Question

When an object is placed at a distance of 60 cm from a convex mirror, the magnification produced is 1/2. Where should the object be placed to obtain a magnification of 1/3?

Answer 1

$u=-60$cm

$\frac{-v}{u}=\frac{1}{2}\Rightarrow \frac{-v}{-60}=\frac{1}{2}$

$\Rightarrow v=30$cm

$\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{f}=\frac{1}{30}+\frac{1}{-60}\Rightarrow \frac{1}{f}=\frac{2-1}{60}=\frac{1}{60}$

$\Rightarrow f=60$cm

Now magnification$=\frac{1}{3}$

$\frac{-v}{u}=\frac{1}{3}\Rightarrow v=\frac{-u}{3}\therefore \frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{\frac{-u}{3}}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{-3}{u}+\frac{1}{u}=\frac{1}{60}\Rightarrow \frac{-2}{u}=\frac{1}{60}$

$\Rightarrow u=-120$cm

$\therefore$ Object should be placed at $120$cm.