Question

An object placed in front of a concave mirror of focal length $0.15m$ produces a virtual image, which is twice the size of the object. The position of the object with respect to the mirror is:
$A)\text{ }-5.5cm$
$B)\text{ }-6.5cm$
$C)\text{ }-7.5cm$
$D)\text{ }-8.5cm$

Answer 1

Hint: This problem can be solved by finding out the relation between the image distance and the object distance for the mirror using the formula for the magnification and then using this relation in the mirror formula to get the value of the object distance in this case.

Formula used:
$m=-\dfrac{v}{u}$
$m=\dfrac{{{h}_{i}}}{{{h}_{o}}}$
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Complete step by step answer:
We will find the relation between the object distance and the image distance for this case from the magnification using the given information about the height of the image and the object. Then we will use this relation in the mirror formula to get the object distance.
The magnification $m$ produced by a mirror when the object distance is $u$ and the image distance is $v$ is given by
$m=-\dfrac{v}{u}$ --(1)
Also, the magnification $m$ produced by a mirror when the height of the image is ${{h}_{i}}$ and the height of the object is ${{h}_{o}}$ is given by
$m=\dfrac{{{h}_{i}}}{{{h}_{o}}}$ --(2)
For a mirror, the focal length $f$, object distance $u$ and image distance $v$ are related as
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$ --(3)
Now, let us analyze the question.
The focal length of the mirror is $f=-0.15m$ (According to sign convention, the focal length of a concave mirror is negative)
Let the object distance, that is, position of the object with respect to the mirror be $u$.
Let the image distance be $v$.
The height of the image is $+{{h}_{i}}$ (positive sign indicates that the image is virtual and erect).
The height of the object is ${{h}_{o}}$.
Let the magnification produced by the mirror be $m$.
Now, according to the question, the height of the image is twice the height of the object.
$\therefore {{h}_{i}}=2{{h}_{o}}$ --(4)
Using (2), we get the magnification $m$ as
$m=\dfrac{{{h}_{i}}}{{{h}_{o}}}=\dfrac{2{{h}_{o}}}{{{h}_{o}}}=2$ --(5) [Using (4)]
Also, using (1), we get $m=-\dfrac{v}{u}$
Putting (5) in the above equation, we get
$2=-\dfrac{v}{u}$
$\therefore v=-2u$ --(6)
Now, according to the mirror formula, that is, (3)
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
Putting (6) in the above equation and using the value of $f$, we get
$\dfrac{1}{-0.15}=\dfrac{1}{-2u}+\dfrac{1}{u}=-\dfrac{1}{2u}+\dfrac{1}{u}=\dfrac{-1+2}{2u}=\dfrac{1}{2u}$
$\therefore 2u=-0.15$
$\therefore u=\dfrac{-0.15}{2}=-0.075m=-7.5cm$ $\left( \because 1m=100cm \right)$
Therefore, the object distance, that is, the position of the object with respect to the mirror is $-7.5cm$.
Therefore, the correct option is $C)\text{ }-7.5cm$.

Note:
Students must be careful about applying the sign conventions properly, especially in optics problems. For example, if in this question, the image was real and inverted, then the height of the image would have had to be taken as negative and then subsequently the magnification would have also turned out to be negative. In fact, a negative magnification is indicative of the fact that the image formed is inverted with respect to the object.