Concentric Meniscus Lenses Design

[Ozen]

Hello Everyone,

I have been working on a lens design that requires a concentric meniscus lens. Initially I was under the impression that r1 should equal r2 for the light to exit at infinity when entering at infinity.

However my ray diagram shows different, it shows the light is being bent quite a bit. Notice how when the image distance is selected at infinity, the light rays are coming towards the lenses diverging; after passing through they are converging. This is not how it should be, they could come parallel from infinity and exit parallel to infinity.

After messing with the lens data I was able to get them very close to how I want them.

My question is on whether the surfaces should have the same radius or like the third image shows, the surfaces should not have the same radius.

Any insight on what should be done here is much appreciated!

 

[Ibix]

"Concentric" doesn't mean "same radius". It means the same center of curvature, which would mean that the radius of curvature of the first surface minus the thickness of the lens is equal to the radius of curvature of the second surface.

I don't know off the top of my head if the resulting lens has the properties you want, but it matches the name rather better than what you describe.

 

[Ozen]

Thank you for the reply!

Perhaps concentric meniscus lens is not the correct name for what I am talking about. I am unsure what its' name would be but I can't find any data on it to solve what I need. I just need the light to pass through this system while staying at infinity, the slight magnification is fine but it has to stay at infinity.

I read on Quora that when a convex and concave meniscus lens is combined, it becomes in arthimic sum of their powers. What my lenses were opposites of each other and clearly the light was being bent. I desperately need help understanding what is the actual truth and how to get the object and image at infinity.

 

[Ibix]

You require the object and image distances, $u$ and $v$ to be $\infty$. That makes $1/f=0$. The lensmakers equation tells you that $$\frac 1f=(n-1)\left(\frac 1{r_1}-\frac 1{r_2}+\frac{(n-1)d}{nr_1r_2}\right)$$Solve for $r_2$ in terms of $r_1$ and $d$ and muck around with those parameters to see what's possible.

It's been a long time since I did any optical design, but my intuition is that the result is likely to be heavily aberrated and off-axis behaviour will be very different.