First slide

Ray optics

Question

A given equiconvex lens is cut into four parts and the parts are rearranged as shown. If the initial focal length is f, then after rearrangement the equivalent focal length is

Moderate

A

f
B

$\frac{f}{2}$
C

$\frac{f}{4}$
D

4f

Solution

Cutting a lens in transverse direction doubles their focal length i.e. 2f.
using the formula of equivalent focal length
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}} + \frac{1}{f_{4}} = \frac{4}{2 f} = \frac{2}{f} \Rightarrow F = \frac{f}{2}$

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