Solving for x in a Combination of Two Convex Lenses

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A combination of two thin convex lenses with equal focal lengths, separated by 20 cm, behaves as a lens system with an infinite focal length. When an object is placed 10 cm from the first lens, the image formed is virtual and located 10 cm from the second lens. The lens equation can be applied to the system as a whole, leading to a calculated image distance of -10 cm, indicating the image is virtual. However, further examples demonstrate that the relationship between object and image distances can vary, suggesting that the initial conclusion may be coincidental. The discussion highlights the complexities of lens systems and the need for careful application of optical principles.
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Homework Statement


A combination of two thin convex lenses of equal focal lengths,is kept separated along the optic axes by a distance of 20 cm between them.The combination behaves as a lens system of infinite focal length.If an object is kept at 10 cm from the first lens,its image will be formed on the other side at a distance x from the second lens.Find x.


Homework Equations


1/f=1/v + 1/u
where f is focal length,u is object distance and v is image distance.


The Attempt at a Solution


Can the above equation be applied for the lens system as a whole,instead of just a single lens at a time? If it can be applied,then I think the problem can be solved in the following way:
Since f of the lens system is infinity,this means that rays initially traveling parallel to the axis before the 1st lens,will continue to do so after passing through 2nd lens.The lens system behaves symmetrically.If we apply the above equation to the system as a whole,we get v= -10 cm and the image is virtual.Therefore x= 1o cm.Is this answer and reasoning correct?
 
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Hello,

I have tried to prove some relation that you want by some simple assumptions: (1)two identical lenses (2)the curvatures of two surface of one lens are the same (3) lenses are thin. But no such a general relation is satisfied. More intuitively, you can consider some special examples with the lenses of the problem:

Example1.
an point object is in front of the first lens with p_1=25\text{(cm)}
(i guess that you can calculate about the image problems with Gauss' formula)
\frac{1}{25}+\frac{1}{q_1}=\frac{1}{10}\quad\Rightarrow\quad q_1=\frac{50}{3}\text{(cm)}
p_2=20-\frac{50}{3}=\frac{10}{3}
\frac{1}{10/3}+\frac{1}{q_2}=\frac{1}{10}\quad\Rightarrow\quad q_2=-5\text{(cm)}
Therefore the image is a virtual one and between the two lenses with a distance 5(cm) from the second lens.
If one calculates again with your point of view, one always gets the same image distance with the distance between the object and the first lens (due to the infinite forcus length).

Example2.
an point object is in front of the first lens with p_1=20\text{(cm)}.
You can try this example without calculations.

Therefore, the problem is just a coincidence.



Hope these helpful.
 
I don't quite understand.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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