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Question: The near point of a hypermetropic eye is \(50cm\). What is the nature and power of the lens required...

The near point of a hypermetropic eye is 50cm50cm. What is the nature and power of the lens required to enable him to read a book placed at 25cm25cm from the lens?

Explanation

Solution

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
1f=1v1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}
Where ff stands for the focal length of the lens, uu stands for the distance of the object and the lens, and vv stands for the distance between the image and the lens.
P=1fP = \dfrac{1}{f}
Where PP stands for the power of the lens and ff stands for the focal length of the lens.

Complete step by step answer:
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 50cm50cm
\therefore The image distance will be,v=50cmv = - 50cm
The object is at a distance u=25cmu = - 25cm
We can write the formula for focal length as,
1f=1v1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}
Substituting the values of uu and vv, we get
1f=150125\dfrac{1}{f} = \dfrac{1}{{ - 50}} - \dfrac{1}{{ - 25}}
This will become,
1f=150+125\dfrac{1}{f} = - \dfrac{1}{{50}} + \dfrac{1}{{25}}
This can be written as,
1f=25+5050×251+250\dfrac{1}{f} = \dfrac{{ - 25 + 50}}{{50 \times 25}} \Rightarrow \dfrac{{ - 1 + 2}}{{50}}
The focal length will be
1f=150\dfrac{1}{f} = \dfrac{1}{{50}}
f=50cm\Rightarrow f = 50cm
The power of the lens will be,
P=1f(m)100f(cm)P = \dfrac{1}{{f(m)}} \Rightarrow \dfrac{{100}}{{f(cm)}}
Substituting the value of focal length,
P=10050=2P = \dfrac{{100}}{{50}} = 2

The power of the lens will be 2D2D

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 25cms25cms. This distance increases as much as 200cms200cms after the age of 6060. Thus when elderly people try to read the image of the book will appear blurred. This defect of the eye is called presbyopia.