The distance of the object from a concave mirror of focal length 10 cm so that image size is four times the size of the object will be

Given that

$\frac{size of the image}{size of the object}=4,f=10cm$Case (i):

We can get image size is four times that of object size as shown

If image is virtual

$m=+4$

$m=\frac{-v}{u}$

u = ?

            $f=-10 cm$

$4=\frac{-v}{u}$

$u=\frac{-v}{4},f=-10 cm,v=-4u$

$We know \frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{-1}{10}=\frac{1}{u}-\frac{1}{4u}$

$\frac{-1}{10}=\frac{3}{4u}$

$\therefore u=\frac{-30}{4}=-7.5cm$

$v=-4u=-4\times 7.5=30cm$

Case (ii):

If image is real

$m=-4$

$-4=\frac{-v}{u}$

$\Rightarrow v=4u$

$f=-10 cm$

$u=?$

We know  $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\frac{1}{u}+\frac{1}{4u}=\frac{-1}{10}$

$\frac{4+1}{4u}=\frac{-1}{10}$

$\Rightarrow 50=-4u$

$\therefore u=\frac{-50}{4}=-12.5 cm$

$v=4u=4\times -12.5=-50 cm$