Question

An object placed in front of a concave mirror at a distance of $x cm$ from the pole gives a 3 times magnified real image. If it is moved to a distance of $(x+5) cm$, the magnification of the image becomes $2 $ The focal length of the mirror is

• A. 15 cm
• B. 20 cm
• C. 25 cm
• D. 30 cm

Answer 1

Answer: (D)

For the first case, $m=-3$
$m=-\frac{v}{u}=-3$ (As the image is real)
$\therefore v=3u$
Here, $u=-x$
$\therefore v=-3x$
According to mirror formula,
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$ ;
$\therefore \frac{1}{(-x)}+\frac{1}{(-3 x)}=\frac{1}{f}\,\,\,\,\dots(i)$
For the second case, $m=-2=-\frac{v}{u}$
$ \therefore v=2 u$
Here, $u=-(x+5)$
$\therefore v=-2(x+5)$
Using mirror formula, $\frac{1}{-(x+5)}+\frac{1}{-2(x+5)}=\frac{1}{f}\,\,\,\,\dots(ii)$
Solving (i) and (ii), we get $f=-30 \,cm$