ displaystyle \ psi _ {P} = {\ frac {a_ {Q} \, k_ {0}} {2 \ pi \, \ mathrm {i}}} \ int _ {\ text {aperture}} \ mathrm {d} S \, f_ {S} \, {e ^ {\ mathrm {i} \, k_ {0} (d + d_ {1})} \ over d \ cdot d_ {1}}.}$
Furthermore, because of the approximation, the denominator can be set. The exponent contains the phase information that is essential for the interference and must not be simplified in this way. It follows
${\ displaystyle d \ cdot d_ {1} \ approx L_ {1} \, L}$
 ${\ displaystyle \ psi _ {P} = {\ frac {a_ {Q} \, k_ {0}} {2 \ pi \, \ mathrm {i}}} {\ frac {1} {L_ {1} \ , L}} \ int _ {\ text {aperture}} \ mathrm {d} S \, f_ {S} \, e ^ {\ mathrm {i} \, k_ {0} (d + d_ {1}) }.}$
The approximation for the expressions and , explicitly carried out up to the 2nd order, results in
${\ displaystyle d}$${\ displaystyle d_ {1}}$
 ${\ displaystyle d = {\ sqrt {L ^ {2} +  {\ vec {s}}  {\ vec {p}}  ^ {2}}} \ approx L \ left (1+ { {\ vec {s}}  {\ vec {p}}  ^ {2} \ over 2L ^ {2}} + \ ldots \ right) = L \ left (1+ {s ^ {2} + p ^ {2 } 2 {\ vec {s}} \ cdot {\ vec {p}} \ over 2L ^ {2}} + \ ldots \ right)}$
such as
 ${\ displaystyle d_ {1} = {\ sqrt {L_ {1} ^ {2} + s ^ {2}}} \ approx L_ {1} \ left (1+ {s ^ {2} \ over 2L_ {1 } ^ {2}} + \ ldots \ right).}$
Expressed by the coordinates and gives that
${\ displaystyle (x, \, y)}$${\ displaystyle (x ', \, y')}$
 ${\ displaystyle d \ approx L \ left (1 + {\ frac {x ^ {2} + y ^ {2} + x '^ {2} + y' ^ {2} 2 (x \, x '+ y \, y ')} {2L ^ {2}}} + \ ldots \ right)}$
and
 ${\ displaystyle d_ {1} \ approx L_ {1} \ left (1 + {\ frac {x ^ {2} + y ^ {2}} {2L_ {1} ^ {2}}} + \ ldots \ right ).}$
Fraunhofer approximation
The Fraunhofer approximation corresponds to a farfield approximation, which means that not only the aperture is assumed to be small, but also the distance to the observation screen is assumed to be large. The Fourier transform of the diaphragm function essentially results as the diffraction integral . This is why one also speaks of Fourier optics in the context of Fraunhofer diffraction .
${\ displaystyle L}$
According to these assumptions, only terms that are linear in and , that is, are considered
${\ displaystyle x}$${\ displaystyle y}$
 ${\ displaystyle d \ approx L \ left (1+ {x '^ {2} + y' ^ {2} 2 (x \, x '+ y \, y') \ over 2L ^ {2}} \ right) = L \ left (1+ { {\ vec {p}}  ^ {2} 2 (x \, x '+ y \, y') \ over 2L ^ {2}} \ right), }$
 ${\ displaystyle d_ {1} \ approx L_ {1}.}$
In this case the diffraction integral is simplified to
 ${\ displaystyle \ psi _ {P} \ approx {a_ {Q} \, k_ {0} \ over 2 \ pi \, \ mathrm {i}} {e ^ {\ mathrm {i} \, k_ {0} (L_ {1} + L + {p ^ {2} \ over 2L})} \ over L_ {1} \, L} \ iint _ {\ mathrm {aperture}} \ mathrm {d} x \, \ mathrm { d} y \, f_ {S} (x, y) \, e ^ { \ mathrm {i} \, k_ {0} {(x \, x '+ y \, y') \ over L}} .}$
If a new wave vector is defined , the integral is
${\ displaystyle {\ vec {K}} = {k_ {0} \ over L} {\ vec {p}}}$

${\ displaystyle \ iint _ {\ mathrm {aperture}} \ mathrm {d} x \, \ mathrm {d} y \, f_ {S} (x, y) \, e ^ { \ mathrm {i} \ , k_ {0} {(x \, x '+ y \, y') \ over L}} = \ iint _ {\ mathrm {aperture}} \ mathrm {d} x \, \ mathrm {d} y \ , f_ {S} (x, y) \, e ^ { \ mathrm {i} \, {k_ {0} \ over L} {\ vec {p}} \ cdot {\ vec {s}}} = \ iint _ {\ mathrm {aperture}} \ mathrm {d} x \, \ mathrm {d} y \, f_ {S} (x, y) \, e ^ { \ mathrm {i} \, {\ vec {K}} \ cdot {\ vec {s}}}}$.
