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[79]
The near point of a hypermetropia eye is $50 cm$ . What is the nature and power of the lens required to enable him to read a book placed $25 cm$ from the eye?

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Convex and

2 D

Convex and

3 D

Concave and

2 D

Concave and

4 D

answer is A.

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Detailed Solution

The near point of a hypermetropia eye is

50 cm

. The nature of the lens is a convex lens and power of the lens required to enable him to read a book placed

25 cm

from the eye is

2 D

.
Given, object distance

u = - 25 cm; v = - 50 cm

Using lens formula,

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

where,

u =

the distance between the object and the pole of the mirror,

v =

the distance between the image and the pole of the mirror,

f =

focal length.

\Rightarrow \frac{1}{f} = \frac{1}{- 50} - \frac{1}{- 25}

\Rightarrow \frac{1}{f} = \frac{1}{25} - \frac{1}{50}

Focal length

f = + 50 cm

= 0.5 m

The nature of the lens is convex.
We know that power is the inverse of focal length.

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

where

P = power

and

f = focal length

\Rightarrow P = \frac{1}{0.5} D

\Rightarrow P = 2 D

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