The situation is that there is an incoming light wave that is transmitted through a lens. The thickness of the lens is chosen in the z-direction and the cross section of the lens is in the x,y-plane. A plane is defined at one edge of the lens at $z = z_1$ and another plane is defined at the other edge of the lens at $z = z_2$. We want to know the phase difference of the light between the moments that the light enters the plane at $z=z_1$ and the moment the light leaves the plane at $z = z_2$. In general the light will travel partly in the vacuum and partly in the medium of the lens. If the light would be incident exactly at $\Delta_0$ there would only be travel in the lens medium, but in all other situations part of the light path will be in the vacuum and the other part will be in the lens medium. The total difference in phase is equal to the contribution caused by the part in the lens plus the contribution caused by the part in the vacuum.
Let's assume the incoming light can be written as an amplitude function times an exponential, so: $A(r)e^{ikr}$. The phase has to do with the complex exponential, namely $kr$.
Let's make the discussion a bit general and let some light travel a distance between two planes. The electric and magnetic fields of the light oscillate with a period $T$ and a angular frequency $\omega$. The light has a certain wave length $\lambda$. If you look a the light at position $x = x_0$ and a position $x = x_0 + \lambda$ you don't see a difference due to the periodicity of the light. So, if the light travels a certain distance $r$ you need to subtract the wave length $\lambda$ as many times as you can and the difference that is left at the end is interesting for the difference in phase.
Let the light enter the first plane at position $r_1$ with a light phase $\phi_1$ and let the light hit the second plane at position $r_2$ with a light phase $\phi_2$. Then $mod(|r_2 - r_1|, \lambda)$ is the fraction of the period we need to shift. Also $\phi$ is just a number in the interval $\left[0, 2 \pi \right)$. 586
The phase at position 2 is $\phi_2 = \phi_1 + mod(|r_2 - r_1|, \lambda) \cdot 2 \pi $. Thus: $\phi_2 - \phi_1 = mod(|r_2 - r_1|, \lambda) \cdot 2 \pi$. Let's put the phase in a complex exponential, because that is the formula for the light. $ e^{i(\phi_2 - \phi_1)} = e^{i( mod(|r_2 - r_1|, \lambda) \cdot 2 \pi)}$. The right hand side can be rewritten using the periodicity of the complex exponential. $ e^{i(\phi_2 - \phi_1)} = e^{i( (|r_2 - r_1|/ \lambda )\cdot 2 \pi)}$.
When light enters a medium with a different optical density, such as a lens, the wave length changes but not the frequency. $\lambda_{medium} = \lambda_{vacuum}/n$. The $\lambda$ used in the previous paragraph is $\lambda_{medium}$, so you can replace that $\lambda$ in the previous paragraph with $\lambda_{vacuum}/n$. In the formula's $k$ is the wave vector in vacuum.
The definition of the wave vector: $k = 2 \pi / \lambda_{vacuum}$ leads to $e^{i (\phi_2-\phi_1)} = e^{i k n |r_2 - r_1|}$. What we know now is that when you let some light of the form $A(r)e^{ikr}$ travel a distance $|r_2 - r_1|$ you can multiply with $e^{i k n|r_2 - r_1|}$ to get the correct phase. Let's call this factor $t$. So $t = e^{i k n|r_2 - r_1|}$.
Back to our set-up. The total distance between the two planes $z_2-z_1 = \Delta_0$. Distance travelled through the lens is $\Delta(x,y)$ and the distance travelled through the vacuum is $\Delta_0 - \Delta(x,y)$. 804