Problems from IIT JEE

Problem (IIT JEE 2010): The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is,

Solution: The lens formula, $1/v-1/u=1/f$, for the two locations of the object gives, \begin{align} \label{oda:eqn:1} \frac{1}{v_1}-\frac{1}{-25}=\frac{1}{20},\\ \label{oda:eqn:2} \frac{1}{v_2}-\frac{1}{-50}=\frac{1}{20}. \end{align} The first equation gives $v_1={100}\;\mathrm{cm}$ and second equation gives $v_2={100/3}\;\mathrm{cm}$. Thus magnification in the first case is $m_{25}=v_1/u_1=-4$ and that in the second case is $m_{50}=v_2/u_2=-2/3$. Hence, $m_{25}/m_{50}=6$.