How do we know that bending of light around stars is due to bending of space-time and not diffraction? One question that popped up during the studies of special and general relativity (which I am forced to take unfortunately) is the following:   

How do we know that this is due to the bending of space-time and not just plain old 3rd semester diffraction. If you find this a silly question, downvote, otherwise consider the following picture.
The Sun is the straightedge, the screen is the earth and the star is the point source.
This is simplified to a great extent but the idea still holds (I think)
 A: *

*We know this because the position of the apparent star is perfectly matching the GR calculations about bent spacetime, depending on a few things including the mass of the star (the one in between that bends spacetime, in your case the Sun).

*What you are describing, interference, would not depend on the same way on the mass, the density, stress-energy and a few more things as GR describes bent spacetime.

*There were numerous calculations and experiments like the Shapiro test and they all perfectly gave the matching numbers according to GR.

*Interference would not depend on the same things, for example interference would react differently on the size/mass ratio or density of the star, whereas in GR it really matters what your star's energy density, for example, is compared to its size, for example, a black hole in your case would have an interference of what? I believe that interference would not even work with a black hole.
A: On one hand, a typical diffraction angle $\theta$ for light with wavelength $\lambda$ by a spherical obstacle with the radius $R$ of the Sun is
$$\theta~\sim~\frac{\lambda}{R}~\sim~~\frac{10^{-6} \text{ m} }{10^{9} \text{ m}}~\sim~10^{-15}\text{ rad};$$
while on the other hand, the gravitational bending/deflection of light by the Sun
$$\theta~=~\frac{2r_s}{R}~\approx~\frac{2\cdot 3 \text{ km} }{7 \cdot 10^5\text{ km}}~\sim~10^{-5}\text{ rad}$$
is a roughly 10 orders of magnitude bigger!
