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# The near point of a hypermetropic eye is $50cm$. What is the nature and power of the lens required to enable him to read a book placed at $25cm$ from the lens?

Last updated date: 12th Sep 2024
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Hint: In the human eye, the light is focused on the retina by the eye lens. The retina will transmit the electrical signals to the brain via optic nerves. Depending on the distance of the object from the eye the ciliary muscles of the eye will adjust the focal length of the lens so that the image is focused on the retina. Hypermetropia is a defect of the eye that affects the vision of a person.
Formula used
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where $f$ stands for the focal length of the lens, $u$ stands for the distance of the object and the lens, and $v$ stands for the distance between the image and the lens.
$P = \dfrac{1}{f}$
Where $P$ stands for the power of the lens and $f$ stands for the focal length of the lens.

If the eye lens focuses the light at a point behind the retina, the image will not be clear. This defect of the eye is known as hypermetropia.
We use a convex lens to correct hypermetropia,
The near point is given as $50cm$
$\therefore$ The image distance will be,$v = - 50cm$
The object is at a distance $u = - 25cm$
We can write the formula for focal length as,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Substituting the values of $u$ and $v$, we get
$\dfrac{1}{f} = \dfrac{1}{{ - 50}} - \dfrac{1}{{ - 25}}$
This will become,
$\dfrac{1}{f} = - \dfrac{1}{{50}} + \dfrac{1}{{25}}$
This can be written as,
$\dfrac{1}{f} = \dfrac{{ - 25 + 50}}{{50 \times 25}} \Rightarrow \dfrac{{ - 1 + 2}}{{50}}$
The focal length will be
$\dfrac{1}{f} = \dfrac{1}{{50}}$
$\Rightarrow f = 50cm$
The power of the lens will be,
$P = \dfrac{1}{{f(m)}} \Rightarrow \dfrac{{100}}{{f(cm)}}$
Substituting the value of focal length,
$P = \dfrac{{100}}{{50}} = 2$

The power of the lens will be $2D$

Note
The closest distance for which the lens can focus light on the retina is called the least distance of distinct vision or near point. This will be $25cms$. This distance increases as much as $200cms$ after the age of $60$. Thus when elderly people try to read the image of the book will appear blurred. This defect of the eye is called presbyopia.