A concave mirror of focal length $10$ cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is __________ cm.

Given focal length of the concave mirror, $f = -10$ cm.

The image is double the size of the object, so the magnification can be $m = \pm 2$.

Using the magnification formula in terms of focal length and object distance:

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

Case 1: For a real image, $m = -2$

$-2 = \dfrac{-10}{-10 - u_1}$

$20 + 2u_1 = -10$

$2u_1 = -30 \Rightarrow u_1 = -15$ cm

Case 2: For a virtual image, $m = +2$

$2 = \dfrac{-10}{-10 - u_2}$

$-20 - 2u_2 = -10$

$-2u_2 = 10 \Rightarrow u_2 = -5$ cm

The distance between the two positions of the object is:

$\Delta u = |u_1 - u_2| = |-15 - (-5)| = 10$ cm

Answer: $10$