Question

The given equi-convex lens is broken into four parts and rearranged as shown. If the initial focal length is $f$, then after rearrangement the equivalent focal length is

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

Answer 1

The correct option is B $\frac{f}{2}$

Let us say that the magnitude of radius of curvature of each side of the lense is $R,$ then from lens makers formula
$\frac{1}{f}\propto \frac{1}{R}-\frac{1}{-R}$
$f\propto \frac{R}{2}$
Now consider each of the part after cutting, it has one side having radius of curvature $R$ and other plane surface.
Hence
$\frac{1}{f^{\prime }}\propto \frac{1}{R}-\frac{1}{\infty }$
$f^{\prime }\propto R\Rightarrow f^{\prime }=2f$
The focal length of each part is $2f$ Effective focal length:
$\frac{1}{F}=\frac{1}{2f}+\frac{1}{2f}+\frac{1}{2f}+\frac{1}{2f}=\frac{4}{2f}$
$F=\frac{f}{2}$