Magnification produced by a convex mirror is $\dfrac{1}{3}$, then distance of the object from mirror is-
$
A. \dfrac{f}{3} \\
B. \dfrac{{2f}}{3} \\
C. {\text{ 1f}} \\
{\text{D. 2f}} \\
 $

Hint- In order to solve this question, firstly we will apply the formula of diverging lens i.e. $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ where f is the focal length, u is distance of object and v is the distance of image respectively from the mirror. Then we will substitute the given value of magnification i.e. $\dfrac{1}{3}$ to get the required answer.

Complete step-by-step solution -
Formula used- $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Magnification is defined as the ratio of the image size to the object size. It can also be expressed as the ratio of the distance of the image(v) to the distance of object (u) from the mirror. Magnification has no unit.
Convex mirror is a diverging mirror.
Now we will use diverging lens formula-
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}..........\left( 1 \right)$
Where u = height of object
v = height of image
f = focal length
Multiplying both sides by u in equation 1,
We get-
$ \Rightarrow \dfrac{u}{f} = \dfrac{u}{v} - \dfrac{u}{u}$
Which means-
$ \Rightarrow \dfrac{u}{f} = \dfrac{u}{v} - 1$
Taking -1 from R.H.S to L.H.S,
We get-
$ \Rightarrow \dfrac{u}{f} + 1 = \dfrac{u}{v}$
Taking LCM of $\dfrac{u}{f} + 1$,
We get-
$ \Rightarrow \dfrac{u}{v} = \dfrac{{u + f}}{f}$
Or we can rewrite it as-
$ \Rightarrow \dfrac{v}{u} = \dfrac{f}{{f + u}}$
$ \Rightarrow m = \left| {\dfrac{{ - v}}{u}} \right| = \dfrac{f}{{f + u}}$
Here we are given that-
Magnification, $m = \dfrac{1}{3}$
So, we get-
$\dfrac{1}{3} = \dfrac{f}{{f + u}}$
By cross-multiplication,
We get-
$ \Rightarrow f + u = 3f $
$u = 2f$
Therefore, we conclude that if the Magnification produced by a convex mirror is $\dfrac{1}{3}$, then the distance of the object from the mirror is two times the distance of the image.

Note- While solving this question, we must know the difference between the diverging lens formula (convex mirror) i.e.$\dfrac{1}{u} - \dfrac{1}{v} = \dfrac{1}{f}$ and converging lens formula (concave mirror) i.e.$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$ as here in this question, we used a diverging lens formula.