Question

An equi-convex lens has power $P$. It is cut into four symmetrical pieces as shown with $(A,B,C$ and $D).$ Then all the four parts are arrangement as shown in the figure (ii). The focal length of the combination.

• A. $\frac{1}{P}$
• B. $\frac{1}{2P}$
• C. $\frac{2}{P}$
• D. $\frac{4}{P}$

Answer 1

The correct option is B $\frac{1}{2P}$

According to the question;
The new power of each piece of lens $P_1=\frac{P}{2}$
The total power of the combination is
$P_2=P_A+P_B+P_C+P_D$
$P_2=4\times \frac{P}{2}$
$P_2=2P$
The focal length of the combination is
$F=\frac{1}{P_2}=\frac{1}{2P}$