Lens focusing a collimated beam into a disk of some material. Focus shift due to movement of material? I encountered a geometric optics problem that gave the example of a lens focusing a collimated beam into a disk of some material with refractive index $n$. It then claimed that, if the disk moves towards the lens a distance $t$, whilst ensuring that the focus still remains inside the material, then the focus shifts by $nt$ inside the material. This assumes the paraxial approximation.
However, no explanation is provided as to why the focus shifts by $nt$, nor is any explanation provided as to how one comes to this result.
I have previously derived the following equation for the transverse shift of a ray when travelling through air and hitting a slab of some material:
$$x = d \sin(\theta) \left[ 1 - \dfrac{\sqrt{1 - \sin^2(\theta)}}{\sqrt{n^2 - \sin^2(\theta)}} \right],$$
where $d$ is the thickness of the material. I then used this to find subsequent focus shift along the optical axis:
$$F_2 - F_1 = \dfrac{x}{\sin(\theta)}$$
It seems to me that these are the relevant results in deriving the focus shift for a problem such as this. However, I've so far been unable to use them to derive $nt$. 
My immediate thought was that I could use $F_2 - F_1 = \dfrac{x}{\sin(\theta)}$ to solve this problem, but, even after making the paraxial approximation, it doesn't seem to get me the desired result (unless I've made an error):
$$\begin{align} F_2 - F_1 &= \dfrac{x}{\theta} \\ &= \dfrac{d \theta \left( 1 - \frac{1}{n} \right)}{\theta} \\ &= d \left( 1 - \dfrac{1}{n} \right) \end{align}$$
And this doesn't seem to account for $t$, the shift of the material towards to lens. 
My sketch of the problem is as follows:

I would greatly appreciate it if people could please take the time to explain this.
 A: First attack: The answer seems independent of disc and lens geometry.    
Consider the following fig1: 

Here instead of a disc we use a vertical half-plane of index $\eta$. $d$ is where the ray would have focused if there wasn't any material. In the presence of material the ray focuses at $d'$
Clearly,
$$d'=d\frac{tan(i)}{tan(r)}\approx d\eta$$
using small-angle approximation.Therefore
$$\Delta d'=\Delta d \eta$$
for displacements along the optical axis. If the material is displaced by $t$ towards the lens i.e. $\Delta d=t$, then focus shifts by $\Delta d'=\eta t$
As long as small-angle approx. holds, observe that
 1. the result is same even if the material wasn't vertical. This is since all the tilt would do is change $i$.
 2. a disc at each point of incidence is just a tilted tangent plane

Second attack: lens equation
For a spherical lens of 


*

*radius $R$ 

*refractive index $\eta$

*object distance $u$

*image distance $v$

*foal length $f$
The following holds :
$$-\frac{1}{u}+\frac{\eta}{v}=\frac{1}{f}=\frac{\eta-1}{R}$$
under


*

*small angle approx &

*cartesian convention &

*the lens extends indefinitely on the right


Rearranging
$$v=\frac{\eta f u}{f+u}$$
therfore for $u'=u-t$,
$$v'-v=\frac{\eta t}{(1+\frac{u}{f})(1+\frac{u-t}{f})}$$
In the regime $t\ll u\ll f$, to first order $\Delta v=v'-v=\eta t$
