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For a mirror, the linear magnification $m$ comes out to be $ + 2$ . What conclusion can be drawn from this?
A) The mirror is concave.
B) The mirror can be convex or concave but it cannot be a plane mirror.
C) The object lies between the pole and the focus.
D) The object lies beyond the focus.

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Last updated date: 17th Jun 2024
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Answer
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Hint: The magnification of a mirror refers to how big or small is the produced image with respect to the object size. It can be defined as the ratio of the negative of the image distance to the object distance or as the ratio of the height of the image to the height of the object. The object is always placed to the left of the mirror and the object distance is always taken to be negative.

Formula Used:
The linear magnification of a mirror is given by, $m = \dfrac{{{h_i}}}{{{h_o}}} = \dfrac{{ - v}}{u}$ where ${h_i}$ and ${h_o}$ are the sizes of the image and object respectively while $v$ and $u$ are the image distance and object distance respectively.

Complete step by step answer:
Step 1: Express the relation for the linear magnification of a mirror to find the nature of the image formed.
The linear magnification of a mirror is given by, $m = \dfrac{{{h_i}}}{{{h_o}}}$ - (1) 
where ${h_i}$ and ${h_o}$ are the sizes of the image and object respectively.
Or it can be defined as $m = \dfrac{{ - v}}{u}$ - (2) where $v$ and $u$ are the image distance and object distance respectively.
Substituting the given value for $m =  + 2$ in equation (1) we get, $\dfrac{{{h_i}}}{{{h_o}}} =  + 2$ 
$ \Rightarrow {h_i} = 2{h_o}$
i.e., the size of the image increased or we say the image got magnified.
Only concave mirrors produce magnified images. 
So option A is correct.
Since the object is always placed in the upright position for the image formation, we have ${h_o} =  + ve$.
So the height of the image ${h_i}$ is also positive i.e., the image is erect.
Substituting the given value for $m =  + 2$ in equation (2) we get, $\dfrac{{ - v}}{u} =  + 2$ 
Since the object distance is negative ( $u =  - ve$ ), the image distance must be positive for the magnification to be positive i.e., $v =  + ve$ .
This suggests that the object is formed behind the mirror and hence it is a virtual image.
The concave mirror forms an erect, magnified and virtual image only when the object is placed between the pole and the focus. The image formation is given below.
seo images

So option C is also correct.
Thus the correct options are (A) and (C).

Note: A concave mirror is a converging mirror and hence will magnify the objects. The images formed by the convex mirror will be diminished in nature. A plane mirror, however, forms an image of the same size as that of the object. Positive $m$ means the image formed is virtual and erect. While constructing the ray diagram for the image formation, the ray passing through the centre of curvature gets reflected along the same path and the ray parallel to the principal axis passes through the focus upon reflection. The reflected rays are then retracted behind the mirror to obtain the image.