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A person can see clearly objects only when they  lies between 25 cm and 200 cm from his eyes. In  order to increase the maximum distance of  distinct vision to infinity, the power of correcting  lens, the person has to use, will be (1) + 0.5 D (2) – 0.5 D (3) + 0.25 D (4) – 0.25 D
Asked by humerachopda143 | 04 Aug, 2021, 08:22: PM
This person is suffering from Myopia. This person can not see far away objects .
Far point of his eye is 200 cm ( 2 m ) .

Inorder to correct this myopic eye concave lens is used so that ligh rays coming from far away objects
that are parallel rays will appear to eye as if these rays are coming from focal point of concave lens as shown in figure.

We have lens equation :-   (1/v) - (1/u) = 1/f

where v is lens-to-image distance, u is lens-to-object distance and f is focal length

According to cartesian sign convention , v = -200 cm , u = ∞ , hence focal length of lens is given as

(-1/200) - (1/∞ ) = 1/f   or   f = -200 cm = -2 m

Power of lens is reciprocal of focal length if focal length is expressed in metre.

Power = -1/2 = -0.5 D
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