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

[Physicsme]

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...

 

[anathema]

Thank you for reaching out for help with your optics problem. It seems like you have a good understanding of the concepts involved, but are running into some confusion with the specific details of this problem.

First of all, it is important to note that in optics, the image position is typically measured from the lens itself, not from the focal length. So in this case, the image position would be measured from the first lens, which is 5.0 cm from the object.

Now, let's look at what happens when the lenses are interchanged. If we use your approach of finding the image through the first lens and then using that as the object for the second lens, we can see that the image through the first lens will be at a distance of 5.0 cm from the second lens. This means that the image will be at the focal point of the second lens, since the focal length is also 5.0 cm.

However, when we use the image as the object for the second lens, we need to remember that the image is inverted. This means that the image will now be on the opposite side of the second lens, and its distance from the lens will be negative. So, the final image position would be -5.0 cm.

In summary, when the lenses are interchanged, the image will be at a distance of -5.0 cm from the second lens. This is because the image through the first lens is at the focal point of the second lens, and when we use that image as the object, it is inverted. I hope this helps to clarify the problem for you.

If you have any further questions or need more explanation, please don't hesitate to ask. Keep up the good work in your physics studies!

 

[thepatientmental]

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