Q142 / 160 · AP EAMCET 2024 (22 May Shift 1) · Physics
A person can see objects clearly when they lie between 40 cm and 400 cm from his eye. In order to increase the maximum distance of distant vision to infinity, the type of lens and power of correction lens required respectively are
- AConvex, 0.25 Dioptre
- Correct answerBConcave, -0.25 Dioptre
- CConcave, -0.5 Dioptre
- DConvex, 0.5 Dioptre
Solution
$\mathrm{u}=-\infty, \mathrm{v}=-400 \mathrm{~cm}$
By lens formula, $\begin{aligned} & \frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-400}-\frac{1}{-\infty} \\ & \therefore \quad f=-400 \mathrm{~cm} \end{aligned}$ $\therefore \quad$ Power of lens is given by $P=\frac{100}{f(\text { in } \mathrm{cm})}=\frac{100}{-400}=-0.25 \mathrm{D}$ $\therefore$ Concave lens, -0.25 D
By lens formula, $\begin{aligned} & \frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-400}-\frac{1}{-\infty} \\ & \therefore \quad f=-400 \mathrm{~cm} \end{aligned}$ $\therefore \quad$ Power of lens is given by $P=\frac{100}{f(\text { in } \mathrm{cm})}=\frac{100}{-400}=-0.25 \mathrm{D}$ $\therefore$ Concave lens, -0.25 D
Hence, the correct answer is option: 'B'.