Spherical aberration in Biconvex lens

[VVS2000]

I was recently looking for proven relations between focal length, radius of curvature, refractive index etc of a convex lens as I was working on an experiment, I did Find a relation, between Height from principal axis and focal length, and it was a huge relation!I did the experiment to verify it, and it holds good. But I still don't know how to even derive such a huge relation. the image is attached.
If anyone know How to solve it, please help
Thanks In advance!

 

[Twigg]

What is your goal here? To get some intuition? To get an idea of where it comes from in the math? To get actual predictions or improve an optical design? It's a broad question.

 

[VVS2000]

What is your goal here? To get some intuition? To get an idea of where it comes from in the math? To get actual predictions or improve an optical design? It's a broad question.

to get an idea of where it comes in from the math. Like whether it's a geometric derived result or some kind of solution obtained out of brute force numerical method

 

[berkeman]

Paging @Andy Resnick

 

[BvU]

to get an idea of where it comes in from the math. Like whether it's a geometric derived result or some kind of solution obtained out of brute force numerical method

Numerical meethods are very unlikely to come up with explicit expressions like in the sheet you posted (source?)

 

[Twigg]

The short version is that you get this expression when you use Snell's law and solving for the focal point of an incident ray where $\theta$ is the angle between the incident ray and the normal axis to the biconvex lens at the point at distance h off the optical axis. Naturally, you have to use Snell's again where the ray exits the lens on the second convex surface.

In practice, virtually no one uses this formula. You can get the Taylor coefficients by using some neat ray tracing tricks, so we tend to use ray tracing software to do aberration analysis. It's far more computationally efficient and general. My point is: don't feel like you need to know the above expression.

 

[VVS2000]

Numerical meethods are very unlikely to come up with explicit expressions like in the sheet you posted (source?)

thomas k gaylord, georgia tech optical engineering notes

 

[VVS2000]

The short version is that you get this expression when you use Snell's law and solving for the focal point of an incident ray where $\theta$ is the angle between the incident ray and the normal axis to the biconvex lens at the point at distance h off the optical axis. Naturally, you have to use Snell's again where the ray exits the lens on the second convex surface.

In practice, virtually no one uses this formula. You can get the Taylor coefficients by using some neat ray tracing tricks, so we tend to use ray tracing software to do aberration analysis. It's far more computationally efficient and general. My point is: don't feel like you need to know the above expression.

That's the thing, How would one apply snell's law and get such a result, if you have any hint to get me started on the right direction I can make an attempt to solve it. I just want to know the math behind it all, like how one would approach such a complex situation

 

[Twigg]

Sorry for the slow reply. Here's a quick outline of steps that would get you started.

1) Consider a collimated ray ($\theta$ = 0) incident on the first spherical surface of the lens at a displacement $h$ off the optical axis. (See the image you attached in the OP for reference.)

2) Calculate the surface normal vector $\hat{n}_1$ to the first spherical surface at displacement $h$. Now calculate the angle $\theta_{inc,1}$ between the incident ray and $\hat{n}_1$. Use Snell's law to find the new orientation of the ray inside the lens.

3) Trace the path of the ray inside the lens to see where it intersects the second spherical surface of the lens, and apply Snell's law once more.

4) Trace the path of the ray in free space (on the image side), and see where the ray focuses (i.e. where it intersects the optical axis). This focal length should be given by the formula you first quoted.

 

[difalcojr]

Alternatively, maybe just use geometry and trig instead for your mechanical drawings. A lot easier.