# Does integrating over a reflection surface disregard diffraction

Let $$g(R, \theta, \phi)$$ be the reflectivity of a surface such that $$g=0$$ where the surface is not present or not in the line of sight. Let $$f(t)$$ be a wave emitted from the origin at $$t = 0$$. Further assume the Born approximation. Hence no multipath effects.

One can integrate over the surface in order to reconstruct the reflected wave received at the origin. Since the propagation delay is the same for a given $$R$$, $$\iiint g(R, \theta, \phi) f(t - 2 R / c) \, \mathrm{d}A \, \mathrm{d}R.$$

Does the above integral account for diffraction? My intuition would say yes (at least in the far field), but is there a more rigorous proof.

After thinking some more about it and doing some calculations, it is quite obvious that the diffraction effects are accounted for in the integral for a spherical wave emitted at the origin. If the wave is not spherical, one needs to modify the delay to $$f(t - \tau(R, \theta, \phi) - R/c)$$, where $$\tau$$ is the time it takes for the transmitter wave to reach the point $$(R, \theta, \phi)$$.
Diffraction occurs when the transmitted wave diffracts by the infinitesimal area element $$\mathrm{d}A$$. The diffracted wave is a spherical wave that takes $$R/c$$ to reach the origin, because $$\mathrm{d}A$$ is infinitesimal. Sine the wave equation is linear, the integral over all the surface area gives the correct result by superposition.