If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm , the focal length of the mirror must be:

• A

  $\frac{10}{3} \mathrm{~cm}$
• B

  $-12 \mathrm{~cm}$
• Correct answer
  C

  $-10 \mathrm{~cm}$
• D

  15 cm

Solution

Since, image is magnified. So, it is a concave mirror. $\therefore \mathrm{m}=2=\frac{-\mathrm{v}}{\mathrm{u}}$ $2=\frac{-(15-u)}{-u}$ $2 u=15-u$ $3 \mathrm{u}=15 \Rightarrow \mathrm{u}=5 \mathrm{~cm}$ $\mathrm{v}=15-\mathrm{u}=15-5=10 \mathrm{~cm}$ Bymirror formula, $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$ $=\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10}$ $\mathrm{f}=-10 \mathrm{~cm}$

Hence, the correct answer is option: 'C'.