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

Question

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?

Answer 1

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 $$n(r)=n_0-\frac{r^2}{2f}$$