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# A real image, equal in size of the object, is obtained when the object is placed at the centre of curvature of aA. Concave mirrorB. Plane mirrorC. Convex mirrorD. None of these Verified
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Hint:Concave mirror is a mirror of spherical; surface whose reflecting part is bent inwards and convex mirror is also a spherical mirror whose reflecting surface is bent outwards. A plane mirror is one who’s reflecting surface is a plane surface.

Let us examine the concave mirror first. It’s given that the position of the object is the centre of curvature which means $u = - R$ where $R$ is the radius of curvature of a spherical concave mirror. Focal length of a concave mirror is always negative let’s say,
$f = \dfrac{{ - R}}{2}$
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Putting the values of $f = \dfrac{{ - R}}{2}$ and $u = - R$ in this equation:
$\dfrac{{ - 2}}{R} = \dfrac{1}{v} - \dfrac{1}{R}$
$\Rightarrow \dfrac{{ - 2}}{R} + \dfrac{1}{R} = \dfrac{1}{v}$
$\therefore v = - R$
From above calculation, we get to know that if an object is placed at centre of curvature of a concave mirror its image forms at centre of curvature and hence also of same size due to the fact that magnification is equals to one as: $m = \dfrac{{ - R}}{R} = - 1$ which shows image is formed of same size but real and inverted.
Note:It is to be remembered that by the new Cartesian sign convention any distances measured from left of the pole of a spherical mirror is taken as negative while distances measured to the right of pole is taken as positive. And in spherical mirrors the relation between focal length and radius of the spherical surface is fixed which is $f = \dfrac{R}{2}$.