Intuition of inclination factor in Kirchhoff's diffraction law

Kirchhoff's diffraction law (optics), \begin{align} U(P_0) &= \int_\Sigma U^{\prime}(P_1) \frac{\exp(i kr_{01})}{r_{01}} dS \\ U^{\prime}(P_1) &=\frac{1}{i\lambda} \cdot A\frac{\exp(i kr_{02})}{r_{02}}\cdot \frac{\cos{(\varphi_{n1})} - \cos{(\varphi_{n2})}}{2} \end{align} where a point-source is located at position $$P_2$$ and we measure the (scalar) field at position $$P_1$$, can be understood as the superposition of spherical waves located inside the region of the aperture $$\Sigma$$ with the phase factor $$U^{\prime}(P_1)$$. I'm interested in the "projection factor" (so called inclination factor) $$\cos{(\varphi_{n1})} - \cos{(\varphi_{n2})} = \cos{(\vec{n}, \vec{r}_{01})} - \cos{(\vec{n}, \vec{r}_{02})}$$ where $$\varphi_{n1}$$ is the angle between the $$\vec{r}_{01}$$ and the normal direction of the aperture, $$\vec n$$ (and analog for $$\varphi_{n2}$$).

I understand how this factor is mathematically derived. I also know that two terms can be combined and we obtain a single $$cos$$-function (Rayleigh-Sommerfeld diffraction). However, I am missing the intuition. How can the inclination factor be motivated? What is the intuitive picture behind the projection?

My wrong intuition is: We have a spherical wave, thus, we don't have to consider directions. The phase due to the exponent $$k r_{01}$$ takes care oft everything.

• Have you read/heard about Fresnel's "inclination factor"? To explain Huygens' principle fully you have to get rid of the spherical wavelets of the wavefront in the slot going in the opposite direction, why or how do they superpose only in the forward direction? Those cosine terms correspond to and is one of possibly several mathematically consistent formulation of Fresnel's inclination factor. See, details in Born & Wolf chapter VIII. Commented Apr 13, 2020 at 15:09
• @hyportnex: Thank's for the comments. I read that Fresnel came up with this factor, Kirchhoff derived it from "first principle" (while imposing wrong boundary conditions), and Sommerfeld corrected the calculation by introducing a mirror source. As I mentioned, I am able to do the math, but I am lacking the intuition. Hope Born & Wolf will help me out, because currently I don't see how wavelets matter. I believed that the theory does not consider the time component, but I will study up. Commented Apr 13, 2020 at 16:29
• i suggest you read pages pp412-425 (7th edition); if you cannot find the book let me know and i will post then on imgur. Commented Apr 13, 2020 at 16:40
• Well, I believe the word "wavelet" is used as synonym for "spherical wave" -- which is unfortunate, because of its use in signal analysis. However, skipping through the pages, I could not find any intuitive explanation for the "inclination factor". Commented Apr 13, 2020 at 18:39
• that you need some sort of "inclination factor" is intuitively explained by Fresnel's construction not by the Kirchhoff formula's derivation. It is not Kirchhoff's or Fresnel's fault that they called elementary spherical waves as wavelets 100 years before Daubechies was born, (you know pig-piglet)... Commented Apr 13, 2020 at 19:07

The contribution of a particular ray from source $$P_2$$ to the aperture $$P_1$$ and from there to image $$P_0$$ depends not just on the angular difference between those two segments ($$\overline{P_2P_1}$$ and $$\overline{P_1P_0}$$), but on the relation of each segment to the normal of the boundary in the aperture. If we chose a different surface to cover the aperture, but still including the point of the original aperture surface, the resulting contribution of the same rays to the same points would be different. Hence, we should not expect to obtain a "strong" argument for the inclination factor (also called obliquity factor). Hence, the original argument, that this factor insures that there are no waves going backwards in space, is the best intuitive "justification" I have found so far. Alternatively, one could agree with Goodman's statement, who writes in his book "Intro to Fourier optics" (section 3.7):
The wave equation requires two initial conditions at $$t=0$$: displacement and velocity (or speed of displacement). If the velocity initial condition is derived from the initial displacement rather than being arbitrarily assigned. D'Alembert's formula shows that there will be no backward wave--and therefore no need for an obliquity factor (or inclination factor).
Note that D'Alembert's formula gives the solution to the 1D wave equation and also to the 3D wave equation if $$1/r$$ spherical spreading attenuation is included--so it is good for both plane and spherical waves.