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. Then :

• A. The distance of the retina from the eye lens is $2.5\text{cm}$.
• B. The distance of the retina from the eyelens is $2\text{cm}.$
• C. His near point is at $10\text{cm}$.
• D. His near point is at $7\text{cm}$.

Answer 1

Answer: (B), (C)

His near point is at $10\text{cm}$.

When eye is fully relaxed, its focal length is maximum and power is minimum.

So, according to question this power is $50\text{D}$ and the focal length of the eye lens is,

$f=\frac{1}{P}=\frac{1}{50}\text{m}=2\text{cm}$

Since, the far point is infinity, the parallel rays coming from the infinity are focused on the retina.
Hence, distance of eye-lens from retina is equal to focal length, which is $2\text{cm}$.

Now, when the 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, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

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

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

$\therefore u=-10\text{cm}$

Hence, options $\left(B\right)$ and $\left(C\right)$ are the correct answers.

Note : When a normal eye looks at the object lying at infinity, then the focal length of the eye lens is equal to the distance between the eye lens and the retina.