A convex lens of focal length f cut into parts first horizontally and then vertically. If the focal length of part A (quarter part) of the lens is  xf ,  then x is 

 

$\begin{array}{l}When cut horizontally, focal length remains same as is \\ clear from the following. \\ Formula: \frac{1}{f}=(\mu -1)[\frac{1}{R}-\frac{1}{(-R)}]=\frac{2(\mu -1)}{R}\\ When cut vertically, focal length is doubled\\ \frac{1}{f^{\text{'}}}=(\mu -1)[\frac{1}{R}-\frac{1}{\infty }]=\frac{(\mu -1)}{R}=\frac{1}{2f}\\ Combining these factors, \\ f_A = 2f\\ Quarter part of the lens will behave as a plano convexlens \\ which has double the focal length of the given convex lens.\end{array}$