When the object is placed 4 cm from the objective of microscope, the final image formed coincides with the object. The final image is at the least distance of distinct vision (24 cm). If the focal length of eye piece is 8 cm then the focal length of objective is

The correct option is B $3.25cm$
Given that:-

$u_o=-4cm$

$v_e=-24cm$

$f_e=8cm$

Since, final image is formed at the point of object.

Length of microscope, $L=|v_e|-|u_o|=20cm$

Now, for eye-piece:-

$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$

$⟹\frac{1}{-24}-\frac{1}{u_e}=\frac{1}{8}$

$⟹\frac{1}{u_e}=\frac{-1}{24}-\frac{1}{8}=\frac{-1}{6}$

$⟹u_e=-6cm$

Now, $v_o+|u_e|=L⟹v_o=(20-6)cm$

$⟹v_o=14cm$

For objective lens:-

$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$

$⟹\frac{1}{f_o}=\frac{1}{14}-\frac{1}{-4}=\frac{16}{42}$

$⟹f_o=3.25cm$