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!

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.

To find the distance of the final image, you need to use the lens formula for the second lens:

1/di = 1/f - 1/do

Where:
di = distance of the final image
f = focal length of the lens (which is the same as the first lens, given as 0.060 m)
do = distance of the object from the second lens (which is the distance of the image formed by the first lens)

From the given information, the image formed by the first lens is located 0.030 m to the left of the lens (negative because it's on the left side). This becomes the object position for the second lens, so do = -0.030 m.

Substituting the values into the formula:

1/di = 1/0.060 - 1/-0.030

To simplify this expression, you need to multiply both the numerator and the denominator by -0.030 * 0.060:

1/di = (-0.030 * 0.060) / (0.060 * -0.030) - (-0.030 * 0.060) / (0.060 * -0.030)

1/di = -0.0018 / (-0.0018)

1/di = 1

Taking the reciprocal of both sides:

di = 1 m

Therefore, the final image is located 1 meter to the right of the second lens.