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# Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect?

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Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

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The required diagram is shown below:

Given: Distance of near point of defective eye, i.e.

v = -1 m = -100 cm

Distance of near point of normal eye, i.e.

u = -25 cm

Applying the formula $$\frac{1}{f}$$ = $$\frac{1}{v}$$ - $$\frac{1}{u}$$, we get,

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

= $$\frac{-1}{100}$$ + $$\frac{1}{25}$$

= $$\frac{-1 + 4}{100}$$ = $$\frac{3}{100}$$

or f = $$\frac{100}{3}$$ = 33.33 cm

∴ Power, P = $$\frac{100}{f}$$ = $$\frac{100}{33.3}$$

= 3.0 D

Hence, the power of the required lens is 3.0 D.

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