Question

A virtual, erect and magnified image of an object is to be produced with a concave mirror of focal length ${\text{12}}$cm. Which of the following object distances should be chosen for this purpose?
\[
  {\text{A}}{\text{.10}}\,{\text{cm}} \\
  {\text{B}}{\text{.14}}\,{\text{cm}} \\
  {\text{C}}{\text{.18}}\,{\text{cm}} \\
  {\text{D}}{\text{.24}}\,{\text{cm}} \\
 \]

Answer 1

Hint: Concave mirror forms virtual, erect, magnified image if object is placed at distance less than focal length. Out of various options for positioning the object, the position of the object at a distance less than the focal length produces a magnified image.

Complete step by step answer: Given that focal length of concave mirror ${\text{ = 12}}\,{\text{cm}}$
Now, seeing the image properties for concave mirrors we find that virtual, erect and magnified images are formed when object distance (d) is less than focal length (f).
\[{\text{d < f}}\]
Now, in the present question, on investigation we find that only one value ${\text{10}}\,{\text{cm}}$ is less than focal length.

So, the correct answer is “Option A”.

Additional Information:
1. The equation connecting the distances of the object and image with the focal length of a lens is called Gaussian form of lens equation or simply lens equation $\dfrac{1}{u} + \,\dfrac{1}{v}\, = \dfrac{1}{f}$
Where $u = $distance of object from lens
$v = $distance of image from lens
$f = $focal length
2. Linear magnification is the ratio of image formed by lens to the size of object.
3. The ability of a lens to converge or diverge the rays of light incident on it is called power of the lens$(P)$.
If $f = 1m$
$P = 1$dioptre where dioptre is unit of power of lens.

Note:
A concave lens always produces virtual and erect images. There are various types of possibilities of output regarding the size of image depending upon position of object with respect to focal point. A convex mirror can never form a magnified image.