Answer
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Hint In a converging lens, all the rays that are parallel to the principal axis will converge to a fixed point on the principal axis. Such a lens is called a convex lens. Here, we have to find where we have to place the object to get a magnified image. We have to use a ray diagram to find the right answer.
Complete Step by step solution
In this case, the magnification of the image will happen when the object is placed between \[F\]and $2F$
The image will be formed on the opposite side of the lens and it will be located beyond $2F$ on the other side. The image will be inverted. The size of the image will be greater than that of the size of the object. This means that the image is magnified. Magnification is the ratio of the image size to the object size. For inverted images magnification will be negative and for real images, magnification will be positive.
Here the correct options are not listed among the given options.
Additional Information
The location of the image is where the light rays are converging. If we place a screen at that point we will get the projection of the image on the screen. When the object is placed on the focus $F$and on the center of curvature $C$the image will be of the same size as that object. So there will not be any magnification.
Note
Alternate answer:
The image will be magnified when the object is placed between the focus $F$and the optic centre $O$of the lens. The image will be formed on the same side as the object. The image will be erect and virtual. Therefore a converging lens is also used as a magnifying glass.
Complete Step by step solution
In this case, the magnification of the image will happen when the object is placed between \[F\]and $2F$
The image will be formed on the opposite side of the lens and it will be located beyond $2F$ on the other side. The image will be inverted. The size of the image will be greater than that of the size of the object. This means that the image is magnified. Magnification is the ratio of the image size to the object size. For inverted images magnification will be negative and for real images, magnification will be positive.
Here the correct options are not listed among the given options.
Additional Information
The location of the image is where the light rays are converging. If we place a screen at that point we will get the projection of the image on the screen. When the object is placed on the focus $F$and on the center of curvature $C$the image will be of the same size as that object. So there will not be any magnification.
Note
Alternate answer:
The image will be magnified when the object is placed between the focus $F$and the optic centre $O$of the lens. The image will be formed on the same side as the object. The image will be erect and virtual. Therefore a converging lens is also used as a magnifying glass.
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