When an object is located in point 3, the image produced by the converging lens is around near point?
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To determine where the image will be located for an object placed at point 3 in front of a converging lens, we can use the lens equation:
1/f = 1/do + 1/di
where:
- f is the focal length of the lens,
- do is the object distance (distance from the lens to the object), and
- di is the image distance (distance from the lens to the image).
From the given diagram, we can see that the object is located before the lens (do > 0), which means the object distance should be positive.
To find the location of the image, we need to know the values of the object distance (do) and the focal length (f). Unfortunately, the values are not directly provided in the diagram, so we can only make an estimation based on the diagram.
Looking at the diagram, we can see that the radius of curvature on both sides of the lens is the same, indicating a symmetrical lens. Therefore, we can assume that the focal length of the lens is positive (+).
Now, let's consider the different cases based on the distance of the object from the lens (do):
1. If the object is placed beyond 2F (twice the focal length):
- In this case, do > 2F, and the image will be real, inverted, and formed on the opposite side of the lens.
- The image distance (di) will be positive (+).
2. If the object is placed between F and 2F (focal length and twice the focal length):
- In this case, F < do < 2F, and the image will still be real, inverted, but formed on the same side of the object.
- The image distance (di) will be positive (+).
3. If the object is placed at F (focal length):
- In this case, do = F, and the image will be formed at infinity.
- The image distance (di) will be infinite (∞).
4. If the object is placed between the lens and F (focal length):
- In this case, 0 < do < F, and the image will be virtual, upright, and formed on the same side of the object.
- The image distance (di) will be negative (-).
Based on the given diagram, it seems that the object is located between the lens and F (focal length), putting it in case 4. Therefore, the image produced by the converging lens would be a virtual image formed on the same side of the object.