Question
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 :-
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Solution
Sol. [Key Concept: Cutting a biconvex lens along its axis doubles each half's focal length; to keep $M = \frac{L}{F_{0}}\frac{D}{F_{e}}$ fixed, the tube length must increase.]
$${M = \frac{L}{F_{0}}\frac{D}{F_{e}} }{\frac{1}{{\text{ }F}_{0}} = (n - 1)\left( \frac{2}{R} \right) }$$But when cut
$$\frac{1}{{\text{ }F}_{0}^{1}} = (n - 1)\left( \frac{1}{R} \right) = \frac{1}{2{\text{ }F}_{0}} $$So, $\frac{L}{F_{0}} = \frac{L^{'}}{F_{0}^{1}} = 2\text{ }L$
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