It is a geometrical problem which I find difficult to solve reading the Fresnel's paper "Memoir on the diffraction of Light". According to the figure Fresnel sets $z$ as the distance of the element $nn'$ from the point $M$---- (I suppose $z=nM$)-----, $a=CA$, $b=AB$, $IMA$ is an arc with center $C$, $EMF$ is an arc with center $P$ tangential to the point $M$ with the first arc. Eventually, Fresnel calculate that the distance $nS=\frac{z^2(a+b)}{2ab}$.( I believe that it is an approximation saying $nS≈\frac{z^2(a+b)}{2ab}$)
(1) How does he find that result?
(2) In my attempt I find that $nS≈\frac{z^2}{2PM}$, pretty close but I cannot find the $PM$ value.Are there any ideas?
(3) Υοu can find the original paper here (page 119): https://archive.org/stream/wavetheoryofligh00crewrich#page/118
There are also the conditions that the circle $C$ is tangent to the screen (the dashed line) and $PN$ is orthogonal to the screen.
The requirement that $z$ is small implies that $PB$ and $r - b$ are also small.
From $\triangle MCN$, $\sin (\angle MCN / 2) = z/(2 a)$. From $\triangle CPN$, $$(r + x)^2 = a^2 + (r + a)^2 - 2 a (a + r) \cos \angle MCN.$$ Dropping the $x^2$ term gives $$r^2 + 2 r x = a^2 + (r + a)^2 - 2 a (a + r) \left( 1 - 2 \left( \frac z {2a} \right)^2 \right), \\ x = \frac {(a + r) z^2} {2 a r} \sim \frac {(a + b) z^2} {2 a b} \quad \text{when } z \to 0.$$