Thin Layer Interference: Why the OPD gets smaller with larger incident angle?

Question

A little easily seeming geometric question, that tickles me:

On Wiki https://en.wikipedia.org/wiki/Thin-film_interference the derivation for the Optical path difference ($\textrm{OPD}$) in a thin layers uses:

$\mathrm{OPD} = n_2 \cdot (\overline{\mathrm{AB}} + \overline{\mathrm{AC}}) - n_1 \cdot \overline{\mathrm{AD}}$

Resulting to the Formula:

$\mathrm{OPD} = 2\cdot\mathrm{d}\cdot n_2\cdot \cos(\theta_2)$

where $\mathrm{d}$ is the thickness of the layer and $\theta_2$ the angle of refraction.

To me this seems paradox: with larger incident angle $\theta_1$ the path $\overline{\mathrm{AB}} + \overline{\mathrm{AC}}$ should $\textbf{increase}$ and thus the OPD!

But the Formula states:

with larger incident angle $\theta_1$ (meaning larger refracting angle $\theta_2$) the OPD will $\textbf{decrease}$. Why is that?

Answer 1

Take the simpler example of two point sources $S_1$ and $S_2$.

For evident geometrical reasons, the OPD is maximal and equal to $S_1S_2$ when you observe along the axis (ring-like configuration). Mathematically, this OPD can be written $S_1S_2cos(θ)$ and the angle $θ$ is zero.

When you move away from the axis $S_1S_2$, the path difference decreases and it is zero in the median plane (Young's holes configuration).

Hope it can help and sorry for my poor english.