means that the field at a point P1 on the screen is given by a surface integral: where the inclination factor which ensures that the secondary wavelets do not propagate backwards is given by and A is the amplitude of the source wave is the wavenumber S is the unobstructed surface. The first term outside of the integral represents the oscillations from the source wave at a distance r0. Similarly, the term inside the integral represents the oscillations from the secondary wavelets at distances r1. In order to derive the intensity behind the circular obstacle using this integral one assumes that the experimental parameters fulfill the requirements of the near-field diffraction regime (the size of the circular obstacle is large compared to the wavelength and small compared to the distances g = P0C and b = CP1). Going to polar coordinates then yields the integral for a circular object of radius a (see for example Born and Wolf): This integral can be solved numerically (see below). If g is large and b is small so that the angle is not negligible one can write the integral for the on-axis case (P1 is at the center of the shadow) as (see ): The source intensity, which is the square of the field amplitude, is and the intensity at the screen . The on-axis intensity as a function of the distance b is hence given by: This shows that the on-axis intensity at distances b much greater than the diameter of the circular obstacle is the same as the source intensity, as if the circular object was not present at all. However at larger distances b, it turns out that the size of the bright spot (as can be seen in the simulations below where b/a is increased in successive images) is larger therefore making the spot easier to discern. Calculation of diffraction images To calculate the full diffraction image that is visible on the screen one has to consider the surface integral of the previous section. One cannot exploit circular symmetry anymore, since the line between the source and an arbitrary point on the screen does not pass through the center of the circular object. With the aperture function which is 1 for transparent parts of the object plane and 0 otherwise (i.e. It is 0 if the direct line between source and the point on the screen passes through the