Focal length of the objective after cutting:
For a symmetric biconvex lens with both radii of magnitude R:
fo1=(μ−1)(R1−−R1)=R2(μ−1)
When the lens is cut vertically (perpendicular to the principal axis), each half becomes a plano-convex lens with one flat surface:
fo′1=(μ−1)(R1−0)=Rμ−1
Comparing the two: fo′=2fo, so the focal length of the objective doubles.
Magnification of a compound microscope:
M≈foL⋅feD
where L is the tube length, fo is the objective focal length, fe 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:
foLinitial⋅feD=2foLfinal⋅feD
foLinitial=2foLfinal
Lfinal=2Linitial
Hence, to maintain the same magnification, the tube length must be doubled.
The correct option is (1) increased two times.