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The near-point of a person suffering from hypermetropia is $50 cm$ from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is $25 cm$ ).

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Convex lens;

- 2 D

Concave lens;

- 2 D

Convex lens;

+ 2 D

Concave lens;

+ 2 D

answer is C.

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Detailed Solution

The nature and power of the lens needed to correct this defect is convex lens;

+ 2 D

.
Given, near point of the normal eye

(u) = - 25 cm

Near point of the defective eye

(v) = - 50 cm

Using lens formula,

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

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

\Rightarrow \frac{1}{f} = \frac{25}{(25) (50)}

\Rightarrow f = 50 cm

⇒

f = 0.5 m

Power of a lens is given by

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

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

⇒

P = 2 D

Therefore, the power of the lens to be used is

+ 2 D

, and as the focal length of this lens is positive, it is a convex lens.

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