JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Geometrical Optics

The magnification of a compound microscope is $M=\frac{L}{f_o}\cdot \frac{D}{f_e}$, where $L$ is the tube length and $f_o$ the objective focal length. For the symmetric biconvex objective, $\frac{1}{f_o}=(n-1)\frac{2}{R}$. Cutting it into a plano-convex lens leaves a single curved surface, so $\frac{1}{f_o^{\prime }}=(n-1)\frac{1}{R}=\frac{1}{2f_o}$, i.e. $f_o^{\prime }=2f_o$. With the object distance unchanged (so $f_e$ and $D$ are fixed), keeping $M$ the same needs $\frac{L}{f_o}=\frac{L^{\prime }}{f_o^{\prime }}$, giving $L^{\prime }=L\cdot \frac{f_o^{\prime }}{f_o}=2L$. The tube length must be increased two times.