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The given 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-
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$f / 2$
Cutting a lens in transverse direction doubles their focal length i.e. $2 f$. Using the formula of equivalent focal length, $\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}+\frac{1}{f_{4}}$
We get equivalent focal length as $f / 2$
We get equivalent focal length as $f / 2$
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