Intuition of inclination factor in Kirchhoff's diffraction law

Question

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.

Answer 1

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 obliquity factor, has no simple "quasi-physical" explanation, but arises in slightly different forms in all the theories of diffraction. It is perhaps expecting too much to find such an explanation. After all, there are no material sources within the aperture; rather, they all lie on the rim of the aperture. Therefore the Huygens-Fresnel principle should be regarded as a relatively simple mathematical construct that allows us to solve diffraction problems without paying attention to the physical details of exactly what is happening at the edges of the aperture.

Answer 2

I do not think the obliquity factor is needed anymore for the cancellation of the backward wave.

See my paper: https://www.researchgate.net/publication/340085346
appendix C ) Cancellation of backward wave using D'Alembert's formula

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.