The far point of a myopic eye is 1.5 m. To correct this defect of the eye, the power of lens is -

Asked by dhairya | 23rd Aug, 2009, 10:23: AM

Expert Answer:

u = - 1.5 m; v = -0.25 m;

Using

(1 / f) = (1/ v) - (1 / u)

1 / f = - 1 / 0.25 + 1 / 1.5 = (- 6 + 1 ) / 1.5 = - 3.33

P = - 3.33 D

Answered by  | 24th Aug, 2009, 12:41: PM

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