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**Hint:**We will use the concept of image formation by a lens. The concept of magnification is also used in order to solve the problem. Magnification produced due to a lens is defined as the ratio of the height of the image to that of the size of the object. The relation of magnification in terms of image distance and object distance will be used.

**Formula Used:**We are using the following formula to find the correct answer:-

$m=\dfrac{f}{f+u}$

**Complete step-by-step solution**

We know that magnification is defined as the ratio of the height of the image to that of an object. Relation of magnification in terms of focal length and object distance will be used, which is given as follows:-

$m=\dfrac{f}{f+u}$………………… $(i)$

Where $m$ is the magnification, $f$ is the focal length and $u$ is the object distance

Value of magnification is given in the problem. For virtual image, $m=+2$ and $u=-10cm$. Object distance is taken as negative by using sign conventions. Putting respective values in $(i)$we get

$2=\dfrac{f}{f+\left( -10 \right)}$

$\Rightarrow 2(f-10)=f$

$\Rightarrow 2f-20=f$

$\Rightarrow 2f-f=20$

$\Rightarrow f=20cm$

Therefore, we get the focal length, $f=20cm$.

Now for real image magnification, $m=-2$, using $(i)$ we get,

$m=\dfrac{f}{f+u}$

$\Rightarrow -2=\dfrac{f}{f+u}$

Putting the values in $(i)$ we get,

$-2=\dfrac{20}{20+u}$

$\Rightarrow -2(20+u)=20$

$\Rightarrow -40-2u=20$

$\Rightarrow -2u=40+20$

$\Rightarrow 2u=-60$

$\Rightarrow u=\dfrac{-60}{2}$

Therefore, $u=-30cm$.

**Hence, option $(B)$ is the correct one.**

**Note:**We have used the formula $m=\dfrac{f}{f+u}$which is just derived from the lens formula, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ . We should use the sign convention properly and without any confusion. We should not get confused between the properties of the lens and mirror. For the mirror, we use properties of the mirror and for the lens, we use properties of the lens respectively. Apart from this, the characteristics of concave and convex lenses are also different from each other and hence sign conventions do change with the use of lenses.

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