JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

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 :-

A
increased two times
B
increased $\frac{3}{2}$ times
C
decreased two times
D
decreased $\frac{3}{2}$ times

Solution

For a symmetric biconvex lens with both radii $R$ , the lensmaker's equation gives $\frac{1}{F _{0}} = (n - 1) \frac{2}{R}$ .

After cutting vertically along its axis, each half is a plano-convex lens with one flat face and one curved face of radius $R$ , so $\frac{1}{F _{0}^{'}} = (n - 1) \frac{1}{R} = \frac{1}{2 F _{0}}$ .

The objective focal length doubles to $F_{0}^{'} = 2 F_{0}$ . Magnification of a compound microscope is $M = \frac{L \cdot D}{F _{0} \cdot F _{e}}$ , so to keep $M$ constant the tube length $L$ must double as well.

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