Physics - Geometrical Optics Question with Solution | TestHub

$\frac{\text{1}}{\text{f}}\text{=(μ-1)}\left(\frac{\text{2}}{\text{R}}\right)\frac{1}{f^{\prime }}=(\mu -1)\left(\frac{1}{R}\right)$

$f^{\text{'}}=2f_{}$

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

$\frac{1}{\text{f}}=\frac{1}{{\text{f}}_1}+\frac{1}{{\text{f}}_2}+\frac{1}{{\text{f}}_3}+\frac{1}{{\text{f}}_4}$