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?
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.
To determine the location of the final image, we can use the lens formula:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance from the lens
u = object distance from the lens
For the first lens:
f1 = 0.060 m
u1 = -0.020 m (negative because the object is to the left of the lens)
1/0.060 = 1/v1 - 1/-0.020
Simplifying the equation, we get:
(1/v1) + (1/0.020) = 1/0.060
(1/v1) = 1/0.060 - 1/0.020
(1/v1) = (1 - 3)/0.060
(1/v1) = -2/0.060
v1 = -0.030 m
The image formed by the first lens is located 0.030 m to the left.
Now, let's consider the second lens:
f2 = 0.060 m
u2 = 0.100 m (positive because the image from the first lens is to the right of the second lens)
We can use the lens formula again:
1/0.060 = 1/v2 - 1/0.100
(1/v2) + (1/0.100) = 1/0.060
(1/v2) = 1/0.060 - 1/0.100
(1/v2) = (10 - 6)/0.0600
(1/v2) = 4/0.060
v2 = 0.015 m
The image formed by the second lens is located 0.015 m to the right.
Therefore, relative to the second lens, the final image is located 0.015 m to the right.