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Two reasons are given to explain the black drop effect here, but I think I came up with a third.

Consider a two-dimensional cross section of the situation just after second contact and just before third contact in a stellar transit.

It has many similarities to the classical single-slit experiment:

                            a   b
============================|---|====================|-----------------------
        Empty Space         ↑       Disk of Planet            Disk of Star
                            |
                   Edge of Disk of Star

Our slit begins at point a and ends at b.

Consider that the diffraction pattern of the single-slit experiment spreads out the total energy of the light which is incident on the range of the slit over some angle. This indicates that the incident intensity of the light behind a slit is less than the intensity would be if no slit was present (and no diffraction occurs).

Could this possibly help account for the Black Drop phenomenon? Or must the "slit" be a lot smaller, such that this could not possibly contribute in any significant way? After all, the aperture in this case would be on the order of hundreds of kilometers.

Another thing that just occurred to me is that the left-edge of the star is not being obscured by anything, only the right side is. There would only be one-sided diffraction affecting our incident light intensity.

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This is not a reasonable explanantion, since the single-slit effects (in this case it's the sharp-edge effect) leads to bending light only for the light which comes in within a few wavelengths of the edge of the planet. If the planet has no edge, there is no diffraction.

The scale of astronomical objects simply means that it is impossible for any of the things that happen on the scale of microns to affect observations, because only the tiniest sliver of the light passed this close to the planet, you aren't going to detect it at any astronomical distance.

This does not require a calculation, it is a pure order of magnitude of interference effects. In order to see diffraction from a planet, you need a wavelength comparable to the planetary sizes.

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Right. I remember now from my class on Physics of Sound that the diffraction effect of structures depends on the pitch of the sound. I'm sure if I work out the expressions and factor in the distances the attenuation factor due to diffraction will work out to an infinitesimal quantity. This quantum effect would probably affect the measurement of light intensity there for a nanosecond. –  Steven Lu Jun 28 '12 at 1:53
    
@StevenLu: It's not necessary to calculate anything, it would be wrong to calculate. It's like asking "If I sneeze, will I break my building's floor?" The order of magnitude is off, you don't need to calculate. One must also say that the diffraction effects don't happen when there are no sharp edges, as in the case of venus. –  Ron Maimon Jun 28 '12 at 4:50
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