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FatCat0
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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?
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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