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The near point of a hypermetropic eye is $1 m$ . Power [[1]] $D$ of the lens required to correct this defect.

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Detailed Solution

The near point of a hypermetropic eye is

1 m

. Power

0.3 D

of the lens required to correct this defect.
The convex lens is required to correct hypermetropia defects.

Object distance, u = - 25 cm

Image distance, v = - 1 m

∴ v = - 100 cm

Using lens formula,

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

\Rightarrow \frac{1}{f} = - \frac{1}{100} - \frac{1}{- 25}

\Rightarrow f = 33.3 cm

Power P = \frac{1}{f}

\Rightarrow

P = \frac{1}{33.3} D

∴ P = 0.3 D

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