This is just the Fourier transform of the aperture function .
${\ displaystyle f_ {S} \, (x, y)}$
Fresnel approximation
The Fresnel approximation corresponds to a nearfield approximation. It also takes account of quadratic terms in the exponent. The diffraction integral, brought into the form of a Fourier transform, can then generally no longer be solved analytically by an additional term, but only numerically .
Taking into account quadratic terms in and results
${\ displaystyle x}$${\ displaystyle y}$
 ${\ displaystyle d \ approx L \ left (1+ {x ^ {2} + y ^ {2} + x '^ {2} + y' ^ {2} 2 (x \, x '+ y \, y ') \ over 2L ^ {2}} + \ ldots \ right),}$
 ${\ displaystyle d_ {1} \ approx L_ {1} \ left (1+ {x ^ {2} + y ^ {2} \ over 2L_ {1} ^ {2}} + \ ldots \ right).}$
In this case the diffraction integral is
 ${\ displaystyle \ psi _ {P} \ approx {a_ {Q} \, k_ {0} \ over 2 \ pi \, \ mathrm {i}} {e ^ {\ mathrm {i} \, k_ {0} (L + L_ {1} + {p ^ {2} \ over 2L})} \ over L \, L_ {1}} \ iint _ {\ mathrm {aperture}} \ mathrm {d} x \, \ mathrm {d} y \, f_ {S} (x, y) \, e ^ {\ mathrm {i} \, k_ {0} ({x ^ {2} + y ^ {2} 2 (x \, x '+ y \, y') \ over 2L} + {x ^ {2} + y ^ {2} \ over 2L_ {1}})}.}$
Introduction of with and then results in the diffraction integral in nearfield approximation
${\ displaystyle L '}$${\ displaystyle {1 \ over L '} = {1 \ over L} + {1 \ over L_ {1}}}$${\ displaystyle {\ vec {K}} = {k_ {0} \ over L} {\ vec {p}}}$
 ${\ displaystyle \ psi _ {P} \ approx {a_ {Q} \, k_ {0} \ over 2 \ pi \, \ mathrm {i}} {e ^ {\ mathrm {i} \, k_ {0} (L + L_ {1} + {p ^ {2} \ over 2L})} \ over L \, L_ {1}} \ iint _ {\ mathrm {aperture}} \ mathrm {d} x \, \ mathrm {d} y \, f_ {S} (x, y) \, e ^ {i \, k_ {0} {x ^ {2} + y ^ {2} \ over 2L '}} e ^ { \ mathrm {i} \, {\ vec {K}} \ cdot {\ vec {s}}}.}$
Derivation
From the source with amplitude at comes the spherical wave , the amplitude of which decreases reciprocally with distance ( ). Wave vector times distance gives the phase shift of the wave at the location , angular frequency times time the phase shift to time . The wave is described by the phase at the location at the time :
${\ displaystyle Q}$${\ displaystyle a_ {Q}}$${\ displaystyle {\ vec {r}} _ {Q}}$${\ displaystyle \ psi _ {Q}}$${\ displaystyle 1 /  {\ vec {r}} }$${\ displaystyle k}$${\ displaystyle  {\ vec {r}} }$${\ displaystyle {\ vec {r}}}$${\ displaystyle \ omega}$${\ displaystyle t}$${\ displaystyle t}$${\ displaystyle {\ vec {r}}}$${\ displaystyle t}$
 ${\ displaystyle \ psi _ {Q} ({\ vec {r}}, t) = a_ {Q} \, {e ^ {\ mathrm {i} (k \,  {\ vec {r}}   \ omega \, t)} \ over  {\ vec {r}} }.}$
At the point at , the wave meets the diaphragm at a distance . Let it be the field distribution of the wave at the point .
