Editing Fraunhofer diffraction (section)

===Diffraction by a narrow rectangular slit===

[[Image:Single_Slit_Diffraction_(english).svg|right|thumb|Graph and image of single-slit diffraction]]

The width of the slit is {{mvar|W}}. The Fraunhofer diffraction pattern is shown in the image together with a plot of the intensity vs. angle {{mvar|θ}}.<ref>{{harvnb|Hecht|2002|loc=Figures 10.6(b) and 10.7(e)}}</ref> The pattern has maximum intensity at {{math|1=''θ'' = 0}}, and a series of peaks of decreasing intensity. Most of the diffracted light falls between the first minima. The angle, {{math|α}}, subtended by these two minima is given by:<ref>{{harvnb|Jenkins|White|1957|p=297}}</ref>

<math display="block"> \alpha \approx {\frac{2 \lambda}{W}} </math>

Thus, the smaller the aperture, the larger the angle {{math|α}} subtended by the diffraction bands.  The size of the central band at a distance {{math|''z''}} is given by

<math display="block">d_f = \frac {2 \lambda z}{W}</math>[[File:Difraction glass.jpg|thumb|225x225px|Diffraction glass with 300 lines per millimeter]]

For example, when a slit of width 0.5&nbsp;mm is illuminated by light of wavelength 0.6&nbsp;μm, and viewed at a distance of 1000&nbsp;mm, the width of the central band in the diffraction pattern is 2.4&nbsp;mm.

The fringes extend to infinity in the {{math|''y''}} direction since the slit and illumination also extend to infinity.

If {{math|W < λ}}, the intensity of the diffracted light does not fall to zero, and if {{math|D << λ}}, the diffracted wave is cylindrical.

====Semi-quantitative analysis of single-slit diffraction====
[[Image:single slit diagram.svg|100px|right|thumb|Geometry of single-slit diffraction]]
We can find the angle at which a first minimum is obtained in the diffracted light by the following reasoning. Consider the light diffracted at an angle {{math|θ}} where the distance {{math|''CD''}} is equal to the wavelength of the illuminating light. The width of the slit is the distance {{math|''AC''}}. The component of the wavelet emitted from the point A which is travelling in the {{math|θ}} direction is in [[phase (waves)#phase shift|anti-phase]] with the wave from the point {{math|''B''}} at middle of the slit, so that the net contribution at the angle {{math|θ}} from these two waves is zero.  The same applies to the points just below {{math|''A''}} and {{math|''B''}}, and so on.  Therefore, the amplitude of the total wave travelling in the direction {{math|θ}} is zero.  We have:

<math display="block">\theta_\text{min} \approx \frac {CD} {AC} = \frac{\lambda}{W}.</math>

The angle subtended by the first minima on either side of the centre is then, as above:

<math display="block">\alpha = 2 \theta_\text{min} = \frac{2\lambda}{W}.</math>

There is no such simple argument to enable us to find the maxima of the diffraction pattern.

==== Single-slit diffraction using Huygens' principle ====
[[File:Singleslithuygens.jpg|thumb|Continuous broadside array of point sources of length ''a''.]]
We can develop an expression for the far field of a continuous array of point sources of uniform amplitude and of the same phase. Let the array of length ''a'' be parallel to the y axis with its center at the origin as indicated in the figure to the right. Then the differential [[Electric field|field]] is:<ref name=":0">{{Cite book|url=https://books.google.com/books?id=NRxTAAAAMAAJ|title=Antennas for all applications|last1=Kraus|first1=John Daniel|last2=Marhefka|first2=Ronald J.|date=2002|publisher=McGraw-Hill|isbn=9780072321036|language=en}}</ref>

<math display="block">dE=\frac{A}{r_1}e^{i \omega [t-(r_1/c)]}dy=\frac{A}{r_1}e^{i(\omega t-\beta r_1)}dy</math>

where <math>\beta=\omega/c=2\pi /\lambda</math>. However <math>r_1=r-y\sin\theta</math> and integrating from <math>-a/2</math> to <math>a/2</math>,

<math display="block">E \simeq A' \int_{-a/2}^{a/2} e^{i\beta y \sin\theta} dy</math>
where <math>A' = \frac{Ae^{i(\omega t-\beta r)}}{r}</math>.

Integrating we then get
<math display="block">E = \frac{2A'}{\beta \sin \theta} \sin\left(\frac{\beta a}{2} \sin \theta\right)</math>

Letting <math>\psi^'=\beta a \sin \theta = \alpha_r \sin \theta</math> where the array length in radians is <math>a_r=\beta a=2\pi a/\lambda</math>, then,<ref name=":0" />
<math display="block">E= A' a \frac{\sin(\psi^'/2)}{\psi^'/2}</math>