To understand the Single Slit example, one must revert to Huygen's Principle of 1678 which describes how a point on a wave-front can be thought of as contributing to a secondary forward travelling spherical wavelet. In the case of the Single slit, any part of the slit can be thought of as such a point. The coherent superposition of these wavelets produced at the slit results in the diffraction pattern seen at the focal length of the lens (the Fraunhofer Plane). This diffraction pattern can be explained by the Fresnel-Kirchoff Formula explained in the introduction to this lab. With the below situation,
L represents the slit length, b the slit width, and y the point of origin of the forward travelling wavelet. Plugging these constraints and variables into the Fresnel-Kirchoff formula yields the diffraction intensity relationship in the following Fourier transform integral:
The Ldy replaces LdA and r is replaced by ro+ySin(theta) (**ro is the value for r when y=0) in the Fresnel-Kirchoff formula. This integral is easily evaluated over this range in a program such as mathmatica. Integration and simplification yields the result
.
If one thinks of the Fraunhofer plane as we
previously did the single slit, then it ALSO represents a source for secondary forward
travelling wavelets. Each point on the diffraction pattern on the plane will cause
these wavelets to superpose and recombine at some other point in the image plane of the
lens. It can also be characterized by a Fourier integral and believe it or not it is
the expression for the diffraction pattern at the Fraunhofer plane 
!!! Performing the Fourier transform of this should then give us a
relationship describing the slit itself. In other words, at some point beyond the
Fraunhofer plane we will see an image of the slit! To see whether or not all of
this actually occurs, check out our data section where we show you....or not!
The screen situation is analogous to several square slits next to each other, both horizontally and vertically. Mathematically, the problem is an extension of the single and double slit examples in two dimensions. All of the variables that we need for a one-dimensional discussion are defined in the picture below:
In the picture of multiple slits above, b represents the slit width and h their separation. It also consists of N identical parallel slits, of course that may be an impossible situation to produce on the production line of the chicken wire, but we theorize for the fun of it! Using the base equation for the single slit and modifying to fit our situation, we are blessed with the following integral:
So, to solve this problem, we integrate over each slit individually and add them together which makes sense intuitively as well. The solution to this integral, with N number of slits of width b and separation h, is also similar to the solution of the single slit problem, with a few extra terms of course, and looks like this:
Above, 
and 
while
N normalizes the expression. The observed intensity is again the square of this expression
times the incident intensity at the slits.
This experiment is basically the equivalent of sending light from a source through a circular aperture, in our case a piece of aluminum foil with a pinhole punched through it. We therefore have the following scenario:
Imagine the square with the circle in it as the front view of the vertical
aperture on the right in the picture above. In this case, we are integrating over y
with dy defined as above. The dA term of the Fresnel-Kirchoff equation now becomes
. Now it is easy to define a relationship that describes our situation by merely
adjusting the Fresnel-0Kirchoff
formula again. Substituting our adjusted dA term into it and simplifying, we arrive at the following relationship:
Even though it was somewhat easily derived, the solution to this integral
is very nasty and involves what are known as bessel functions. Thus, a change of
variables is performed to simplify our anguish a little. We define two new
variables, mu and rho, which are expressed as 
and 
. Our integral now becomes 
which
can be integrated using mathmatica:
The solution reveals a first order Bessel function (oh the joy!) which then gives us a function describing the diffraction in the Fraunhofer plane. The function is plotted below using mathmatica. The picture on the right is the diffraction amplitude as mathmatica says it will appear in three dimensions (we go from rho=0 to rho=15). The actual intensity observed at the Fraunhofer plane is seen in the picture on the right (Intensity is simply the square of the solution)
The central fringe, which is appears as a circle, can be seen very clearly in relationship to the smaller outer fringes.
With each of the previous apertures we used some sort of spatial filter at the Fraunhofer Plane to prohibit certain components of the diffraction pattern from recombining at the image plane.
Consider the diffraction aperture a function of the spatial variable. Expanding what was previously discussed, Fourier transforming the aperture function should yield the diffraction pattern (as a function of a 'spatial frequency') observed at the Fraunhofer Plane. Taking another Fourier transform should return the aperture function.
If one includes a spatial filter in the system, a representative function of 'spatial frequency' must be included in the second Fourier integral. This filtering results in information blocked and prevented from reaching the image plane. Thus, 'holes' appear in the image. As will be seen in the data and conclusions section, when we blocked the central peak of the diffraction pattern and allowed the outside fringes to pass to the image plane, this information did little to build the central portion of the image, but constructed the edges of the image.
A helpful analogy to this is in considering the composition of a square wave g(y).
It's Fourier transform demonstrates that it is composed of many sine waves,
with one dominant frequency.
When the central peak (or, in the Fourier transform analogy, the dominant
frequency) is removed, only the fringes (or the other frequencies composing the square
wave) remain to form the image.
Thus,
when the dominant frequency is removed, only the edges of the image will appear clearly at
the image plane.
Likewise, when the fringes surrounding the central bright spot are removed, information carried in the diffraction pattern about the edges of the object is lost. Thus, at the image plane, the object's edges do not appear clearly.
Click here to see an example of how this is modeled, via. mathematica, for a single-slit aperture.
Pictures and derivation adapted from:
Fowles, Grant R. Introduction to Modern Optics. Dover Publications: New York. 1975.