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:

enter image description here

I would greatly appreciate it if people could please take the time to explain this.

  • $\begingroup$ Perhaps this will help: en.wikipedia.org/wiki/Snell%27s_law $\endgroup$
    – S. McGrew
    Commented Nov 22, 2019 at 18:14
  • $\begingroup$ @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... $\endgroup$ Commented Nov 22, 2019 at 18:15
  • $\begingroup$ 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. $\endgroup$
    – S. McGrew
    Commented Nov 22, 2019 at 18:20
  • $\begingroup$ Isnt the paraxial approximation that sin theta= tan theta= theta? $\endgroup$
    – lalala
    Commented Nov 22, 2019 at 18:24
  • $\begingroup$ @lalala That is correct. $\endgroup$ Commented Nov 22, 2019 at 18:25

1 Answer 1


First attack: The answer seems independent of disc and lens geometry.

Consider the following fig1: ray diagram: a ray approaches a vertical half-plane of index eta

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 $d'$


$$d'=d\frac{tan(i)}{tan(r)}\approx d\eta$$ 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 :



  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 $\Delta v=v'-v=\eta t$

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

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