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I've read about GRIN (Gradient Refractive Index) lens but nowhere have I come across a derivation of its focal length (in the paraxial limit, approximately). Here is a typical GRIN lens -

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Let it's refractive index be given by $n(r) = n_0(1-\frac{kr^2}{2})$, where $r$ is the distance from the axis, and is small in the paraxial limit, $d$ is the thickness and $k$ is the gradient constant. I've seen that the focal length $f$ for such a setup is given by $\frac{1}{n_0\sqrt{k}sin(d\sqrt{k})}$. But they don't derive this result, for any reason whatsoever. So I tried to use the fact that rays would meet at the focus at the same time, and used the paraxial limit approximations and ended up with $$f=\frac{1}{n_0dk}$$ which means that it holds only for very small thickness of the lens. I believe that I need to calculate the optical path when a ray at a distance $r$ from the axis is passing through the lens with the help of Snell's law and Fermat's principle, which essentially states- $$\delta \int n(r) \,ds = 0$$ where $ds=\sqrt{dx^2+dy^2}$. But I'm not really able to see how to set up the path in order to calculate the path integral in this case. Can someone help me out in this? Is there a better and easier way to derive this result? If so, it would be of great help if someone could show me the way.

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The ray propagates perpendicular to the wavefront. The speed of the wavefront is $$v(r) = c/n = \frac{2c}{(1 - kr^2)}$$

In time $dt$, the wave propagates $$dx(r) = v(r)dt$$ and $$dx(r+dr) = (v(r) + \frac{dv}{dr}dr)dt$$ The wavefront tilts. $$d\theta \approx tan(d\theta) = \frac{\frac{dv}{dr}drdt}{dr}$$ or $$\frac{d\theta}{dt} = \frac{dv}{dr}$$ Finally, $$\frac{d\theta}{dx} = \frac{d\theta}{dt} \frac{dt}{dx} = \frac{dv/dr}{v(r)}$$

If you make a small angle assumption that the ray does not descend too far as it traverses the lens, you can use the lens thickness to figure out the tilt at any height.

$$\Delta \theta = \frac{d\theta}{dx} \Delta x$$

I have left evaluating the derivative for you, and the work of showing how well the rays come to a focus.

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  • $\begingroup$ I'm really sorry I didn't get your point. I mean, I can understand everything that you've done here, but how does it help me to show that the rays focus at a point? I can calculate the angle at which the ray entered the lens using your equation and considering it to be a right-angled triangle from the small-angle argument, but I really have no idea what to do with this information. Maybe my mind is too messed up to process this, but I was hoping for a way to calculate the time spent by the ray inside the lens, find the height at which it leaves the lens, and equate the total time taken. $\endgroup$
    – AntMan
    Aug 18, 2022 at 21:05
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    $\begingroup$ All rays enter parallel to the axis. That is, $\theta = 0$. So the exit angle is $\Delta \theta$. $\Delta \theta$ is a function of $r$, so you know it at each height. Can you see that the rays come to a focus if $tan(\theta) \propto r$? Or in the small angle approximation, if $\theta \propto r$ $\endgroup$
    – mmesser314
    Aug 19, 2022 at 2:03
  • $\begingroup$ Firstly, thanks for elaborating. I could get the exiting height with the help of $\Delta \theta$. But, if I use the small angle approximation, I can get the distance travelled inside the lens as the hypotenuse of a right-angled triangle, but that deprives me of a $sin$ term, and I can't get it elsewhere. Also, when I go on to equate the times taken, the equation is all messy due to a lot of terms and I don't think I would end up with the focal length expression as mentioned, as I simply can't simplify it. Do you have any suggestion as to how should I proceed from here? $\endgroup$
    – AntMan
    Aug 20, 2022 at 14:25
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    $\begingroup$ For small angles, $sin(\theta) \approx \theta$. Does that help? $\endgroup$
    – mmesser314
    Aug 20, 2022 at 20:58
  • $\begingroup$ Thanks a lot for helping! I was able to derive the expression for the focal length. $\endgroup$
    – AntMan
    Aug 21, 2022 at 19:57

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