The objects named $K, L$ and $M$ are placed in front of a concave mirror as shown in the figure below correctly showing the images of the objects?

A.

B.

C.

D.

Hint: We are given three objects placed at three different positions in front of a concave mirror and we need to find the positions of the image formed of each object. We need to be clear with concepts of image formation by a concave mirror for doing such problems.

Complete step by step answer:
The object $L$ is placed at the center curvature of the mirror which is equal to twice the length of the focal length of the mirror. Therefore using mirror formula for the given situation
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Where $u$ is the object distance from the mirror, $v$ is the image distance and $f$ is the focal length
Here $u = 2f$, therefore substituting the value we get,
$ \Rightarrow \dfrac{1}{v} + \dfrac{1}{{2f}} = \dfrac{1}{f}$
$ \Rightarrow v = - 2f$
This means the object will be formed at the same distance from the mirror where the object is placed but will be inverted.

Now the magnification can be calculated using $m = - \left( {\dfrac{v}{u}} \right)$.
Substituting the value we get,
$ \therefore m = - 1$
This shows that the object will be the same as that of the object. Now for object $K$, the value of the object distance from the mirror will be more than $2f$ which will give a minimized image and the image distance will be less than $2f$ and more than $f$. For object $M$ the value of the object distance from the mirror will be between $2f$ and $f$ which will give a magnified image and the image distance will be more than $2f$.

Hence, option D is correct.

Note: We should note that the concave mirror produces both real and virtual images. When the object is placed very close to the mirror, a virtual and magnified image is obtained and if we increase the distance between the object and the mirror, the size of the image reduces and real images are formed.