NOTE 2: Don't forget that these diffraction integrals apply to electric-field distributions, not just amplitudes nor irradiance distributions. That scenario is just not as useful since the object/image planes are virtual (i.e., not physically accessible). NOTE: The math still works with a negative lens. Modeling the lens as parabolic, too, kind of seems to in a way cancel out the parabola-ness of these waves, returning the Fourier relationship observed from treating these as plane waves. but I do know Fraunhofer diffraction approximates wavefronts as a sum of plane waves while Fresnel diffraction approximates them as a sum of parabolic waves.
Things start to cancel out in the phase term of the kennels.Ĭonceptually why does this work? I don't have a great explanation other than the math. So, this didn't click until I sat down and wrote out the Fresnel diffraction integral for: 1) propagating a field a focal length's distance f, then 2) adding a quadratic phase corresponding to a thin lens of focal length f, then 3) propagating the result a focal length's distance again. These are approximations, of course, and one should familiarize themselves with the derivations and assumptions involved.Įveryone's got their own way of understanding me, I lean towards more of a mathematical approach. Mathematically, this is because the quadratic phase approximation for the phase induced by a (thin) lens cancels with the quadratic phase that shows up in the Fresnel kernel, simplifying to a Fraunhofer / Fourier kernel. The Fresnel diffraction integral simplifies down to a Fourier transform when observing the image at the lens' back focal plane from an object at the lens' front focal plane.
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