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)

30 cm

Answer 2

Given $v_{1}=3 x$ and $v_{2}=2(x+5)$
So here, Ist case
$\frac{1}{-x}+\frac{1}{-3 x}=\frac{1}{f}$ ...(i)
$\frac{1}{-(x+5)}+\frac{1}{-2(x+5)}=\frac{1}{f}$ ...(ii)
On solving Eqs. (i) and (ii) we get $f=30\, cm$