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# The focal length of a concave mirror is $50{\text{ cm}}$ where an object is to be placed so that its image is two times magnified, real and inverted:A. $75{\text{ cm}}$B. $72{\text{ cm}}$C. ${\text{63 cm}}$D. ${\text{50 cm}}$

Last updated date: 13th Jun 2024
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Hint A concave mirror is a spherical mirror whose reflecting surface is its bent surface. There is a relation between the focal length of a mirror $f$, distance of an object $u$ and distance of the image formed $v$, known as Mirror formula and given by $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ .
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object. If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .

Let us first know about a concave mirror.
A concave mirror is a spherical mirror whose reflecting surface is its bent surface.
As given in the question that the focal length of a concave mirror, $f = 50{\text{ cm}}$ .
We are asked to find the distance of the object $u$ such that the image formed is two times magnified, real and inverted.
If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .
Let us discuss the linear magnification of a mirror.
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object.
So, according to the question,
$\dfrac{{ - v}}{u} = - 2$ which simplifies to $v = 2u$
As we know that there is a relation between the focal length of a mirror $f$, distance of an object $u$ and distance of the image formed $v$, known as Mirror formula and given by $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ .
So, substituting the values in the mirror formula we have
$\dfrac{1}{{50}} = \dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{3}{{2u}}$
On simplifying we have
$u = 75cm$
Therefore, the object is to be placed at distance $75{\text{ cm}}$ from the mirror so that its image is two times magnified, real and inverted.

Hence, option A is correct

Note Here, we take negative sign for the magnification because the image formed is inverted i.e. below the central axis of the mirror.
Distances of any point are measured along the central axis by convention, as the direction towards the object in mirror is taken as positive and away from the object is taken as negative. Hence the radius of curvature and the focal length are negative for convex mirrors and positive for concave mirrors.