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A person wears glasses of power – 2.0 D. The defect of the eye and the far point of the person without the glasses will be

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Farsighted, 50 cm

Astigmatism, 50 cm

Nearsighted, 250 cm

Nearsighted, 50 cm

answer is A.

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Detailed Solution

The defect is myopia (nearsightness)
As we know for myopic person f = – (defected far point)
$\Rightarrow Defected far point = - f = - \frac{1}{P} = - \frac{1}{(- 2)} = 0 .5 m$
= 50 cm

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