I Aberration in Lenses: Formula & Relation Explained

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Aberration in lenses can be complex, with no straightforward formula for calculating it, though various coefficients like Seidel and Zernike can be used. Plano concave lenses exhibit negative aberration, while plano convex lenses can correct this, but the relationship depends on the specific optical design and application. Experimental methods, such as using an iris diaphragm to adjust the numerical aperture, can help quantify aberration by observing focal point shifts. Calculations often involve ray-tracing methods and can quickly become complicated, requiring specialized software for accurate modeling. Understanding these principles is crucial for effective lens design and aberration management.
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Is there any formula or a mathematical relation to find aberration in lenses? I read recently that plano concave lens has a negative aberration and plano convex lens is used to correct it. I am not sure what those statements mean. Is there some type of relation that explains these statements or is it like an experimental fact?
 
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Wiki does it for the thin-lens approximation.
 
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VVS2000 said:
Is there any formula or a mathematical relation to find aberration in lenses? I read recently that plano concave lens has a negative aberration and plano convex lens is used to correct it. I am not sure what those statements mean. Is there some type of relation that explains these statements or is it like an experimental fact?
The short answer is "no, but you can calculate the various aberration coefficients given the optical design". There are 'stop-shift formulas', but I don't think that's what you mean. It's important to note that there are aberrations caused by the shape(s) of the lens element(s) and different aberrations (chromatic) caused by the material(s) the element(s) are made of. Then, of course, there are aberrations caused by misalignment :).

The calculation gets extremely complicated very quickly, and there aren't any simple references, either. In general, aberration theory is rooted in either Seidel coefficients or Zernike coefficients, and converting between the two is non-trivial.

Kingslake's book is a standard:
https://www.amazon.com/dp/012374301X/?tag=pfamazon01-20

And Buchdahl's book is the most comprehensive:
https://www.abebooks.com/Optical-Ab...MI9fPZ7N6q9QIVzGtvBB2NGAfrEAQYASABEgJWvvD_BwE
 
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If you have an optical design in hand, the best route to go is to simulate your optical design in an appropriate software product. There are many, but the most popular in research is Zemax. The problem is it's expensive as heck and their anti-pirating security measures make it a real pain in the butt for legitimate paying users (hardware keys, remote access versions, etc.).

There are software packages that offer limited free trial versions. I'm partial to WinLens Basic from QIoptiq, but I'm sure you could find others out there. You can find WinLens Basic tutorial videos on youtube.

If you're curious, these software products use ray-tracing methods to calculate aberrations. There are also analytical formulas, which I'm sure you could find in either of the books @Andy Resnick listed.

VVS2000 said:
I read recently that plano concave lens has a negative aberration and plano convex lens is used to correct it.
This is highly dependent on what you're trying to accomplish. There's no one-size-fits-all solution.

Now, if you're trying to build understanding about spherical aberration (which is what it sounds like you're talking about in the first post), then I recommend you try this exercise:

Consider a converging beam of light initially with a convergence angle of ##\theta## that passes through a slab of glass at normal incidence. Calculate the focal point as a function of ##\theta##. Notice that the answer depends on ##\theta##. That means the center of the beam has a different focal point than the outer rim of the beam. This results in a caustic beam profile (as opposed to a Gaussian beam).

If you solve the above problem but approximate the ##\sin \theta## in Snell's law by ##\theta##, you get the answer in the absence of all aberrations. If you solve it by approximating Snell's law by ##\theta + \frac{1}{3!} \theta^3 ##, you get the answer including 3rd order spherical aberration, and so on. I hope that gives you a little insight into what's going on.
 
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Andy Resnick said:
The calculation gets extremely complicated very quickly, and there aren't any simple references, either. In general, aberration theory is rooted in either Seidel coefficients or Zernike coefficients, and converting between the two is non-trivial.
Well not soo deep at the theoretical level but The thing I want to understand is on more of an experimental level. Coz I was doing experiments with lenses which have 10cm diameter and 15 cm focal length, and all rays parallel to principal axis don't focus on the focal point. So aberration does exist but I cannot or don't know how to quantitatively put it across
 
Twigg said:
Consider a converging beam of light initially with a convergence angle of θ that passes through a slab of glass at normal incidence. Calculate the focal point as a function of θ. Notice that the answer depends on θ. That means the center of the beam has a different focal point than the outer rim of the beam. This results in a caustic beam profile (as opposed to a Gaussian beam
I am so sorry but do you have like a diagram that shows what you're trying to say, I might not able to get which angle and rays that you might be reffering to
 
VVS2000 said:
Coz I was doing experiments with lenses which have 10cm diameter and 15 cm focal length, and all rays parallel to principal axis don't focus on the focal point.
The diameter of the lens is more or less irrelevant (assuming you have a separate stop that isn't the lens itself). It's the radial distance of the marginal rays (or equivalently the numerical aperture (NA)) that matters.

If you want to quantify the amount of aberration in your system, try using an iris diaphragm to vary the NA of your system. Measure how much the focal point moves when you close the iris vs when you fully open the iris. The more the focal point moves, the worse your aberration is.

VVS2000 said:
I am so sorry but do you have like a diagram that shows what you're trying to say, I might not able to get which angle and rays that you might be reffering to
Yeah, sorry I was being lazy. Here you go:

1641969090110.png

Two rays incident at angle ##\theta## on a rectangular slab of glass. When they hit the glass, they're separated by a distance of ##2h##. Slab has a thickness ##t##. Find the position of the focal point from ##\theta##, ##h##, ##t##, and ##n## (index of the slab). Expand your answer as a series for small ##\theta##. The lowest term should be cubic and it refers to 3rd order spherical aberration.

Notice that your optical design contains a range of values of ##\theta## from 0 up to your aperture stop. That means your focal point will be smeared out over a range, giving you a caustic.
 
Twigg said:
The diameter of the lens is more or less irrelevant (assuming you have a separate stop that isn't the lens itself). It's the radial distance of the marginal rays (or equivalently the numerical aperture (NA)) that matters.

If you want to quantify the amount of aberration in your system, try using an iris diaphragm to vary the NA of your system. Measure how much the focal point moves when you close the iris vs when you fully open the iris. The more the focal point moves, the worse your aberration is.Yeah, sorry I was being lazy. Here you go:

View attachment 295381
Two rays incident at angle ##\theta## on a rectangular slab of glass. When they hit the glass, they're separated by a distance of ##2h##. Slab has a thickness ##t##. Find the position of the focal point from ##\theta##, ##h##, ##t##, and ##n## (index of the slab). Expand your answer as a series for small ##\theta##. The lowest term should be cubic and it refers to 3rd order spherical aberration.

Notice that your optical design contains a range of values of ##\theta## from 0 up to your aperture stop. That means your focal point will be smeared out over a range, giving you a caustic.
Ok, will do it
Thanks for the image!
 
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