JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

In a microscope the objective is having focal length $f_{0} = 2$ cm and eye-piece is having focal length $f_{e} = 4$ cm. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____________.

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Solution

For a compound microscope in normal adjustment (final image at the near point $D = 25$ cm):

$m = \frac{L}{f _{0}} \cdot \frac{D}{f _{e}} = \frac{32}{2} \cdot \frac{25}{4} = 16 \times 6.25 = 100$ .

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