While studying about Fresnel and Fraunhofer diffraction, I came across a statement which says that the fringes disappears and the image would take on the limiting shape of the aperture when wavelength goes to zero, which is the same as predicted by geometrical optics. I don't understand the meaning of this, especially the usage 'wavelength goes to zero'. Can anyone help me?

  • $\begingroup$ As wavelength goes to zero, the energy goes up so that the radiation will just blast thru whatever is in its way with refraction no longer being relevant. Even xrays are hard to focus for this reason, and there is a long way from xrays to zero wavelength. $\endgroup$ – Olin Lathrop Dec 7 '13 at 21:03

In the double slit experiment, "wavelength" is the distance between two successive "peaks", at one moment in time, of the wavefront before it reaches the slits. When wavelength goes to zero, they are referring to "taking a limit". Since it's impossible to actually produce zero-wavelength waves, we reduce the wavelength to almost zero, measure the system behaving "almost geometrical optics", and infer that if we could make it precisely zero we would have perfect geometrical optics.

So why does this happen? Suppose you had a short line of point sources, all in phase with each-other (this is one way we model the slit experiment, each slit as a line of sources). The strongest part of the beam is sent perpendicular to the line. However, if you are at a skew angle, the distance travelled from each source to you is different, so the phases of the waves when they get to you will don't reinforce each-other. The shorter the wavelength, the more sensitive the waves are to slight differences in path-length, and the more energy is concentrated in a narrow beam.

An excellent wave simulator (java applet) that can demonstrate diffraction and other wave phenomena is Paul Falstad's (falstad.com) "ripple tank".


Diffraction in many kinds of waves is a consequence of the relationship between the wavelength of the wave, and the total size of the wave front, which may relate to the size or aperture of the radiating "antenna". Diffraction effects are small, when the wave front aperture is large compared to the wavelength, so the larger the aperture or the smaller the wavelength, the less significant diffraction effects are. So allowing the wavelength to reduce without limit (to zero) shows us what the simplified diffraction free result would be.

Every hi-fi sound buff knows that those tiny two inch cube "loud speakers" are not the right way to build a "sub contra bass woofer" to play a 32 foot organ pipe sound through. In that case, not only is the sound spread almost isotropically by diffraction, but the source is extremely poorly coupled to the air medium to generate the waves.

In light applications, wave fronts are most often many orders of magnitude larger than wavelengths (not in fiber optics), so the waves follow simpler geometrical optics laws rather well. Diffraction effects ultimately limit the resolution of optical systems, and in the largest of optical telescopes, the diffraction effects come as close to the zero wavelength approximation, as we are likely to get.


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