The focal lengths of the objective and the eyepiece of a compound microscope are $2.0$ cm and $3.0$ cm, respectively. The distance between the objective and the eyepiece is $15.0$ cm, The final image formed by the eyepiece is at infinity. The two lenses are thin. The distance(in cm) of the object and the image produced by the objective, measured from the objective lens are respectively.

The correct option is A $2.4$ and $12.0$
Here, $f_v=2\text{cm}$ and $f_e=3\text{cm}$.
Using lens formula for eyepiece,
$-\frac{1}{u}+\frac{1}{v_1}=\frac{1}{f_e}$
put the values
$\Rightarrow \frac{-1}{u}+\frac{1}{\propto }=\frac{1}{3}\Rightarrow u_1=-3\text{cm}[\therefore i=0]$
But the distance between objective and eyepiece is $15$ cm (given).
Therefore, distance of image formed by the objective,
$v=15-3=12cm$.
Let u be the object distance from the objective, then for objective lens $\frac{1}{u}+\frac{1}{v}=\frac{1}{f_0}\frac{-1}{u}+\frac{1}{12}=\frac{1}{2}$
$\Rightarrow \frac{-1}{u}=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}u=-\frac{12}{5}=2.4\text{cm}$
(a) is the correct option.