${\ displaystyle S}$${\ displaystyle {\ vec {r}} _ {S}}$${\ displaystyle d_ {1}}$${\ displaystyle \ psi _ {1}}$${\ displaystyle S}$
 ${\ displaystyle \ psi _ {1} (t) = \ psi _ {Q} (d_ {1}, t) = a_ {Q} \, {e ^ {\ mathrm {i} (k \, d_ {1 }  \ omega \, t)} \ over d_ {1}}}$
According to Huygens' principle , the point is the starting point of an elementary wave, the secondary wave .
${\ displaystyle S}$${\ displaystyle \ psi _ {S}}$
 ${\ displaystyle \ psi _ {S} ({\ vec {r}}, t) = a_ {S} \, {e ^ {i (k \,  {\ vec {r}}   \ omega \, t)} \ over  {\ vec {r}} }.}$
The amplitude of is proportional to the source amplitude and the aperture function . The aperture function indicates the permeability of the aperture. The simplest case is when the shutter is open and when the shutter is closed. is the infinitesimal surface element of the aperture at the point .
${\ displaystyle \ psi _ {S}}$${\ displaystyle a_ {Q}}$${\ displaystyle f_ {S}}$${\ displaystyle f_ {S} = 1}$${\ displaystyle f_ {S} = 0}$${\ displaystyle \ mathrm {d} S}$${\ displaystyle S}$
 ${\ displaystyle a_ {S} (t) \ sim f_ {S} \, \ mathrm {d} S \, \ psi _ {1} (t)}$
The secondary wave generates the wave intensity at point at on the screen . It is infinitesimal because only the contribution from and not all other points on the diaphragm are considered.
${\ displaystyle \ psi _ {S}}$${\ displaystyle P}$${\ displaystyle {\ vec {r}} _ {P}}$${\ displaystyle \ mathrm {d} \ psi _ {P} (t)}$${\ displaystyle \ mathrm {d} S}$
 ${\ displaystyle \ mathrm {d} \ psi _ {P} (t) = \ psi _ {S} ({\ vec {d}}; t) = a_ {S} \, {e ^ {\ mathrm {i } (k \, d \ omega \, t)} \ over d}}$
The time dependency can be neglected, as it disappears later when calculating with intensities anyway due to the time averaging. By inserting you get:
${\ displaystyle e ^ { \ mathrm {i} \, \ omega \, t}}$
 ${\ displaystyle \ mathrm {d} \ psi _ {P} \ sim f_ {S} \, \ mathrm {d} S \, \ psi _ {1} (t) \, {\ frac {e ^ {\ mathrm {i} \, k \, d}} {d}} = f_ {S} \, \ mathrm {d} S \, a_ {Q} \, {\ frac {e ^ {\ mathrm {i} \, k \, d_ {1}}} {d_ {1}}} \, {\ frac {e ^ {\ mathrm {i} \, k \, d}} {d}} = f_ {S} \, \ mathrm {d} S \, a_ {Q} \, {\ frac {e ^ {\ mathrm {i} \, k \, (d_ {1} + d)}} {d_ {1} \, d}} }$
A secondary wave emanates from every point on the diaphragm. The intensity in the observation point is generated by the superposition of all individual contributions:
${\ displaystyle \ psi _ {P}}$${\ displaystyle P}$
 ${\ displaystyle \ psi _ {P} \ sim a_ {Q} \ int _ {\ text {aperture}} \ mathrm {d} S \, f_ {S} \, {\ frac {e ^ {\ mathrm {i } \, k (d + d_ {1})}} {d \ cdot d_ {1}}}.}$
This equation is already very reminiscent of the diffraction integral given above. With the proportionality factor we get (inclination angle neglected):
${\ displaystyle k / (2 \ pi \, \ mathrm {i})}$
 ${\ displaystyle \ psi _ {P} = a_ {Q} \, {\ frac {k} {2 \ pi \, \ mathrm {i}}} \, \ int _ {\ text {aperture}} \ mathrm { d} S \, f_ {S} \, {e ^ {\ mathrm {i} \, k (d + d_ {1})} \ over d \ cdot d_ {1}}.}$