Question

A person is unable to see objects nearer than$50cm$. He wants to read a book placed at a distance of $25cm$. find the nature, focal length, and power of the lens he needs for his spectacles.

Answer 1

We know that a normal eye has the near point at $25cm$or $0.25m$ and it is up to an infinite distance. Hence, in this range, if an object is placed the image of it is created at the retina of the eye. If the range of an eye is less or greater than this range, we call it the defective eye. The important sight-based defects are long sight or hypermetropia, short-sighted or myopia, presbyopia, and astigmatism.
The focal length can be found from the lens equation. The power can be calculated from the formula of power in terms of the focal length. The lens that is needed to cure is either convex or concave as per the defects.

Formula used:
The lens equation:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where, $u$is the object’s distance,
$v$ is the image’s distance,
$f$ is the focal length of the lens.
The power of the lens $P = \dfrac{1}{f}{\text{dioptre}}$

Complete answer:
Given that, A person is unable to see objects nearer than$50cm$. So, his eye has a near point of distance$50cm$
The book is placed at a distance $25cm$.
Hence, the near point of the given eye is greater than that of a normal eye.
This type of defect is called long sight or hypermetropia.
To reduce this defect, the glass of the convex lens is needed.
To find the focal length of the convex lens the lens equation has to be used i.e.
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where, the object’s distance $u = - 25cm$
The image’s distance $v = - 50cm$
Negative signs imply the directions of the measurements.
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{ - 50}} - \dfrac{1}{{ - 25}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{25}} - \dfrac{1}{{50}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{50}}$
$\Rightarrow f = 50cm$
The power of the lens $P = \dfrac{1}{f}{\text{dioptre}}$
$P = \dfrac{1}{{50}} = 0.02{\text{dioptre}}$

The nature of the lens is the Convex lens.
The focal length of the lens is $\Rightarrow f = 50cm$
The power of the lens is $P = 0.02{\text{dioptre}}$

Note:
The near point distance of a normal eye to$25cm$.
If the power of the lens for the near point of the defective eye$(d)$ is greater than the near point of the normal eye,
$P = 4 - \dfrac{{100}}{d}{\text{dioptre}}$
This is for long-sighted defect reduction.
To reduce the short sight defect, the glass of negative power made up of a concave lens is needed. The power of the concave lens is, $P = - \dfrac{{100}}{d}{\text{dioptre}}$.
To reduce presbyopia, a glass of the bifocal lens is used. A bifocal lens is made up of both concave and convex lenses.
To reduce astigmatism, the glass of the toric lens is used.