Question
If the distance between object and its two times magnified virtual image produced by a curved mirror is $15 \mathrm{~cm}$, the focal length of the mirror must be :
✓
Answer
$ \mathrm{m}=2=\frac{-\mathrm{v}}{\mathrm{u}} $
$ 2=\frac{-(15-\mathrm{u})}{-\mathrm{u}} $
$ 2 \mathrm{u}=15-\mathrm{u} $
$ 3 \mathrm{u}=15 \Rightarrow \mathrm{u}=5 \mathrm{~cm} $
$ \mathrm{v}=15-\mathrm{u}=15-5=10 \mathrm{~cm} $
$ \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}$
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