Question

A young boy can adjust the power of his eye lens between $50\text{D}$ and $60\text{D}$. His far point is at infinity. What is the distance of the retina from the eye-lens and his near point?

• A. $2.5\text{cm}$ and $5\text{cm}$
• B. $2\text{cm}$ and $12\text{cm}$
• C. $2\text{cm}$ and $10\text{cm}$
• D. $2.5\text{cm}$ and $7.5\text{cm}$

Answer 1

The correct option is C $2\text{cm}$ and $10\text{cm}$

When eye is fully relaxed, its focal length is maximum and power is minimum.
This power is $50\text{D}$, so the focal length of eye-lens is,
$f=\frac{1}{P}=\frac{1}{50}\text{m}=2\text{cm}$

Since, far point is infinity, the parallel rays coming from the infinity are focused on the retina is equal to focal length which is $2\text{cm}$
Now, when eye is focused at near point, the power is maximum, which is $60\text{D}$

Focal length, $f=\frac{1}{P}=\frac{1}{60}\text{m}=\frac{5}{3}\text{cm}$

Image is formed on retina $\therefore v=2\text{cm}$

Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{2}-\frac{1}{u}=\frac{3}{5}$

$\Rightarrow \frac{1}{u}=\frac{1}{2}-\frac{3}{5}$

$\therefore u=-10\text{cm}$
So, near point is $10\text{cm}$

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