Question

A young boy can adjust the power of his eye lens between 50D 60D. His far point is infinity.
1)What is the distance of his retina from the eye lens ?
2)Whay is near point?

Answer 1

Answer:

10 cm.................

Answer 2

Answer:

(a). The distance of his retina from the eye lens is $2cm$.

(b). The near point is at $10cm.$

Explanation:

(a)

We know that when the eye is fully relaxed, its focal length will be the largest and the power of the eye lens will be at its minimum.

This power is given to be $50D$ in the question.

Then the focal length in the relaxed case will be $\frac{1}{50}m =2cm$

As the far point is at infinity, the parallel rays coming from there are focused on retina which is in its fully relaxed condition.

Hence, the distance of retina from the lens $=$ focal length which is $2cm.$

When the eye is focused at the near point, we know that the power will be maximum which is given to be $60D$.

The focal length in the focused case is :

$f=\frac{1}{60} m=\frac{5}{3} cm$

The image is formed on the retina, thus distance $v=2cm$

(b)

We know that : $\frac{1}{v} -\frac{1}{u}=\frac{1}{f}$

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

$\frac{1}{2} -\frac{3}{5}$

That is, $u=-10cm$

Therefore the near point is at $10cm.$