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

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

Question:

The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Here, the person suffering from myopic defect, so he is not able to see objects which are lying more than 80 cm away from his eyes as the far point of the eye is 80 cm. So, the corrective lens used to correct the problem should form the image of the distant objects at the far point of his eye, which is 80 cm in this case.

So, $v=-80$ cm and $u=\infty$

Using the lens formula,

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

we get,

$\frac{1}{f} = \frac{1}{(-80)}-\frac{1}{\infty}$

$\frac{1}{f} = \frac{1}{(-80)}$

$f = -80 cm$

$f = 0.8 m$

We know,

$P = \frac{1}{f(in\ m)}$

$P = \frac{1}{(-0.8)} = -1.25\ D$

Since the power of the lens is negative.

Therefore, the concave lens of power -1.25 D should be used to rectify the eye problem.

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