Consider diffraction from a single slit. Assume that the length of the slit is much larger than the height a of the slit so that the height along the slit is not considered. Assume that the distance D to the “screen” is much larger than the slit width. If the incident light is a monochromatic plane wave, then there are some angles in which the overall light from the slit constructively interferes and others that destructively interfere giving rise to light and dark bands on the screen. This is illustrated in Single-Slit Diffraction.

Now let us replace the flat screen with a cylindrical screen centered on the slit and let the distance to the screen be R, the radius of the semicircle. Then for every angle to the screen we can project back to the plane of the slit such that we can now map every angle to x on a line such that the distance from the origin is given by $\mathrm{x}=\mathrm{R}\mathrm{ }\times \mathrm{s}\mathrm{i}\mathrm{n}\left(\mathrm{\theta }\right)$ ⁽¹⁾. For ease, R = 1 so that $\mathrm{x}=\mathrm{s}\mathrm{i}\mathrm{n}\left(\mathrm{\theta }\right)$ . With this mapping to angular space, x ranges from [-1 to 1] as θ ranges from [-90° to 90°]. All values outside of that range are non-physical.

The intensity profile of the bands is proportional to

which, with the mapping chosen becomes

To determine how much light goes into each a subtended angle, all of the light that enters the slit must be accounted for. The profile function accommodates any x such that

but x is restricted to

as previously noted.

When the far field is integrated,

All the light goes somewhere into the space restricted by +/- 90°. This normalization allows to determine the proportionality constant A that gives the intensity at each angle. All of the angles outside +/- 90° must be zero, therefore integrating the intensity of all the bands calculates A.