The first diagram shows all rays emerging from a single source S and being reflected from the plate back to the screen. The second diagram shows the same rays "unfolded" as though they are coming from 2 "virtual sources" S1 and S2 behind the plate similar to "virtual images" behind a mirror.
Why are there 2 virtual sources? This is because rays travel a different optical path length OPL from the pinhole source S back to screen, depending on whether they are reflected from the front or rear face of the plate. The extra OPL travelled by rays reflected close to the normal from the back face is approx. $2nd$ where $n$ is the refractive index of the plate. So the distances of virtual sources S1 and S2 from the screen are $2\ell$ and $2\ell+2nd$. Not quite as given in the diagram.
Waves from sources S1 and S2 are emitted out of phase by $\pi$ radians because there is a change of phase on reflection from the front face of the plate but not from the rear face. So rays emitted from S1 and S2 at angle $\theta$ to the axis will be in phase if the difference in their OPL (which is S2N = S1S2$\cos\theta$ in the 2nd diagram) is equal to an odd integer $2k-1$ of half-wavelengths, :
$$2nd\cos\theta=\frac12\lambda (2k-1)$$
For small angles $\theta$ we can use the approximations $\cos\theta \approx 1-\frac12 \theta^2$ and $\theta \approx \frac{r}{2\ell}=\frac{D}{4\ell}$ where $D$ is the diameter of a fringe, which is a ring. Setting $\frac{\lambda}{2nd}=x$ we can write
$$2-\theta^2 \approx (2k-1)x$$
We can find $\lambda$ by plotting a graph. Measure the diameters $D$ of fringes of order $k=1, 2, 3$ etc. Calculate corresponding values of $\theta$. Plot values of $\theta^2$ against corresponding values $2k-1$. This should give a straight line of slope $-x$ from which $\lambda$ can be calculated - assuming that we know the values of $n, d$.
