### Chapter 2: The Human Eye and the Colourful World

Q
##### The Human Eye and the Colourful World

Question:

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Hypermetropia is the defect of the human eye in which the eye can see far off objects or distant objects clearly but cannot see the nearby objects clearly. Hypermetropia eye defect arises because the focal length of the eye lens becomes too long or if the eyeball has become too small. If the focal length of the eye lens is larger than the normal, the image of the object is formed behind the retina.

Correction of hypermetropia: Convex lens of appropriate power is used so that the light ray converges more and the image form at the retina.

The diagram below shows the correction of hypermetropia.

The near point of the normal human eye is 25 cm which means the near objects are visible clearly minimum at this distance.

It is given that the near point of a hypermetropic eye is 1 m, which means the person can see clearly beyond 1 m but cannot see if the object lies between 25 cm and 1 m. So, a corrective lens needed here which can form the image at 1 cm of the object placed at 25 cm (near point of the normal human eye).

So, the position of object, $u = -2\ cm$

The position of image, $v= -1\ m$

$v = -100\ cm$

Using the lens formula,

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

We get,

$\frac{1}{f} = \frac{1}{(-100)}-\frac{1}{(-25)}$

$\frac{1}{f} = -\frac{1}{100} + \frac{1}{25}$

$\frac{1}{f}=\frac{-1+4}{100}$

$\frac{1}{f}=\frac{3}{100}$

$f=\frac{100}{3} = 33.3\ cm$

We know,

$P=\frac{100}{f(in\ cm)}$

$P = \frac{100}{33.3} = +3\ D$

$P = +3\ D$

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