JEE Main 2025 Jan 24 Shift 1, Physics Q2: Thin Lens and Mirrors

JEE Main2025Jan 24, Shift 1Physics

What is the relative decrease in focal length of a lens for an increase in optical power by $0.1$ D from $2.5$ D ? ['D' stands for dioptre]

Solution

Power and focal length are reciprocals: $f = 1/ P$ .

Initial focal length $f = 1/2.5 = 2/5$ m. New focal length $f^{'} = 1/2.6 = 5/13$ m.

Relative decrease:

$\frac{f - f ^{'}}{f} = \frac{2/5 - 5/13}{2/5} = 1 - \frac{5/13}{2/5} = 1 - \frac{25}{26} = \frac{1}{26} \approx 0.04$

Equivalent quick check: for small changes, $\frac{Δ f}{f} \approx \frac{Δ P}{P} = \frac{0.1}{2.5} = 0.04$ .

$∴$ Option (1).

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.