# Deriving the focal length of a graded index lens (GRIN)

I want to find a closed expression of the focal length of a graded index since I don't manage to find any on the internet. I already checked this out: Determining the focal length of a gradient index lens

But they didn't seem to finalize the result and the technique of their derivation is unfamiliar to me.

The lens has a radius of $$r'$$. Assume that the index of refraction at $$r = 0$$ is $$n_0$$ and at $$r = r'$$ is $$n_r$$. We know that the refractive index varies parabolically, meaning that $$n(r) = n_0 - (n_0-n_r)\frac{r^2}{r'^2}.$$

Also, assume that the lens has a thickness of $$t$$. Then the phase change when light is transmitted through the lens at a distance $$r$$ from the axis of symmetry is given by $$\phi(r) = k_0 n(r) t = k_0 t(n_0 - (n_0-n_r)\frac{r^2}{r'^2}).$$

If we compare this with the phase chage of a general lens with focal length $$f$$, which is given by $$\phi(r) = -k_0 r^2/(2f)$$ (with some arbitrary phase constant), comparing both expressions yields us:

$$k_0t(n_0 - (n_0-n_r)\frac{r^2}{r'^2}) = -k_0 r^2/(2f) + \text{const}.$$

For which we can rearrange for all constants and set them to $$0$$ since they're arbitrary to finally get the focal length:

$$f = \frac{r'^2}{2t(n_0-n_r)}.$$

Am I in the wrong for comparing it with the phase given by a general lens of focal length of $$f$$, and is my assumption about setting the constants to $$0$$ right as well? In that case, why / why not?

Use Fermat's prinicple. Let the thickness of the lens be denoted by $$d$$, then the optical path length from axial point $$\mathcal P$$ to another axial point $$\mathcal Q$$ on the other side of the lens is $$OPL[{\mathcal P \mathcal Q}]=\sqrt{p^2+r^2}+n(r)d+\sqrt{q^2+r^2}\\ \approx p+\frac{r^2}{2p} + n(r)d + q+\frac{r^2}{2q}$$ For imaging this must be independent of $$r$$, that is, $$\frac{r^2}{2}\big(\frac{1}{q}+\frac{1}{p}\big) + n(r)d =\frac{r^2}{2f}+n(r)d=n_0$$ where $$\frac{1}{q}-\frac{1}{p}= \frac{1}{f}$$ with $$n(r)=n_0-\frac{r^2}{2f}$$

• I don’t know Fermats principle, but using your method, would this mean our results will match or not? Apr 11, 2023 at 15:12
• In your formula the focal lens depends on the radius $r'$ of the lens, I have difficulty believing that. So forget Fermat, just accept that all rays must have the same optical length, ie., the same phase shift from starting point to their endpoint. Apr 11, 2023 at 17:00
• In my case $r’$ is a constant and doesn’t vary radially. Or did I misunderstand you? Either way I’ll try with what you suggested. Thanks. Apr 12, 2023 at 9:18
• Yes it is a constant but it appears that in your last formula the focal length is proportional to $r'^2$, now imagine that you place a diaphragm that lets only the middle half of the light in and now the focal length goes down by a factor of 4! Does this sound reasonable? What kind of lens is that? Apr 12, 2023 at 9:46
• Oh yes that’s indeed not reasonable. Thank you! Apr 12, 2023 at 10:09