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

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?

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).