Question

A concave mirror of focal length f produces an image n times the size of the object. If the image is real then the distance of the object from the mirror is
(A) $(n + 1)f$
(B) $\dfrac{{(n + 1)}}{n}f$
(C) $\dfrac{{(n - 1)}}{n}f$
(D) $(n - 1)f$

Answer 1

Hint : Use the magnification formula to find the image distance and use mirror formula to find out the distance of the object from the mirror. Magnification of an object is given by, $\dfrac{{{h_i}}}{{{h_o}}} = - \dfrac{v}{u} = m$ Where, $m$ is the magnification of the object The mirror formula is given by, $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ where $v$ is the image distance, $u$ is the image distance and $f$ is the focal length of the mirror.

Complete Step By Step Answer:
We know that the mirror equation is given by $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ where $v$ is the image distance, $u$ is the image distance and $f$ is the focal length of the mirror.
Here, we are given that the magnification of the object is $m$ times. Now we know, magnification of an object is given by, $\dfrac{{{h_i}}}{{{h_o}}} = - \dfrac{v}{u} = m$ Where, $m$ is the magnification of the object, ${h_i}$ is the image height, ${h_o}$ is the object height, $v$ is the image distance and $u$ is the image distance . Now, here magnification is given as $m = n$ .(All the distances are measured from the optical centre of the mirror, any length measure to the left is negative and to the right is positive) .
So, given $\dfrac{{{h_i}}}{{{h_o}}} = - \dfrac{v}{u} = n$
From this we can find image distance as,
$v = - nu$
Here it is given that the image is real . therefore, $v$ must be negative so $n$ is negative.
So, we put $v = - v$
Therefore, $v = nu$
Putting the value of $v$ in mirror’s formula we get,
$\dfrac{1}{{nu}} + \dfrac{1}{u} = \dfrac{1}{f}$
$\Rightarrow \dfrac{{(1 + n)u}}{{nuu}} = \dfrac{1}{f}$
$\Rightarrow \dfrac{{(1 + n)}}{{nu}} = \dfrac{1}{f}$
On simplifying we get,
$\Rightarrow u = \dfrac{{(n + 1)f}}{n}$
$\therefore u = \dfrac{{(n + 1)f}}{n}$
Hence, for a real image of $n$ times magnification objection will be at $\dfrac{{(n + 1)f}}{n}$ distance.
Hence, option ( B ) is correct.

Note :
Negative Magnification refers that the image is inverted, and positive magnification refers that the image is erect. Here, For real images we have found that $v$ must be negative. Now, here is an interesting fact that if $n = 1$ (sign is negative as sign is already taken into account in the before calculation) , object distance $u = 2f$ . If $n = - 1$ (sign is positive) image distance $u = 0$ . which is impossible so you can see for a concave mirror with real image $n$ must be negative. If $f = \infty$ , then only $u$ can take any value, but the mirror is not concave anymore, rather it is a plane mirror. then also the value of the object distance is the same as the image distance.