Question

The near point and far point of a person are 40 cm and 250 cm respectively. Determine the power of the lens he/she should use while reading a book kept at distance 25 cm from the eye.
(A) 2.5D
(B) 5D
(C) 1.5D
(D) 3.5D

Answer 1

Hint If the book is kept at 25 cm, it means that the person needs to make an image of the object with the object length = 25 cm. The image distance is the length of the retina from the lens. We will calculate power of lens using formula $P = \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ where, $v$ is image distance and $u$ is object distance.

Complete step by step answer
Near point:
It is the closest point at which an object can be placed such that the image formed at retina within the accommodation range.
Far point:
It is the farthest point at which an object can be placed such that the image formed at retina within the accommodation range.
Focus of lens:
It is that point at which all parallel rays coming from an object coincide or meet.
Power of lens:
It is reciprocal of focal length. It is also defined as the tendency of light to deviate. More is the focal length, more is the deviation.
It is positive in case of convex lens and negative in case of concave lens.
It is measured in a Diopter.
The Nearest point is 40 cm and the farthest point is 250 cm. Only near points are used in this case. Image will shift from 25 cm to the near point of the defected eye of a person.

$u = 25\,cm = 0.25m = $ object distance
$v = 40\,cm = 0.4\,m = $ image distance
$P = \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
$\dfrac{1}{f} = \dfrac{1}{{ - 0.4}} - \dfrac{1}{{ - 0.25}}$
$P = \dfrac{1}{f} = 1.5D$

Therefore, The power of the lens is 1.5D. Hence, option C is correct.

Note
If far point is used instead of near point then solution, we will get the wrong solution. This is because for a normal eye, it is easy to view objects placed at a distance of 25 cm and therefore object distance is known in this case and it is easy to shift it to a near point.