Question

The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(A) Between principal focus and centre of curvature
(B) At the centre of curvature
(C) Beyond the centre of curvature
(D) Between pole of mirror and its principal focus

Answer 1

Hint
A concave mirror is a spherical mirror that can produce real or virtual, magnified or diminished, upright or inverted based on the distance of an object from the pole of the mirror. The focus of a concave mirror is real and lies at half the distance of the pole from the centre of curvature.

Complete step by step answer
When the object is placed between the focal point and mirror, the pole of the mirror. The image formed is virtual, upright and magnified. As we approach the focus (moving leftwards), the image gets smaller with increasing distance, as the focus is reached, the rays become parallel which means that the image is formed at infinity. After this, all images that form are real and diminished in size.
This table sums up the behaviour of a concave mirror-

Object’s position (S) at-	$S<F$	$S=F$	$F<S<2F$	$S=2F$	$S>2F$	$S\to \mathrm{\infty }$
Image formed is-	Virtual	Not formed	Real	Real	Real	Real
Orientation-	Upright	-	Inverted	Inverted	Inverted	Inverted
Size-	Magnified	-	Magnified	Same size	diminished	Highly diminished
Image formed at-	Beyond mirror (inside)	Infinity	After $2F$	At $2F$	Between $F\text{ and 2}F$	$\to F$(approaches focal point)

Here S shows the position of the object,$\text{F}$ and $2\text{F}$are used as references which tell about the distance at which various images formed,
 F= focal length of the mirror,
$\therefore 2\text{F}$= Centre of curvature (C).
From this table it is clear that a concave mirror forms a virtual, upright and magnified image only when the object is kept between its pole and the Focus.
So, the correct answer is option (D).

Note
The image that is formed by a concave mirror when $S=F$is assumed to be highly magnified, It is real when $S>F$ and virtual when $F>S$. At exactly $S=F$ the image never forms because the rays are parallel, not diverging or converging.