Question

A concave mirror gives virtual, erect and enlarged image of the object but image of smaller size than the size of the object is
(A) At infinity
(B) Between F and C
(C) Between P and F
(D) At F

Answer 1

Hint: To answer this question we need to use the magnification formula. From there we can find the correct answer from the options. The formulae used to solve this question are given by
- ${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
- $m=-{\displaystyle \frac{v}{u}}$ .

Complete Step-by-Step solution:
We know that the magnification produced by a mirror is given by
 $m=-{\displaystyle \frac{v}{u}}$ ......................(1)
Now, from the mirror formula we have
 ${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
Multiplying with $-u$ on both the sides, we get
 ${\displaystyle \frac{-u}{v}}+{\displaystyle \frac{-u}{u}}={\displaystyle \frac{-u}{f}}$
 $\Rightarrow -{\displaystyle \frac{u}{v}}=1-{\displaystyle \frac{u}{f}}$
Taking LCM on the RHS, we have
 $-{\displaystyle \frac{u}{v}}={\displaystyle \frac{f-u}{f}}$
Taking the reciprocals of both the sides, we get
 $-{\displaystyle \frac{v}{u}}={\displaystyle \frac{f}{f-u}}$
From (1)
 $m={\displaystyle \frac{f}{f-u}}$ ......................(2)
Now, we know that according to the Cartesian sign convention, both the object distance and the focal length of a concave mirror are taken as negative. This means that we can write
 $f=-\left|f\right|$ ......................(3)
Also
 $u=-\left|u\right|$ ......................(4)
Putting (3) and (4) in (2) we get
 $m={\displaystyle \frac{-\left|f\right|}{-\left|f\right|+\left|u\right|}}$
Multiplying the numerator and the denominator by $-1$ we get
 $m={\displaystyle \frac{\left|f\right|}{\left|f\right|-\left|u\right|}}$
According to the question, the image formed is virtual, erect and enlarged. Since the virtual and erect, the magnification is positive. This means that
 ${\displaystyle \frac{\left|f\right|}{\left|f\right|-\left|u\right|}}>0$
 $\Rightarrow \left|f\right|>\left|u\right|$
Or
 $\left|u\right|<\left|f\right|$
So the object distance is less than the focal length of the mirror.
This means that the object must be kept between the pole and the focus of the concave mirror.
Hence, the correct answer is option (C).

Note:
We could also attempt this question by using the ray diagram for the concave mirror. The case considered in the above question is the special case of the concave mirror where it produces virtual images. But remember, mathematical analysis is always more reliable.