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A person cannot read a book placed nearer than $150 cm$ . What will be the power of the corrective lens required?

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+ 1.25 D

+ 2.50 D

+ 2.00 D

+ 3.33 D

answer is D.

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Detailed Solution

Power of the required corrective lens is

+ 3.33 D

.
Given,
The near point of the hypermetropic eye

= 150 cm

.
The corrective lens will be a convex lens, and should form an image of the distant objects at the near point of the hypermetropic eye.
Then, the image distance,

v = - 150 cm

.
The object distance,

u = - 25 cm

Focal length of the corrective lens,

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

\Rightarrow \frac{1}{f} = \frac{1}{- 150} - \frac{1}{- 25}

\Rightarrow \frac{1}{f} = \frac{- 1 + 6}{150}

\Rightarrow \frac{1}{f} = + \frac{5}{150} = + \frac{1}{30}

\Rightarrow f = + 30 cm

We know that the power,

P = \frac{1}{f}

\Rightarrow P = \frac{1}{+ 0.30} = + 3.33 D

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