transform frequency, and the function is here defined as The Fourier transform frequency here is , giving Note that the function is here defined as to maintain consistency. Intensity The intensity is proportional to the square of the amplitude, and is therefore Apertures Rectangular aperture When a rectangular slit of width W and height H is illuminated normally (the slit illuminated at the normal angle) by a monochromatic plane wave of wavelength , the complex amplitude can be found using similar analyses to those in the previous section, applied over two independent orthogonal dimensions as: The intensity is given by where the and axes define the transverse directions on the plane of observation or the image plane (described in the above figure), and is the distance between the slit center and the point of observation on the image plane. In practice, all slits are of finite size so produce diffraction on the both transverse directions, along the (width W defined) and (height H defined) axes. If the height H of the slit is much greater than its width W, then the spacing of the vertical (along the height or the axis) diffraction fringes is much less than the spacing of the horizontal (along the width or axis) fringes. If the vertical fringe spacing is so less by a relatively so large H, then the observation of the vertical fringes is so hard that a person observing the diffracted wave intensity pattern on the plane of observation or the image plane recognizes only the horizontal fringes with their narrow height. This is the reason why a height-long slit or slit array such as a diffraction grating is typically analyzed only in the dimension along the width. If the illuminating beam does not illuminate the whole height of the slit, then the spacing of the vertical fringes is determined by the dimension of the laser beam along the slit height. Close examination of the two-slit pattern below shows that there are very fine vertical diffraction fringes above and below the main spots, as well as the more obvious horizontal fringes. Circular aperture The aperture has diameter . The complex amplitude in the observation plane is given by Solution by integration Using the recurrence relationship to give If we substitute and the limits of the integration become 0 and , we get Putting , we get Solution using Fourier–Bessel transform We can write