Question

The near point and far point of a child are at $10\text{cm}$ and $100\text{cm}$ . If the retina is $2.0\text{cm}$ behind the eye-lens, what is the range of the power of the eye-lens?

• A. $50\text{D}$ to $57\text{D}$
• B. $51\text{D}$ to $60\text{D}$
• C. $56\text{D}$ to $63\text{D}$
• D. $60\text{D}$ to $70\text{D}$

Answer 1

The correct option is B $51\text{D}$ to $60\text{D}$

For clear vision, the image should always form on the retina. So, for all situations, $v=2\text{cm}$

Using lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

For object placed at the near point, $u=-10\text{cm}$, $v=2\text{cm}$

$\frac{1}{f}=\frac{1}{2}-\frac{1}{(-10)}=\frac{1}{2}+\frac{1}{10}$

$\therefore \frac{1}{f}=\frac{6}{10}{\text{cm}}^{-1}$

$\therefore P=\frac{1}{f\left(\text{in m}\right)}=\frac{6}{10}\times 100=60\text{D}$

For object placed at the far point,
$u=-100\text{cm}$, $v=2\text{cm}$

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\Rightarrow \frac{1}{f}=\frac{1}{2}+\frac{1}{100}=\frac{51}{100}{\text{cm}}^{-1}$

Now, $P=\frac{1}{f\left(\text{in m}\right)}=\frac{51}{100}\times 100=51\text{D}$

Hence, $\left(B\right)$ is the correct answer.