Why Does an Object at a Lens's Focal Point Create an Image at the Same Location?

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An object positioned at the focal point of a converging lens produces an image at infinity, which reflects off a flat mirror, resulting in a second image that also appears at infinity. This second image acts as the object for the lens again, leading to an image that coincides with the original object's location. The discussion emphasizes that the mirror can be conceptually replaced by another converging lens at a specific distance, maintaining the same optical results. Additionally, the conversation highlights the importance of understanding the behavior of rays from both lenses and mirrors, rather than treating them in isolation. Overall, this insight simplifies the analysis of combined optical systems.
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Homework Statement



A small object is located on some distance from a converging lens. At some distant behind a converging lens lies a flat mirror. The resulting image is exactly at the same location as the small object. Why can we deduce that the object is located in the focal point of the converging lens?

The Attempt at a Solution



The first image created by the converging lens will be at infinity if the initial object is at the focal point of the lens. That image at the infinity is located in front of the flat mirror and when used as an object for the mirror, will produce another image at the infinity behind the mirror. This second image is then used as an object for the converging lens. Because this object is at infinity, the resulting image will exactly be on the focal length of the converging lens. Therefore the overall image is at the same place as the focal length.

Does this sound reasonable to you?
 
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That's quite reasonable - note, the mirror provides an image of the converging lens ... so the a mirror at distance d from the lens can be replaced by a converging lens a distance 2d away and you will get the mirror image of the results.

If the object is distance o in front of the 1st lens, and the image is a distance i from the 2nd, then the problem is to find the condition so that o=i.
 
Simon Bridge said:
That's quite reasonable - note, the mirror provides an image of the converging lens ... so the a mirror at distance d from the lens can be replaced by a converging lens a distance 2d away and you will get the mirror image of the results.

If the object is distance o in front of the 1st lens, and the image is a distance i from the 2nd, then the problem is to find the condition so that o=i.

Wow I have never thought it like that, thanks for pointing that one out!
 
There's also a matrix approach, but I suspect that you are only expected to give a qualitative account.
If you sketched the equivalent situation you should be able to see the only way you get o=i is if the rays from the 1st lens are parallel ... and that only happens when...

I think it is too easy to get used to treating these systems in the abstract.
Mirrors and lenses do not just deal with rays from the object, but from everywhere, and it is not just the principle rays that they deal with either. Worth bearing in mind.

This should now make a bunch of combined element problems suddenly easier.
Enjoy.
 
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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