JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

In a microscope of tube length 10 cm two convex lenses are arranged with focal length of 2 cm and 5 cm. Total magnification obtained with this system for normal adjustment is $(5)^{k}$ . The value of k is_____

A
$2$
B
$5$
C
$3.5$
D
$4$

Solution

For a compound microscope in normal adjustment (final image at infinity), the total magnification is

$M = \frac{ℓ}{f _{o}} \cdot \frac{D}{f _{e}}$

with tube length $ℓ = 10$ cm, objective $f_{o} = 2$ cm, eyepiece $f_{e} = 5$ cm, and the near point $D = 25$ cm.

$M = \frac{10}{2} \cdot \frac{25}{5} = 5 \cdot 5 = 25 = 5^{2}$ .

$∴ k = 2$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →