A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be

Focal length of the objective after cutting:
For a symmetric biconvex lens with both radii of magnitude $R$:
$\dfrac{1}{f_o} = (\mu - 1)\left(\dfrac{1}{R} - \dfrac{1}{-R}\right) = \dfrac{2(\mu - 1)}{R}$

When the lens is cut vertically (perpendicular to the principal axis), each half becomes a plano-convex lens with one flat surface:
$\dfrac{1}{f_o'} = (\mu - 1)\left(\dfrac{1}{R} - 0\right) = \dfrac{\mu - 1}{R}$

Comparing the two: $f_o' = 2f_o$, so the focal length of the objective doubles.

Magnification of a compound microscope:
$M \approx \dfrac{L}{f_o} \cdot \dfrac{D}{f_e}$

where $L$ is the tube length, $f_o$ is the objective focal length, $f_e$ is the eyepiece focal length and $D$ is the least distance of distinct vision.

Since the eyepiece and $D$ are unchanged, equating the initial and final magnifications:
$\dfrac{L_{\text{initial}}}{f_o} \cdot \dfrac{D}{f_e} = \dfrac{L_{\text{final}}}{2f_o} \cdot \dfrac{D}{f_e}$

$\dfrac{L_{\text{initial}}}{f_o} = \dfrac{L_{\text{final}}}{2f_o}$

$L_{\text{final}} = 2\,L_{\text{initial}}$

Hence, to maintain the same magnification, the tube length must be doubled.

The correct option is $(1)$ increased two times.