JEE Main 2026 Jan 21 Shift 1, Physics Q25: Magnification

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 ____________.

Your answer

Solution

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

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

Concept behind this question

MagnificationNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Jan 21 Shift 1, Physics Q25: Magnification

Concept behind this question

More previous year questions on Magnification

Practice more Physics