Hi Andreas
No problem and don’t hesitate to do so.
There are things that require clarifications in your request.
So in first approximation you just have topographic and deformation phase components.
By surface variation, you mean the deformation, right ? It is a bit ambiguous. But in the same time you say you did not remove the topography. And deformation is not dependent of incidence angle.
\phi=\dfrac{4\pi}{\lambda}\cdot d_{\mathrm{LOS}}
Still for the deformation right? This is almost correct except that you need to divide by 2.
Set \phi=2\pi, i.e.1 cycle in the above equation, and you’ll find 2.78 centimeters. This 2.78 cm displacement will induce a 2\pi shift in your interferogram.
I have the feeling that what you are missing is that the optical path difference in interferometry is way more sensitive to displacements than surface topography. Topography and displacements are really different terms in interferometry. With Sentinel-1 and its narrow orbital tube (which reduce the perpendicular baseline), height differences of more than 150-200 meters (highly varying) are required to induce a 2\pi shift. From Ramon Hanssen (Radar Interferometry) but also Antonio Pepe (see below), the topography is related to the phase by
\phi_{\mathrm{topo}}=- \dfrac{4\pi}{\lambda}\cdot \dfrac{b_{\mathrm{perp}}}{r\cdot\sin{\theta}}\cdot H
r is the range, H teh height, \theta the incidence angle, … Set the \phi_{\mathrm{topo}} to 2\pi to find the “amount of height” required to produce a 2\pi shift.
H_{2\pi}= -\dfrac{\lambda\cdot r\sin{\theta}}{2\cdot b_{\mathrm{perp}}}
Take theta = 30°, the range of 1000 kilometers (first order approximation) et b_perp of 100 meters, and you’ll need 140 meters of height to produce a 2\pi shift. A hill of few meters high will never induce several cycles in your interferogram. Fortunately for us otherwise we would need digital elevation models with crazy high accuracy to perform DInSAR.
I greatly encourage you to carefully read this review, which helped a lot : https://www.mdpi.com/2076-3417/7/12/1264/htm, and re-do all calculations with the author (p2 to 7 / 39).
Cheers !
Quentin
NB : it seems in your figure that you resampled your interferogram without resampling the real and imaginary components but directly the phase, causing errors in the boundary 0rad/2pi rad (or -pi/+pi).