The near point of the eye of a person is 50\ ...

CBSE Class 10 Science 2020 Delhi Set 3

Question : 8 of 10

Marks: +1, -0

The near point of the eye of a person is

50\ \text{cm}

. Find the nature and power of the corrective lens required by the person to enable him to see clearly the objects placed at

25\ \text{cm}

from the eye.

Solution:

(a) It is the case of hypermetropia.

For a hypermetropic eye,

u=-25\ \text{cm}

and

v=-50\ \text{cm}

Using lens formula

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

\;\;\frac{1}{f}=-\;\frac{1}{50}+\;\frac{1}{25}

\;\;\frac{1}{f}=\;\frac{1}{50}

We get focal length is

f=50\ \text{cm}=0.5\ \text{m}

From the formula,

P=\;\frac{1}{f}

P=\;\frac{1}{0.5}=2\ \text{D}

So the power of the lens is 2 dioptre and the nature of lens is a converging lens or a convex lens.

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