What Shifts When Lenses Are Swapped in a Dual-Lens System?

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SUMMARY

The discussion focuses on the optical behavior of a dual-lens system comprising a converging lens and a diverging lens, both with a focal length of 5.0 cm, placed 5.0 cm apart. When the object is positioned at the focal point of the converging lens, the image formed is at infinity, leading to an undefined image position when calculating through the lens system. The concept of infinite conjugate is crucial, as it indicates that interchanging the lenses alters the image position significantly, necessitating the use of the thin lens equation to derive the final image location.

PREREQUISITES
  • Understanding of optics principles, specifically lens behavior.
  • Familiarity with the thin lens equation: 1/f = 1/o + 1/i.
  • Knowledge of converging and diverging lenses and their focal lengths.
  • Concept of infinite conjugate in lens systems.
NEXT STEPS
  • Study the thin lens equation in detail and its applications in various lens configurations.
  • Explore the concept of infinite conjugate and its implications in optical systems.
  • Learn about ray diagrams for both converging and diverging lenses to visualize image formation.
  • Investigate practical applications of dual-lens systems in optical devices like cameras and microscopes.
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Students studying optics, physics educators, and anyone interested in understanding the principles of lens systems and image formation in optical devices.

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Physics Homework Help Needed: Optics--Lens

Here's the physics problem that I need help:

A converging and diverging lens are placed 5.0 cm apart, and an object O is placed 5.0 cm from the first lens. Each lens has a 5.0 focal length. What happens to the image position if the lenses are interchanged?

Here's How I solved it:
My plan was to find the image of the object going through the first lens and then use that image as the object for the second lens. Interestingly, I kept getting the final image as 1/0, which is undefined? My first instinct was that there's something wrong with the problem!

So I went to a physics teacher for help. He said that when you have an object being at the position of the focal length, there's NO converging. He asked if my professor lectured on infinite conjugal. I said "no", I've never heard of it.

So in order to solve this problem, what do you do? Do you use limit? Do you neglect the image of the object going through the first lens and just use the object as it is going through the second lens?

I need help...I don't want the answer...I just want to know how you'd go through solving this problem...
 
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Hi there,

Thank you for reaching out for help with your physics homework. I will do my best to guide you through the problem and help you understand the concepts involved.

First, let's review the basics of optics and lens. A converging lens has a positive focal length and can form real, inverted images. On the other hand, a diverging lens has a negative focal length and can only form virtual, upright images.

In the problem, we have two lenses, one converging and one diverging, placed 5.0 cm apart. The object is placed 5.0 cm from the first lens and both lenses have a focal length of 5.0 cm. This means that the object is at the focal point of the first lens, which is why you were getting 1/0 as the final image position.

To solve this problem, we need to understand the concept of infinite conjugate. This occurs when the object is placed at a distance equal to the focal length of a converging lens. In this case, the image formed by the lens will be at infinity. This is important because when we interchange the lenses, the object will now be at infinity, and we can use this as our starting point for the second lens.

So, to solve this problem, we will first find the image of the object formed by the first lens, which will be at infinity. Then, we will use this image as the object for the second lens and calculate the final image position.

To find the image formed by the first lens, we can use the thin lens equation: 1/f = 1/o + 1/i, where f is the focal length, o is the object distance, and i is the image distance. Since the object is at infinity, we can simplify the equation to 1/i = 1/f. Plugging in the values, we get i = f = 5.0 cm. This means that the image formed by the first lens will be at 5.0 cm from the lens, which is at infinity.

Now, for the second lens, we can use the same equation, but this time our object distance will be the image formed by the first lens, which is at 5.0 cm. Plugging in the values, we get 1/i' = 1/f + 1/o'. Since the object distance is now the image formed by the first lens, which is at infinity, we can
 

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