Gaussian theory for the clever , the insane or the insanely clever

[marcuslucas]

Hi guys ,
I am having difficulty with a Gaussian optics question and was wondering if anybody is able to offer me any kind of insight.

A real object 20cm from the first focal point of a coaxial system of two thin lenses forms a real image 45cm from the second focal point. If the focal length of the first lens is +12.5cm and the lens separation is 4cm, find the power of the second lens ?.

The answer given for this question is the second lens F2 is -6.86D.

My own answers are nowhere near this and I must be making some conceptual mistakes, my approach to this question is as follows:

the real object is placed -0.325m (-12.5cm + -20cm) from the front vertex of the front lens and has a vergence of -3.07D
the image formed right of the second lens is +0.575m (12.5cm+45cm) and has a vergence of +1.73D
the lens seperation d is 0.04m
then applying the conjugate foci equations
1/l = -3.07D=L1
L1+F1= L1' = -3.07 + 8.00D = +4.92D
L2 = L1'/1-dL1' = +6.125D
L2' = 1/0.575 = +1.73D
L2'-L2 = +1.73D - +6.125D = -4.39D = F2 hence the wrong answer

Any advice or insight would be greatly appreciated
Thanks in advance.

[Darryl Meister]

I suspect the problem is in the potential ambiguity of the question. When dealing with the total focal power of multiple lenses in conjunction with the individual focal lengths of those lenses, describing the object distance as "20 cm from the first focal point" becomes ambiguous.

Assuming that the object is 20 cm away from the primary focal point of the first lens (i.e., 12.5 cm), as you have, is very different from assuming that the object is 20 cm away from the primary focal point of the entire system. The same goes for image the distance relationship relative to the secondary focal point.

After doing a quick step-along vergence calculation using their specified answer, -6.86 D, I calculated the primary focal length of the entire system as -38.21 cm from the first lens and the secondary focal length as 20.39 cm from the second lens.

Assuming that the object distance was measured relative to the primary focal point of the entire system, I confirmed that an object distance of -38.21 - 20 = - 58.21 cm produces an image distance of 65.34 cm with these two lenses, which is 65.34 - 20.39 = 44.95 cm, almost certainly within rounding errors of the specified value of 45 cm.

On a side note, since you are given "extra-focal" distances for the object and image positions, you may also trying solving this problem using Newton's relationship, although this will require working with the equivalent power of the lenses. I will leave that as an exercise for the reader, as they say.


Best regards,
Darryl

[marcuslucas]

Thanks Darryl for your time and input , definitely food for thought , I will certainly try approaching this problem with a different perspective.
Thanks for the in depth explanation.

[Darryl Meister]

Glad to help. Just remember that, if you continue solving the problem using step-along vergence, you can calculate the primary and secondary focal lengths of the system using the front and back vertex power equations:



and



where t is the separation (in meters) and F₁ and F₂ are the front and back lenses, respectively. Taking the reciprocal of these powers will provide you with the distances of the primary and secondary focal points from the lenses. Add your "extrafocal" distance to these values, and you can then calculate the conjugate foci for each lens.

Best regards,
Darryl