What should be a very simple optics question is giving me problems (two lenses)

In summary, the conversation is about a problem involving two converging, thin lenses and finding the final image location. The approach of using the image from the first lens as the object for the second lens is incorrect and leads to nonsensical answers. The correct approach is to treat the final image as a virtual object with a distance of -10 cm from the second lens. This results in a final image location of 5 cm to the right of the second lens.
  • #1
FatCat0
32
0
I'm studying for the GREs, and I ran across this problem:

Object located at x=0
Lens 1 located at x=40cm
Lens 2 located at x=70cm (30cm from lens 1)

f1 = 20cm
f2 = 10cm

Both lenses are converging, thin lenses. So I have to find the final image location.

The understanding I have of multiple lens problems is that you find the image from the first lens and treat that as the object for the second lens. I'm guessing this approach is wrong because it's leading to nonsensical answers:

[itex]\frac{1}{f1}[/itex] = [itex]\frac{1}{o}[/itex] + [itex]\frac{1}{i}[/itex]

[itex]\frac{1}{20}[/itex] = [itex]\frac{1}{40}[/itex] + [itex]\frac{1}{i}[/itex]

[itex]\frac{2}{40}[/itex] - [itex]\frac{1}{40}[/itex] = [itex]\frac{1}{i}[/itex]

40 = i = o' (this is 10 cm to the right of the second lens; on the focal point which means that I'm going to get the second image at x=-infinity)

[itex]\frac{1}{f2}[/itex] = [itex]\frac{1}{o'}[/itex] + [itex]\frac{1}{i'}[/itex]

[itex]\frac{1}{10}[/itex] = [itex]\frac{1}{40-30}[/itex] + [itex]\frac{1}{i'}[/itex]

[itex]\frac{1}{10}[/itex] - [itex]\frac{1}{10}[/itex] = [itex]\frac{1}{i'}[/itex]

i' = infinity

The answer is actually i' = 5 cm to the right of the second lens. I have no idea how to get to this though. Where'd I go amiss?
 
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  • #2
FatCat0 said:
The answer is actually i' = 5 cm to the right of the second lens. I have no idea how to get to this though. Where'd I go amiss?
Since the image is 10 cm to the right of the second lens, it must be treated as a virtual object: Use an object distance = -10 cm (not + 10 cm, which would represent a real object 10 cm to the left of the lens).
 
  • #3
That fixes the side (thank you), but won't that still result in +infinity as the location of the final image?

Edit: Nevermind, I actually redid it and realized that I jumped the gun. That worked, thanks!
 
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1. How do I determine the focal length of a lens system with two lenses?

The focal length of a lens system with two lenses can be determined by using the formula: f = (f1f2) / (f1 + f2 - d), where f1 and f2 are the focal lengths of the individual lenses, and d is the distance between the two lenses.

2. How do I determine the magnification of a lens system with two lenses?

The magnification of a lens system with two lenses can be determined by using the formula: M = -(f2 / f1), where f1 and f2 are the focal lengths of the individual lenses. This formula assumes that the object being viewed is placed at the focal point of the first lens.

3. How do I determine the power of a lens system with two lenses?

The power of a lens system with two lenses can be determined by using the formula: P = (P1 + P2 - d/f1f2), where P1 and P2 are the powers of the individual lenses, and d is the distance between the two lenses.

4. What is the difference between a converging and diverging lens?

A converging lens is one that causes light rays to bend towards a central point, while a diverging lens causes light rays to spread out. Converging lenses have a positive focal length, while diverging lenses have a negative focal length.

5. How do I determine the image distance of a lens system with two lenses?

The image distance of a lens system with two lenses can be determined by using the formula: di = (d1f2 - d2f1) / (f2 - f1), where d1 and d2 are the distances between the first lens and the object, and between the second lens and the image, respectively.

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