Question

The near point and far point of a person are $40\text{cm}$ and $200\text{cm}$ respectively. Find the power of the lens he/she should use while reading at $25\text{cm}$

• A. $1.5D$
• B. $-1.5D$
• C. $3.5D$
• D. $-3.5D$

Answer 1

The correct option is A $1.5D$

If an object is placed at $25\text{cm}$ from the correcting lens, it should produce the virtual image at $40\text{cm}$
So, $u=-25\text{cm}$, $v=-40\text{cm}$
$\Rightarrow \frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{f}=-\frac{1}{40}+\frac{1}{25}$
$\therefore f=\frac{200}{3}\text{cm}=\frac{2}{3}\text{m}$
$\therefore \text{Power,}P=\frac{1}{f}=\frac{3}{2}=1.5D$
$$\begin{array}{c}\text{Why this Question ?}\\ \text{Concept : To see the nearby objects clearly, image of the object should form at near-point and to see far-away objects clearly, image should form at far-point.}\end{array}$$