d

If initially the objective ( focal length $F$) forms the image at distance v_o​ then

$v_0 = \frac{u_0.f_0}{u_0 - f_0} = \frac{3 \times 2}{3 - 2} = 6cm$

$\text{For lenses in contact } \frac{1}{F_0} = \frac{1}{f_1} + \frac{1}{F'_0}$

$\text{If one of the lenses is removed then Focal length of remaining system is } $

$ \frac{1}{F'_0} = \frac{1}{F_0} -  \frac{1}{f_1} = \frac{1}{2} - \frac{1}{10}$

$\Rightarrow F'_0 = 2.5 cm$

$\text{This lens will form the image of same object at a distance }$ v₀′​ $\text{such that}$

$v'_0 = \frac{u_0.F'_0}{u_0 - F'_0} = \frac{3 \times 2.5}{3 - 2.5} = 15 cm$

So to refocus the image, eye-piece must be moved by the same distance through which the image formed by the objective has shifted i.e. $15-6=9cm$.