Question

The near point and far point of a person are $40\text{cm}$ and $250\text{cm}$ respectively. Find the power of the lens he/she should use while reading at $25\text{cm}$.

• A. $-1.5\text{D}$
• B. $+1.5\text{D}$
• C. $-2.0\text{D}$
• D. $+2.0\text{D}$

Answer 1

The correct option is B $+1.5\text{D}$

If an object is placed at $25\text{cm}$ from the correcting lens, it should produce virtual image at $40\text{cm}$.
Thus, $u=-25\text{cm}$, $v=-40\text{cm}$
We know, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{f}=\frac{1}{-40}-\frac{1}{-25}$
$\Rightarrow f=\frac{200}{3}\text{cm}=\frac{2}{3}\text{m}$
Now, $P=\frac{1}{f}=1.5\text{D}$