A person can see objects clearly when they lie between 40 cm and 400 cm from his eye. In order to increase…

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
  1. Convex, 0.25 Dioptre
  2. Concave, -0.25 Dioptre
  3. Concave, -0.5 Dioptre
  4. Convex, 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

Asked in: AP EAMCET 2024 (22 May Shift 1)

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