If the image formed by the first lens is 0.03 m left of that lens, you add 0.10 m to get the object distance measured to the left of lens 2.
I get for the image location di second lens
1/di = 1/0.06 - 1/0.13
di = 0.1114 also
With two significant figures, that would be 0.11 to the right of lens 2.
Chack for math errors. The formula and method should be correct.
A converging lens has a focal length of 0.060 m. An object is located 0.020 m to the left of this lens. A second converging lens has the same focal length as the first one and is located 0.100 m to the right of it. Relative to the second lens, where is the final image located?
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Physic please help! - drwls, Saturday, October 27, 2007 at 2:37am
The image formed by the first lens is virtual, and is located at X given by
1/0.02 + 1/X = 1/0.06
1/X = -33.33
X = -0.03
The minus sign indicates it is 0.03 m to the LEFT of lens 1. That becomes the object position for the second lens, and it is 0.13 m to the left of lens 2. For the location of the final image, solve
1/0.13 + 1/Di = 1/0.06
Do is the image distance, measured to the right of lens 2.
I understand the workingout all the way up to the object distance to the second lens is 0.13m. I am unclear how to get to the distance of the final image. I know to use the formula
1/di= 1/f - 1/d0
when i use this formula I get 0.11143m but the system is saying that this is incorrect. Please help!
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