# 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.

• Perhaps this will help: en.wikipedia.org/wiki/Snell%27s_law Commented Nov 22, 2019 at 18:14
• @S.McGrew If I derived the transverse shift $x = d \sin(\theta) \left[ 1 - \dfrac{\sqrt{1 - \sin^2(\theta)}}{\sqrt{n^2 - \sin^2(\theta)}} \right]$, then I don't think not knowing Snell's law is the problem... Commented Nov 22, 2019 at 18:15
• A straightforward way to estimate the location of the focus is to use Snell's law to trace the two most extreme-angle rays and see how their crossing point changes as the surface of the medium is shifted. Commented Nov 22, 2019 at 18:20
• Isnt the paraxial approximation that sin theta= tan theta= theta? Commented Nov 22, 2019 at 18:24
• @lalala That is correct. Commented Nov 22, 2019 at 18:25

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

1. radius $$R$$

2. refractive index $$\eta$$

3. object distance $$u$$

4. image distance $$v$$

5. foal length $$f$$

The following holds :

$$-\frac{1}{u}+\frac{\eta}{v}=\frac{1}{f}=\frac{\eta-1}{R}$$

under

1. small angle approx &
2. cartesian convention &
3. 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 $$\Delta v=v'-v=\eta t$$

• Can you please show how you got $d'=d\frac{tan(i)}{tan(r)}$? The trigonometry used here is not clear to me. Commented Dec 29, 2019 at 12:12
• @ThePointer $\tan i=\frac hd$, $\tan r=\frac {h}{d'}$, where $h$ is the height of the point of refraction from the principal axis Commented Dec 29, 2019 at 13:23