Question

A concave mirror forms an inverted image of size double than that of that object. The focal length of the mirror is 15 cm. Find the position of the image and object. Thanks

Answer 1

Data:

f (focal length) = 15 cm 
i (mage) = 2*object
p (position of the object) = ? 
p' (position of the image) = ? 

If: The focal length (f) of a spherical mirror is equal to half its radius of curvature (R).
$f = \frac{R}{2}$
But: A concave mirror forms an inverted image of double size that of the object, thus, we will have:
$\frac{f}{2} = R$
$\frac{15}{2} = R$ 
$7,5\:cm = R$
$\boxed{R = 7,5\:cm}$

Solving:

$\frac{i}{o} = \frac{p'}{p}$
$\frac{2o}{o} = \frac{p'}{p}$
$\frac{2\diagup\!\!\!\!o}{\diagup\!\!\!\!o} = \frac{p'}{p}$
$2 = \frac{p'}{p}$
$2p = p'$
$\boxed{p' = 2p}$

Mirror concave real image (because if it is inverted is always real)
$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
Replace the data found:
$\frac{1}{15} = \frac{1}{p} + \frac{1}{2p}$
$\frac{1}{15} = \frac{2+1}{2p}$
$\frac{1}{15} = \frac{3}{2p}$
$multiply\:cross$
$1*2p = 3*15$
$2p = 45$
$p = \frac{45}{2}$
$\qquad\quad\checkmark\boxed{\boxed{p = 22,5 cm}}\end{array}} \longleftarrow\: The\:position\:of\:the\:object\:in\:the\:mirror$

Solving:

$2p = p'$
$2*22,5 = p'$
$45 cm = p'$
$\qquad\quad\checkmark\boxed{\boxed{p' = 45 cm}}\end{array}} \longleftarrow\: The\:position\:of\:the\:image\:in\:the\:mirror$

Answer 2

a more easier way......