Question

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

Answer 1

Define the image formation process in case of concave mirror. The properties of the image formed depend on the position of the object. Find the different conditions for which we obtain different kinds of image. Also, confirm your conclusion using the mirror formula and magnification formula.

Complete answer:
When an image is formed in a concave mirror, we can get both the virtual, erect and real, inverted image. Depending on the position of the object in front of the mirror, the property of the image formed will change. Let us check the nature of the image formed for particular positions.
If the position of the object is beyond the centre of curvature of the mirror, the image formed will be real, inverted and diminished.
If the position of the object is in between the focal point of the mirror and the centre of curvature of the mirror, the image formed will be real, inverted and magnified.
For the object that is kept at the centre of curvature of the mirror, then the image formed will be real, inverted and of the same size as the object.

If the object distance is less than the focal length of the mirror, we will get a virtual erect and magnified image.
In this question, the focal length of the concave lens is 12cm. So, for a virtual, erect and magnified image to be formed, the object distance must be less than the focal length of the mirror. Among the given options only the 10cm distance is less than the focal length of the mirror. So, for this purpose we will choose object distance as 10cm.
Now recall the mirror formula and substitute these values in it.
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
$\Rightarrow v=\dfrac{uf}{u-f}$
We are given focal length to be, f=-12cm and assumed the object distance to be, u=-10cm.
$v=\dfrac{-10\times -12}{-10-\left( -12 \right)}=\dfrac{120}{2}$
$\therefore v=60cm$
So we found that the image formed will be 60cm behind the mirror, that is a virtual image.
Now from magnification formula,
$m=-\dfrac{v}{u}$
$\Rightarrow m=-\dfrac{60}{-10}$
$\therefore m=6$
So we found the magnification to be positive 6. So, the image formed will definitely be enlarged. Another formula for magnification is given as the ratio of height of the image to the height of the object.
$m=\dfrac{{{h}_{i}}}{{{h}_{0}}}$
Since the height of the object is known to be positive by convention, for the magnification to be positive, the height of the image should also be positive. So, the image formed is also erect.

Hence, the correct option is (A).

Note:
In case of concave mirrors we get different kinds of images depending on the position of the object and so we have discussed the nature of the image formed corresponding to each position. But in case of convex mirrors, the image properties are always the same. For any position of the object, the image formed in the convex mirror is always virtual, erect and diminished.