Q 12-09-070JEE MainJEE Main 2025 (24 Jan, Shift 1)Easy
What is the relative decrease in focal length of a lens for an increase in optical power by $0.1\ \text{D}$ from $2.5\ \text{D}$? ('D' stands for dioptre)
Answer: (B) $0.04$
$f = 1/P$: $f_1 = \dfrac{1}{2.5} = 0.4\ \text{m}$, $f_2 = \dfrac{1}{2.6} \approx 0.385\ \text{m}$.
$$\frac{f_1 - f_2}{f_1} = 1 - \frac{2.5}{2.6} = \frac{0.1}{2.6} \approx 0.04$$
(Using differentials: $\dfrac{|\Delta f|}{f} = \dfrac{\Delta P}{P} = \dfrac{0.1}{2.5} = 0.04$.)
Solution by Sreeraj P, M.Sc Physics