Two thin lenses one with 12 cm focal length and the other with a -6.0 cm focal length are placed 18 cm apart.Locate the final image of an object placed on the axis 18 cm to the left of the positive Lena.
To locate the final image formed by the two lenses, we can use the lens formula:
1/f = 1/v - 1/u
where:
- f is the focal length of the lens,
- v is the image distance from the lens, and
- u is the object distance from the lens.
Given:
Lens 1: f1 = 12 cm
Lens 2: f2 = -6.0 cm (negative focal length indicates a diverging lens)
Distance between the lenses: d = 18 cm
We'll start by finding the image formed by Lens 1 and then use that as the object for Lens 2.
1. Image formed by Lens 1:
Given that the object is placed 18 cm to the left of Lens 1 (positive lens), the object distance u1 = -18 cm.
Using the lens formula, we have:
1/f1 = 1/v1 - 1/u1
Substituting the given values:
1/12 = 1/v1 - 1/(-18)
1/12 = 1/v1 + 1/18
Combining the fractions:
1/v1 = 1/12 - 1/18
1/v1 = 3/36 - 2/36
1/v1 = 1/36
Taking the reciprocal:
v1 = 36 cm
So, the image formed by Lens 1 is 36 cm to the right of Lens 1.
2. Object distance for Lens 2:
As Lens 2 is located 18 cm to the right of Lens 1 and the image formed by Lens 1 is 36 cm to the right of Lens 1, the object distance u2 for Lens 2 will be:
u2 = v1 - d
u2 = 36 - 18
u2 = 18 cm
The image distance v2 formed by Lens 2 can be found similarly using the lens formula:
1/f2 = 1/v2 - 1/u2
Substituting the given values:
1/-6 = 1/v2 - 1/18
Combining the fractions:
1/v2 = -1/6 + 1/18
1/v2 = -3/18 + 1/18
1/v2 = -2/18
Taking the reciprocal:
v2 = -9 cm
The negative sign indicates that the image formed by Lens 2 is virtual and on the same side as the object.
Therefore, the final image of the object placed on the axis 18 cm to the left of Lens 1 can be located 9 cm to the left of Lens 2.