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I have two positive thin lenses that are separated by a distance of $5 cm$. The focal lengths of the lenses are $F_1 = 10 cm$ and $F_2 = 20 cm$. I placed an object 2 cm to the left of the front focal point and calculated the image by using the equation $\dfrac{1}{f} = \dfrac{1}{s_o} + \dfrac{1}{s_i}$ twice, so that the image of the first lens became the object of the second lens.

My system looks like this:

enter image description here

(credit to www.livephysics.com)

I am then told to insert two thick bi-convex lenses of $10 cm$ and $20 cm$ effective focal lengths into the system instead of the thin optical lenses, so that I get the same image with the same object $2 cm$ to the left of the front focal point. These thick lenses have all of the typical information available that one would find in a lens catalogue (radius of curvature, principal plane distances, refractive index, etc.), and I can provide this information if requested.

After insertion of the thick lenses, my system looks like this:

enter image description here

(From Optics, Fifth Edition, by Hecht.)

(Note that $d$ in the thick lens picture above is different to the distance that I am describing below.)

I'm now trying to find the distance between the back surface of the first lens and the front surface of the back lens.

I have two problems:

  1. I'm not sure that I'm correctly interpreting what is meant by inserting the thick lenses, so that one gets the same image with the same object $2 cm$ to the left of the front focal point.

  2. Despite a tremendous amount of research, I don't understand how it is possible to infer the the distance between the two thick lenses that were inserted instead of the thin lenses. I wondered if I was just misunderstanding what is exactly meant by "inserting" thick lenses instead of thin lenses in a system, but I have looked through a lot of optics resources and found nothing that indicates that the lenses must be "inserted" in a specific way that allows for deduction of the distance. I am told that the distance is somewhere between $4 - 5cm$, but I don't understand how such a thing can be calculated. Since this value is so close to the original thin lens distance of $5cm$, this leads me to believe that I am completely misunderstanding something about the nature of thick lens systems and how they are "inserted" into a thin lens system. Could it be that, to "insert" a thick lens instead of a thin lens means to align the secondary principal plane of the first thick lens with the first thin lens, and align the primary principal plane of the second thick lens with the second thin lens, so that the distance between the secondary principal plane of the first lens and the primary principal plane of the second lens is $5 cm$?

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ Maybe add a homework tag? At least it sounds like it. In that case, people can help you get to the answer but shouldn't provide it directly. $\endgroup$ – relatively_random Dec 4 '19 at 8:27
  • $\begingroup$ @relatively_random tag added. $\endgroup$ – The Pointer Dec 4 '19 at 9:03
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At this level of approximation (parallax approximation/Gaussian optics), a thick lens is nothing but a thin lens with rays "skipping" or "teleporting" through the distance between the two principal planes.

So if you use a thick lens with the same focal distance as the thin one, just place its front principal plane where the thin lens used to be. Then for the following components, measure all the same distances you had for the thin lens, but measure them from the back principal plane this time. In effect, you're just adding the thickness of the lens to the whole system - nothing more and nothing less.

The thickness is just the distance between the principal planes that is getting "skipped over". If the thickness is positive, the rear principal plane is behind the front one and the rays skip forward. If the thickness is negative, the rear principal plane is in front of the front one (maybe a bit confusing) and the rays jump back.

As for what the question meant by "same image", I don't know but I'd guess it means same orientation, same magnification.

This sounds a bit like a homework question so I won't post any specific numbers, but hopefully I helped you understand the nature of the thick lens to see it's almost as simple as dealing with thin ones.

EDIT:

Here's what transforming a thin lens into a thick lens means. It's still the same lens, just has a gap in between. Where the rays disappear is the front principal plane, where they reappear is the rear principal plane.

enter image description here

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  • $\begingroup$ Thanks for the answer! "So if you use a thick lens with the same focal distance as the thin one, just place its front principal plane where the thin lens used to be." Can you please explain the reasoning behind why this is done? Why is it that, for this case, if the thick lenses have the same effective focal length as the thin lenses, then we are placing them in this way? $\endgroup$ – The Pointer Dec 4 '19 at 9:03
  • $\begingroup$ If the focal length is the same, all the ray-bending properties will be the same. So you can design the system with the thin lenses and add the thicknesses at the end, if you wish. To do that, you set up the distances so the rays don't realize anything's changed. Thick lens will teleport the rays from the front plane to the rear one. If you don't wish the rays to realize anything's changed, you put the entrance into the lens (front p.p.) at the same spot as before. Then if you set all subsequent elements relative to the exit (rear p.p.), the rays "see the same world" when they leave the lens. $\endgroup$ – relatively_random Dec 4 '19 at 9:19
  • $\begingroup$ Hmm, I understand the "teleporting" nature of the rays through the gap between the principal plane, but what I'm struggling to understand is how one uses what you have described to find the distance. Some additional information: For the thin lenses, I combined them into a single system by drawing in their principal planes (so, reducing a compound thin lens system into a single thick lens), and then I calculated the image distance (using the method I described about using the thin lens equation twice, treating image 1 as object 1 for the second lens) [...] $\endgroup$ – The Pointer Dec 4 '19 at 9:31
  • $\begingroup$ [...] to be 36cm (that is, 36cm from the back principal plane of the combined thin lens system, or $36 - 6 = 30cm$ from the back focal point of the combined thin lens system, since $6cm$ is the back focal length). $\endgroup$ – The Pointer Dec 4 '19 at 9:31
  • $\begingroup$ I'm not sure I understand what you are struggling with. What "same" image have you decided to try and find? (1) The image which is at the same physical position as the previous one was, or (2) the image that has the same size and orientation? The second option should be trivial if you understand the teleporting nature correctly, since the image has to be the same size but not necessarily at the same position. The first would probably take a bit of exploration and equation fiddling, which I haven't tried myself. BTW I added a dirty animation, not sure if it helps. $\endgroup$ – relatively_random Dec 4 '19 at 9:50

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