Question

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

• A. $f$
• B. $f/2$
• C. $f/4$
• D. $4f$

Answer 1

Answer: (B)

$\frac{\text{1}}{\text{f}}\text{=(μ-1)}\left(\frac{\text{2}}{\text{R}}\right)\frac{1}{f^{{}^{\prime }}}=(\mu -1)\left(\frac{1}{R}\right)$
$f^{{}^{\prime }}=2f_{}$
So cutting a lens in transverse direction doubles their focal length but if a lens is cut across its principle axis, then there is no change in focal length. So all the $4$ parts will have same focal length i.e. $2f$
Using the formula of equivalent focal length, we can find net focal length
$\frac{1}{\text{f}}=\frac{1}{{\text{f}}_1}+\frac{1}{{\text{f}}_2}+\frac{1}{{\text{f}}_3}+\frac{1}{{\text{f}}_4}$