Question

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

Answer 1

Use the expression for the mirror formula. This expression gives the relation between the focal length of mirror, object distance and image distance from the mirror. Also use the formula for the magnification of the mirror. Recall the sign of the magnification term if the image formed is real.

Formula used:
The expression for the mirror formula is given by
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ …… (1)
Here, $f$ is the focal length of the mirror, $u$ is the distance of the object from the mirror and $v$ is the distance of the image from the mirror.
The formula for the magnification of the mirror is
$m = - \dfrac{v}{u}$ …… (2)
Here, $m$ is the magnification of the mirror, $u$ is the object distance from the mirror and $v$ is the image distance from the mirror.

Complete step by step answer:
We have given that the focal length of the mirror is $f$.
Also the magnification of the image is $x$ times the size of the object.
The image formed by the mirror is a real image. If the image formed by the mirror is real, then the magnification of the mirror should be negative.
Determine the relation between the magnification of image, object distance from the mirror and image distance from the mirror using equation (2).
Substitute $- x$ for $m$ in equation (2).
$- x = - \dfrac{v}{u}$
$\Rightarrow x = \dfrac{v}{u}$
Rearrange the above equation for the image distance $v$ from the mirror.
$v = ux$
Now let us determine the relation between the distance of an object from the mirror, magnification of the mirror and the focal length of the mirror.
Substitute $ux$ for $v$ in equation (1) and solve it for $u$.
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{{ux}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{{ux + u}}{{{u^2}x}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{{u\left( {x + 1} \right)}}{{{u^2}x}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{{\left( {x + 1} \right)}}{{ux}}$
$\Rightarrow u = \dfrac{{\left( {x + 1} \right)f}}{x}$
Therefore, the distance of the object from the mirror is $\dfrac{{\left( {x + 1} \right)f}}{x}$.

Hence, the correct option is B.

Note:
The students should not forget to use the negative sign for the magnification of the real image formed by the mirror. This is because if the image formed by the plane mirror is real then it is also inverted which means that the height of the image should be measured on a negative axis.