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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?

Q.6.    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?
 

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The myopic person is suffering from the near-sightedness and can be corrected by putting the concave lens in spectacles.

Given the far point up to which a myopic person can see is 80cm in front of the eye.

Here, this person can see the distant object (kept at infinity) clearly if the image of this distant object is formed at his far point.

So, we have

The object distance, u = \infty \ (infinity)

The image distance, v = -80\ cm (far point, in front of the lens)

and the focal length, f = to find.

Hence substituting the values in the lens formula: we obtain

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

\Rightarrow \frac{1}{-80} - \frac{1}{\infty} = \frac{1}{f}

\Rightarrow \frac{1}{f} = \frac{1}{-80}

Or, \Rightarrow f = -80\ cm

Therefore, the focal length of the required concave lens will be 80\ cm.

Now, Power,

Power = \frac{1}{f(m)} = \frac{1}{-0.8} = -1.25\ D

Hence the power of concave lens required is -1.25\ D.

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