How far should one hold an object from a concave mirror of focal length $40\,cm$ so as to get an image twice the size of the object?

Hint: In order to answer this question, first we will rewrite the given focal length of the mirror, and then we will write the image distance in terms of object. And then we will apply the lens's formula to find the required object distance in front of the mirror.

Complete step by step answer:
Given focal length, $f = 40\,cm$. It’s the shaving-mirror situation, so expect an answer less than $f$ (focal length) , less than $40\,cm$.Let the object distance be $u$.
Then the image distance, $v = - 2u$ (as per the question).
\[2u\] because the magnification is 2, minus because it is a virtual distance from the mirror to a virtual picture. In addition, the image is upright; inverted images have negative magnification.

Applying the Len’s formula:- we can use the Lens formula to get the image distance. It is the formula, or the equation, that links the focal length, object distance, and image distance for a lens.
$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$
$\Rightarrow \dfrac{1}{u} + \dfrac{1}{{ - 2u}} = \dfrac{1}{{40}} \\
\Rightarrow \dfrac{1}{{2u}} = \dfrac{1}{{40}} \\
\Rightarrow 2u = 40 \\
\therefore u = 20\,cm $

Hence, an object is $20\,cm$ in front of the mirror.

Note: Convex lenses are also called converging lenses because the rays converge after passing through them, whereas concave lenses are called diverging lenses because the rays diverge after passing through them. These lenses can produce real or virtual images, depending on their distance from the lens, and they can also be of varying sizes. The image distance can be computed using the lens formula and knowledge of the object distance and focal length.